A Drunkard’s Walk Through Baseball Stats
Posted: March 29th, 2009 | Filed under: Baseball | 83 Comments »
Several brilliant readers insisted that I pick up Leonard Mlodinow’s wonderful book The Drunkard’s Walk, which delves into the probabilities of our time, and all I can say is that you people really know me: This is EXACTLY the kind of book I love. I have a funny relationship with numbers. I have mentioned here about 40,000 times that I have no math skills — other than a reasonable talent for adding numbers in my head — and yet, I would guess it’s no secret here to say that I love enjoy baseball statistics.
I also love a good math problem. I have spent way, way, way too much of my life considering the Monty Hall math problem, which is prominent in The Drunkard’s Walk (and also in another book I really enjoyed: The Curious Case of the Dog in the Nighttime) … I’m sure you know the problem, but I’ll go through it here again because i find it so fascinating.
Basically the problem became somewhat famous because Marilyn Vos Savant in her Ask Marilyn column was asked a version of this question. She created a huge stir with her answer.
Here’s the question: You are a contestant on the old “Let’s Make A Deal.” We’re going to assume here that you’re NOT one of those contestants who dresses up like a ketchup bottle or the Queen of Hearts from Alice in Wonderland or whatever — you’re a normal contestant.*
*I was going to go into my “Let’s Make A Deal” bit … and then I remembered that I wrote it once already, like, a year and a half ago. This blog is getting so old that I’m running out of bits. Here is the old bit, if you’re curious.
OK, so, there are three curtains. Behind one is a a great prize — let’s say a season baseball pass for every stadium in the big leagues. And behind the other two are lousy prizes — let’s say a four-hour mandatory lunch with Bud Selig and Gene Orza.
OK, you choose your curtain — for simplicity, we’ll say that you choose what’s behind Curtain No. 1. And Monty goes: Well, before we see what’s behind your curtain, let’s show you what’s behind Curtain No. 3! And they pull back the curtain and there are Selig and Orza, and Bud is yapping about how he wanted mandatory drug testing but the player union would not allow it, and Orza is blabbing about players rights and the irresponsibility of owners and so on.
Now, Monty gives you an option. He says that you can stay with Curtain No. 1 or switch to Curtain No. 2.
So what do you do?
A: Stick with Curtain No. 1.
B: Switch to Curtain No. 2
C: It doesn’t matter because there is an equal chance the prize is behind either curtain.
I know the right answer to this question — as I’m sure you do — but no matter how well I know it, no matter how deeply I understand it, no matter how many times I have gone over it, my wee little brain tells me that the correct answer is C. I simply cannot shake that notion. There are two curtains, one has the prize, you have a 50-50 shot.
As Mlodinow quotes a Harvard professor: “Our brains are just not wired to do probability problems very well.”
The correct answer is B. You ALWAYS switch. If you switch, you are twice as likely to get the big prize. Like I say, I have thought about this and thought about this, and it has twisted me up. I finally have come up with two ways to explain this so that it makes sense to me:
First way to explain: When you chose Curtain No. 1, there was a one-third chance you were right and a two-third chance you were wrong. That’s easy to understand. And it’s also easy to know what to do if, a moment after you chose curtain No. 1, Monty Hall had said to you (without opening up any of the curtains): “OK, now I’ll give you an option: You can either stay with Curtain No. 1 or you can switch and get BOTH Curtain No. 2 and Curtain No. 3.”
Everybody would make that switch. Assuming the game is not rigged, you would have twice as good a chance of winning.
Well, this is the point: You STILL have twice as good a chance of winning. See, Monty Hall knows where the grand prize is, and he is purposely showing you a curtain where it is not. This does not change the odds. This is the most important thing to remember — Monty Hall is tricking up the game. In our scenario, he will always, always, always, always show you a losing curtain. What he does is absolutely irrelevant to your choice. You still have a two-third chance of being right if you make the switch.
There are only three possibilities here:
Pass is behind Curtain No. 1: You pick Curtain No. 1 and are correct.
Pass is behind Curtain No. 2: You pick Curtain No. 1 and are wrong … Monty shows you Curtain No. 3.
Pass is behind Curtain No. 3: You pick Curtain No. 1 and are wrong … Monty shows you Curtain No. 2.
You are twice as likely to be right if you switch curtains.
Second way to explain: One of the reasons this is so hard for me — and millions of others — to wrap our minds around the puzzle is because it’s only three curtains. Imagine, instead, that there were TEN curtains. You choose Curtain No. 1 and then Monty Hall, one by one, shows you eight other curtains that are wrong until there are only two left — your curtain and one other one. He gives you the option to switch. The larger number changes the dynamic, at least in my mind. It now makes sense to my brain that switching would make sense … after all you only had a 10% chance of being right at the start. There’s a 90% chance chance that the prize is behind the other curtain.
You can do the same thing with100 curtains (you would have a 1% chance of being right at the start, and a 99% chance of being right if you switch) and 1,000 curtains (you would have a .01% chance of being right at the start) and so on.
It’s the same thing … you ALWAYS switch.
Marilyn Vos Savant got the answer right. And at that point she was inundated with thousands and thousands of letters, many from people with advanced math degrees, professors, mathematicians — many calling her an idiot and telling her to stop misleading America.
I bring all this up for a reason — I love this stuff. I absolutely love little things that help me see the world is a slightly different way. I have never been able to explain it better than that … I love baseball statistics, many of them, because they help me see baseball in a whole new way.
I have a good friend who is my age and has little use for the advanced stats. He likes homers, RBIs, batting average, pitcher wins. He says that’s all he needs and, more than that, that’s all he wants. He says that when it comes to even some of the mainstream ones like OPS or OPS+ or ERA+ or VORP, his eyes glaze over.
He loves the crack of the bat and the smell of the grass and making his own judgments about players. He believes Derek Jeter is a great defensive shortstop because he has seen him make a lot of great plays. He believes that A-Rod is a choking dog who plays his best when nothing is at stake. He believes Jack Morris belongs in the Hall of Fame because he (unlike Blyleven) was a winner. He believes that Jim Rice should have gone to the Hall of Fame a long time ago because he was the most feared hitter of our youth. He thinks teams should bunt more.
If I gave him these options …
Curtain 1: Three 120 ERA+ starters and a fourth with a 100 ERA+, a versatile bullpen, a high OBP lineup that would finish Top 5 in slugging …
Curtain 2: A pitching staff of winners and a team with hustlers, leaders and excellent chemistry …
… he would ALWAYS choose Option No. 2. He would not switch either, even when I showed him the goat behind that curtain.
You may ask — knowing my own feelings on these subjects — how I maintain my friendship with said person and the answer is simple: We don’t talk.
No, I’m joking — we talk all the time, and we talk baseball all the time. We just have an understanding. We share a lot of common ground. We both love to watch baseball, to break down players swings, to think with pitchers, to consider where the outfield is playing, to see a player stretch a single into a double, and a double into a triple. We both love to see how well the outfielders throw, to see if this reliever has real gas, to admire players who get their uniforms dirty, to stand up when a third baseman makes a diving stop, gets to his feet and fires a low throw to first (and then seeing the first baseman scoop the ball out of the dirt). We both love baseball.
And just because he’s baffled by my interest in boring numbers, and I’m baffled by his plain disinterest in digging deeper into the game, doesn’t prevent us from enjoying the games together. We’re baseball fans. We have much more in common than we have different.
Still, it does frustrate me that I have never been able to explain to him why I’m so fascinated by the numbers. It’s funny: I had a friend growing up who used to love taking apart telephones. These were old rotary phones, of course, and he loved to take them apart and look at the wires and try to figure out how the whole thing works. I was never like that. I liked TALKING on phones, but that was about it. As far as I knew, there was an itty bitty little operator inside the phone who plugged wires into the wall, sort of like a miniature Lily Tomlin.
But here I am now … and when it comes to baseball and many other things I’m that guy tearing apart phones. I have no great for it. But I want to know how it works. For instance, while reading The Drunkard’s Walk, I became fascinated by this formula the author used to explain probability in a seven-game series. He used various statistical tools that I immediately that I immediately had to try myself.
For instance, I wondered: What is the statistical advantage for a team taking the lead in a 7-game series. Well, you can use Pascal’s Triangle (man, this book is making me nerdy) to help figure it out. If both teams are equal, then at the start of the series both teams have exactly a 50% chance of winning. That makes obvious sense. And if they are tied at any point in the series they each have a 50% chance of winning the series.
And here, best I can tell, are how those percentages move when one team takes a lead over the other.
Team A Team B
Teams are tied. 50% 50%
Team A leads 1 games to 0 66% 34%
Team A leads 2 games to 0 81% 19%
Team A leads 3 games to 0 94% 6%
Team A leads 2 games to 1 69% 31%
Team A leads 3 games to 1 88% 12%
Team A leads 3 games to 2 75% 25%
These percentages are obviously flawed: They don’t include countless factors such as home ballpark, pitcher on the mound, etc. But that’s OK: My friend long ago stopped reading this post. He does not care what the probabilities are … he just wants to watch the game. He wants to enjoy the tension. But for me, this is PART of enjoying the tension.
And I’m off now. I start to wonder what are the odds that a team would win a series if they are NOT evenly matched? Let’s take the Royals and the Yankees. I don’t know how much better the Yankees are than the Royals*,
*Bill James actually came up with a formula for this too … by using two teams’ willing percentages there is a way to determine what percentage of the time a team should defeat another team.
Last year’s Yankees had a .549 winning percentage, and the Royals had a .463 winning percentage. Punching that into the formula, it comes out of that the Yankees should be the Royals 58.5% of the time.
What are the chances the Royals would win a seven-game series? So I play around with another formula — now I am trying (and no doubt failing) to use what are called Bernoulli Trials. Here’s what I come up with (and the line on the left is just an estimate of how often you think the Royals would beat the Yankees):
Best of seven series:
Royals win Yankees win
Royals-Yankees are even 50% 50%
Royals win 45% of time 39% 61%
Royals win 42% of time 32% 68%
Royals win 40% of time 29% 71%
Royals win 35% of time 20% 80%
Royals win 30% of time 13% 87%
Now, obviously I’m using the most simplistic formula … one that does not even consider things like home-field advantage. But it’s still interesting to me. Even terrible teams — teams that would be expected to lose two out of every three — would still win a best-of-seven series roughly one out of five times. Last year’s Royals would be expected to win roughly one out of three seven game series from the Yankees.
That does seem to indicate that the postseason is a bit of crapshoot.
Of course, a best-of-five series is even MORE of a crapshoot.
Best of five series:
Royals win Yankees win
Royals-Yankees are even 50% 50%
Royals win 45% of time 41% 59%
Royals win 42% of time 34% 66%
Royals win 40% of time 32% 68%
Royals win 35% of time 24% 76%
Royals win 30% of time 16% 84%
And, why not, we’ll give you probabilities for a best of three series too:
Royals win Yankees win
Royals-Yankees are even 50% 50%
Royals win 45% of time 43% 57%
Royals win 42% of time 37% 63%
Royals win 40% of time 35% 65%
Royals win 35% of time 28% 72%
Royals win 30% of time 22% 78%
Now, this stuff is just fun for me. I can understand why it wouldn’t be fun for everyone — or anyone. But it adds all these new layers of baseball for me. Of course, statistics will never replace the sheer joy of watching Jim Thome get a hold of one or the thrill of seeing Carlos Beltran trying to score from first on a double or the absolute kick I get out of seeing David Eckstein try to throw the ball from hole at shortstop (you can almost hear him grunt “Uh” as he throws a 20-foot high lollipop). There are no statistics that can recreate the smell of beer and cotton candy and cut grass, the way your shoes stick to the concrete in a ballpark, the perfect throw to nab a base stealer, the perfectly painted outside fastball for strike three looking.
But it does make it more fun and interesting to watch a game knowing that walks are valuable, that just because I saw a guy make two diving defensive plays does not make him a great defensive player, that I’m not crazy for thinking that more often than not it’s stupid to give you outs, that using RBIs and wins and batting average to determine a ballplayers ability is a bit like coming into a movie with 10 minutes left and then using those minutes to determine exactly what the movie was about. You’ll get some things right, but you’ll miss a whole lot along the way.
I’m 2,400 words into this, and I don’t really know if I explained why I love baseballs statistics. It doesn’t matter anyway: I’m now reading about the probability of streaks. So I have some more math to do.
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I’m that first vote on the new poll ” less than 500 baseball card”
it’s really only one – my Tom Seaver card late 1970’s.
wow ! he was great pre trade
I never was a math guy. In fact, I went to college and was a journalism major precisely because I hated math.
But I do love baseball statistics, and one of my favorite times of the year is getting my Baseball Prospectus.
It is interesting that for some of us, statistical analysis improves our enjoyment of the game, while others are not only not interested, but angry when numbers are mentioned. I do think this is an important difference in the way we approach not only baseball, but life. I think we all probably have particular topics about which we would prefer not to see the columns of numbers behind the curtain.
Religious faith might be one of those… lots of statistical studies have been done on prayer. One of the larger, more recent ones showed that prayer had no effect whatsoever on the recovery of patients in a hospital. What does that mean? I think a lot of people who believe in prayer would rather not hear about the results of that study. They don’t want to have to argue with it, or come up with reasons why they don’t believe it or don’t agree with the results. They just want to allow their beliefs to improve their lives, and the lives of those they love.
I don’t know what I think about it all… I can appreciate the value of both, and I think each has its pitfalls. Probably a balance between them is best.
Two thoughts…
1) I actually laughed out loud when you started this post with the logic problem. I took a chance and watched “21″ last night when I was bored (despite having heard it was really bad…wasn’t as bad as I expected, but it wasn’t great either) and this problem made an appearance early in the movie. They didn’t do a great job explaining it, so I spent the next 45 minutes trying to figure out why it was better to switch, and finally gave up, paused the movie and tried to find a better explanation online. Makes sense now, and your explanation (from what I can tell at least) is spot on…at least, it’s the same as the one I found last night.
2) Talking about the probabilities of a team winning a best-of-seven series sounds like the end of Moneyball, how the A’s had found a way to defeat teams over the course of a long season, but the postseason suffers from being a small-sample size, so basically any advantage Billy Bean’s squad had is somewhat negated. I still have a hard time believing the Royals would have won a best-of-seven series against the Yankees one out of three times*, but I agree, it is interesting to think about.
*Well, maybe last year’s Yankees, they weren’t that great either…but certainly not Tampa or Philadelphia or Boston.
That hurts my head.
I’m a smart guy and I see where you’re going with the Lets Make a Deal thing, but I’m just having a hard time getting from A to C. B is in the middle waving its hand saying “Yeah but now you know Curtain #3 is not the grand prize and thus you are being granted a brand new choice of either Curtain #1 or Curtain #2, with one having the grand prize and one not”. I mean I get what you’re saying from the perspective of the odds not changing, but it seems to me that you’re making a brand new choice, that while the odds of the ORIGINAL choice wouldn’t change just because something was removed, by asking you to switch, you’re being granted a brand new choice with only two options, and that third curtain may as well not exist. Same with “10 curtains, 8 are removed, do you want to switch?”, that still seems like a brand new choice with only two options.
I’m certainly not saying you’re wrong, I think I’m in your boat Joe…good with adding numbers in my head, love statistics, love a good math problem, don’t have the “higher” math skills. So if the smart people say that’s correct, I’ll take their word for it, I just don’t see why “Okay, Curtain #3 has a bad prize, would you like to switch” isn’t a brand new choice with new probabilities.
As a math grad student, I can more than sympathize with Joe’s attitude towards numbers and statistics. Indeed, thinking back, my initial interest in baseball may have been largely doe to the numbers in the game!
The Monty Hall incident is one of the most embarrassing episodes for the mathematical community in quite some time. Part of the back story is that Vos Savant had made a number of comments about mathematics, especially Wiles’ proof of Fermat’s Last Theorem, that were highly misleading and uninformed, and as such people were primed to expect nonsense out of her. Even though, that of course does not excuse the rather serious error a lot of people made.
joe, i’m with you. i love watching baseball. i love baseball statistics. and, i would also add to this, i love the history of the game (the players, the evolution of the game. i think it’s the stats and history of the game that set it apart from any other sport.
That solution to the Monty Hall problem still bothers the hell out of me. It may be that I never watched Let’s Make a Deal (before my time), but why aren’t there 4 possibilities:
1) You pick 1 and you are correct, but Monty shows you Curtain 2
2) You pick 1 and you are correct, but Monty shows you Curtain 3
3) You pick 1 and you are wrong, Monty shows you Curtain 2
4) You pick 1 and you are wrong, Monty shows you Curtain 3
It seems like the solution assumes that he’ll only show you a curtain if you’re wrong, which shouldn’t be the case (otherwise it’d be a pretty easy game to win).
Additionally, you present the “if he gave you the option of switching #1 out for #2 AND #3″…” explanation. However, couldn’t he also say “You can switch to #2, or I will let you keep #1 AND you get #3″. Same logic, different solution by this logic.
I’m fully willing to accept that there’s probably something I’m overlooking on this, I just can’t figure out what it is. Someone end my frustration please.
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The key thing to remember about the Monty Hall problem is that he’s only going to open a curtain that doesn’t have the prize behind it.
So, in other words, think of it this way: You pick curtain 1. There’s a 1/3 chance the prize is behind curtain 1, and a 2/3 chance it’s not behind curtain 1. We can all see this easily. If it’s not behind curtain 1, it’s either behind curtain 2 or curtain 3.
And if that’s the case (here’s the important part), Monty basically tells you which curtain it’s behind. ‘Cause if you pick curtain 1 and the prize is behind curtain 2, he’s not going to open up curtain 2 and say, “Oh, here’s the prize! You were wrong; you lose.” He’s going to open up curtain 3, and say, “Look, curtain 3 doesn’t have the prize!”
So again, if we’ve accepted that there’s a 1/3 chance it’s behind curtain 1, and a 2/3 chance it’s behind curtains 2 or 3, Monty has essentially told us, “Well, if you were thinking of picking curtain 2 or 3, don’t pick 3, ’cause the prize isn’t there.”
A perhaps clearer way to put it is to recognize that the only way you can lose out by switching is to have picked correctly the first time. If you didn’t pick correctly the first time, the prize is behind one of the other two curtains, and Monty shows you which curtain it’s not behind, so it can only be behind the remaining curtain. The only way it’s not behind the remaining curtain is if you picked correctly the first time, and there’s only a 1/3 chance of you picking correctly the first time, so there’s a 2/3 chance you get it if you switch.
off topic, but speaking of numbers…i recently read an article in a magazine (espn or wired, i can’t remember) about an australian bowler who rolled two-handed. i could care less about professional bowling (i’m a candlepin man myself), but the thing i found interesting was that the the guy who rolled two-handed puts more spin on the ball than an average one-handed bowler. i forget how many revolutions it was but the reason it stuck with me was that it involved numbers.
today, running on the treadmill at the gym, i watched the guy win his first pba championship.
“why aren’t there 4 possibilities”
Short answer: there are 4 possibilities – but they are not all the same weight.
Analogy: in any given at bat, a batter gets a hit or he doesn’t – 2 possibilities. But somehow it’s been a long time since we’ve seen a .500 hitter.
The piece of information you are overlooking is that Monty (a) knows where the prize is and (b) never chooses to reveal it.
If Monty doesn’t know the result before opening a curtain, then the math is changed a lot.
For those who want a real brain bender, try to figure out if her other child is also a girl.
or better yet, try to figure out how to explain the answer….
That’s an awful lot of trouble for a washer and dryer, or whatever was behind the right curtain.
Greg’s explanation of the Let’s Make a Deal dilemma is excellent. I get it now. Thanks!
I was interested in philosophy for several years before and during college. As often happens with philosophic inquiry, my interest drowned in a pool of semantic sludge.
Game theory is like philosophy for the modern man (namely, me). The wisdom lurks beneath the statistics. At least, that’s how I explain my fascination with it.
Response to Greg:
Do not think of it as Curtain 2 and Curtain 3. Monty will simply pick a curtain that does not conceal the great prize.
By deliberately choosing the curtain that is not concealing this great prize, Monty is “moving” all 2/3rds probability that you initially chose incorrectly to the other curtain you did not choose.
Re-stress: He’s not randomly picking a curtain, he’s definitely going to show you one that is concealing a bad prize, as there always will be one available that you did not choose. If you did pick the correct curtain, then, yeah, he’ll choose one of ‘em randomly. But that will happen only 1/3rd of the time!
Two words that sum it all up: Conditional probability
Teams are tied 50% 50%
Team A leads 1 games to 0 66% 34%
Team A leads 2 games to 0 81% 19%
Team A leads 3 games to 0 94% 6%
Team A leads 2 games to 1 69% 31%
Team A leads 3 games to 2 75% 25%
Yeah, but what about when Team B leads?
I think I like it better when Joe was flippin’ his quarter in his Phoenix hotel room – Much more straight forward for me to figure the head or tails dilemma. Once Monty Hall gets involved with his three curtains and the gorgeous prize ladies I’m just not worth a damn.
Joe – If we are going to go “Game Show”, why not play “The Price Is Right” and use baseball player stats, personal info like age, general health, # of yrs in MLB? Either you or that very old horn-dog Bob Barker gives out the above info and then we have to decide the appropriate salary. Then you or Bob tells us who the player is that you were describing and the winner is the guy who comes closest to the salary with out going over.
Hey, I’m wingin’ this but it might be fun….
Greg:
The problem with your explanation, as someone stated above, is that your given probabilities do not all have the same weight. The probability that you are right on the first choice is 1/3. The probablility that you are wrong on the first choice is 2/3. You seem to be assuming that the probability of your choices 1, 2, 3, and 4 are all equal. In fact, the probability of choices 1 and 2 added together is 1/3. The probability of choices 3 and 4 added together is 2/3. You need to assume that you were wrong on the first choice, because no matter whether you were right or wrong, he will ALWAYS show you a curtain that has the bad prize behind it. It doesn’t change the odds; it’s extraneous information that he’s using to distract you.
This problem seems to have applications regarding “Deal or No Deal” as well, though I’m not sure what the implications would be since there are many “good choices”. I think we can safely say that if the player makes it to the end, there are two cases left, and the $1,000,000 is still out there, the player should always switch.
I was incredulous at the start when I read this, too. I had never heard of the problem before now. Joe’s “1000 curtains” explanation convinced me, though. Marilyn Vos Savant’s Wikipedia page has some helpful information on the question as well.
Actually, now that I think about it, I’m not sure about the Deal or no Deal part, since Howie Mandel doesn’t pick the cases– the player picks them randomly. I guess that would be a different problem altogether.
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I love the Monty Hall problem. Of course, there’s another caveat: you have to assume that Monty ALWAYS shows you a curtain. In the real world it’s possible that Monty would be more likely to show a curtain when you have picked correctly to throw you off, and this throws all the probability out of whack.
PS
Longtime blog reader but I just read Soul of Baseball for the first time. Great book! Does Joe have anyhthing else coming out soon (ie in the next 6 months or so)?
Here is how I am able to rationalize the Monty Hall problem in my head: run the problem 100 times (or 99 to make it even), where the prize is evenly distributed. So, 33 times it is behind A, 33 times it is behind B, and 33 times it is behind C. If you choose A every time and Monty shows you an empty curtain every time, 33 times you would have chosen correctly, but 66 times you would get the good prize if you switch.
There is 0% chance that Monty will show you the winning curtain before you pick. Or in other words, there is a 100% chance that the winner will be among the curtains you can choose from. So these two cases _are_ exactly alike:
Three curtains, Monty removes one bad curtain, you choose one of the remaining two.
Three curtains, you choose one, Monty removes one bad curtain.
Your chances of winning are 50/50 either way.
If you have 100 curtains, you are still certain that Monty will always remove 98 bad curtains before leaving you to choose between two remaining curtains.
Your initial chance of getting the right curtain is 1 in 3 (or 1 in 100) BUT THOSE CONDITIONS NO LONGER EXIST!
Another take: Three curtains. First you pick a curtain. From the two remaining curtains (A and B) Monty will always show a wrong curtain. He has these choices:
Monty shows A, B is also wrong
Monty shows B, A is also wrong
Monty shows A, B is right
Monty shows B, A is right
He has these four choices, no matter what you choose initially. So while you have a 1 in three chance of being right initially, that makes no difference to Monty. Whatever you choose, he will always have those four remaining choices. So what you choose when your chances are 1 in 3 (or 1 in 100) doesn’t matter.
Once you have chosen your initial curtain, there is no “probability” left. You are either right or wrong. Probability applies to future events.
The simple answer (in this case, that you have a 50/50 chance after Monty shows you a bad curtain) is usually the correct one. Occam’s razor is a handy tool.
And yes, I know that people disagree. But I have yet to be convinced.
“The simple answer (in this case, that you have a 50/50 chance after Monty shows you a bad curtain) is usually the correct one. Occam’s razor is a handy tool.”
Unfortunately, this is one of the cases where Barbie’s Razor (“Math is hard”) wins out.
The simple answer is “of course you switch. Keeping the same door is no better than 50/50. Therefore, switching is no worse than 50/50… but it may be better than that, if these other schnooks are on to something.”
In this case, the problem you are running into is assuming that Monty’s four choices are equally likely – which is equivalent to the assumption that you are can pick the correct door (out of three) 50% of the time.
It is sometimes helpful to notice that if we introduce a third party to the action, who is allowed to pick a curtain after Monty’s first reveal, but it not told which curtain you have chosen, that player does have a 50/50 chance. You, however, have more information.
Joe –
The New York Times did an article on this about 5-8 years ago, shortly after the Vos Savant piece in Parade magazine. The problem is, there is no “Monty Hall” problem.
I was doing some research on this and wanted to get all of the old “Let’s Make a Deal” shows to actually record the data and work through the problem myself. However, when I called the Hall/Hatos production company a very nice man told me not to bother because the show didn’t work that way. On the show, when the first door was revealed you got the choice of the prize or $100; you simply didn’t get the choice to switch to another door. I was told that the NYT article elaborates on this.
The nice man who spoke with me was Monty Hall. He answered the phone because it rang in his kitchen; claiming it was the first time in 30 years his secretary had been late for work so the call was forwarded to his house. I spent 45 mins on the phone with him and he shared some of his memories of the show. Pretty random.
Ray: “So these two cases _are_ exactly alike:
Three curtains, Monty removes one bad curtain, you choose one of the remaining two.
Three curtains, you choose one, Monty removes one bad curtain.
Your chances of winning are 50/50 either way.”
This is completely false.
reach on error
The big difference in these 2 situations is the time at which the player picks the curtain. Many others have already explained clearly why it is always in the player’s interest to switch. But I think the point you make in the beginning is why so many people get tripped up. Those 2 cases ARE NOT exactly alike. And those 4 events that you have described don’t have equal likelihood.
The second problem with your argument is your list of Monty’s choices…
Your examples assumes the player picked C. In this case your first 2 scenarios have a 1/3 chance combined. (i.e. the chance that the player picked the correct curtain out of three choices). The last 2 scenarios have a 2/3 chance combined, and these are the times where it benefits the player to switch.
Its a huge fallacy to believe that just because you can list a number of different events that they all have the same probability of happening.
For example an 0-2 pitch to Juan Pierre, here are the possible outcomes:
1) strikeout
2) ball
3) hit by pitch
4) single
5) double
6) triple
7) homerun
Now just because I can list each event it does not mean they have the same probability of happening. I think we can assign much higher probabilities to outcomes 1 or 2 compared to numbers 7 or 8.
Just to add it, here’s Cecil Adams’ (The Straight Dope) take on the Monty Hall problem: http://www.straightdope.com/columns/read/916/on-lets-make-a-deal-you-pick-door-1-monty-opens-door-2-no-prize-do-you-stay-with-door-1-or-switch-to-3
Ah, the Monty Hall problem. I’ve had discussions of this problem before that threatened to come to blows.
Joe’s explanation is great, although I wish he’d stated up front the key assumption, which other readers have mentioned: that Monty will ALWAYS reveal a non-winning curtain, regardless of whether you picked correctly or not.
Here’s my favorite way of illustrating what’s going on: pick three cards out of a deck, one Ace and two Kings. The Ace represents the “grand prize.” Lay the three cards face down and have a friend pick select one card. (you’re playing the role of Monty here)
Now, you get to look at the other two cards, which will either be two Kings, or one Ace and one King. Either way, you’re going to turn over a King. Then give your friend the option of switching to the remaining (not turned over) card, or sticking with his original choice.
If you try this for a while, I don’t think it will take long to convince you that switching is better.
I’m with Ray C. To my handicapped mind, the odds reset when Monty shows you what’s behind curtain number three. These other explanations haven’t yet made me see the two-thirds thing.
I understand if you want to say Curtain 1 is a one-third chance. Before I pick, as far as I know, the same is true of Curtain 2 and of Curtain 3. Let me ask you crazy people who, due to your mental strangeness, understand math a question:
What if we’re playing against each other? I pick Curtain 1, you pick Curtain 2. Now we BOTH have 1-in-3 chances of being right, right? Then Monty shows us goats behind Curtain 3. Are my odds still 1-in-3? How about yours? Is it the right statistical play for *each* of us to switch?
The odds have got to reset once the choice effectively is no longer between three curtains but between two. Let me rephrase it again: Let’s say there are three of us playing, you, me, and Joe. I pick 1, you pick 2, Joe picks 3. We each have a 1-in-3 chance. Then Monty shows us the goats, behind Curtain 3. Which of us went from 1-in-3 odds to 2-in-3 odds? Surely Joe doesn’t still have 1-in-3 odds. How didn’t Joe go to 0-in-whatever odds, and each of us to 1-in-2 odds?
I think the disconnect for me is that the statistics seem to be about my chances of winning, but I feel the question is about the chances the prize is behind a given curtain. It’s a 1-in-3 chance to start, but once Curtain 3 is out of the loop, the odds have to redistribute. No?
All these explanations that there’s still a two-thirds chance it’s not behind Curtain 1 seem to stress that Monty’s going to show you the curtain that a) isn’t the one you picked and b) isn’t the curtain with the prize. Why does this mean the odds of it being Curtain 2 somehow double? Monty’s showing me goats doesn’t double Curtain 2’s odds any more than it doubles Curtain 1’s odds.
Not-so-brilliant reader mike
Oops. I’ve shown the opposite of what I was typing. Need coffee.
There’s a equal chance that Monty will show one or the other of the two remaining curtains, assuming he always offers you the switch, he always shows a bad curtain, and he chooses randomly if they’re both bad. But that doesn’t affect the initial odds, which remain 1 in 3 that you were initially correct.
Self-chastened, and wishing there was a delete option,
-R
Here’s a slightly more technical (but still avoiding all those math equations) explanation…
Assume for simplicity’s sake that the player selects Curtain A. Now, I think we can all agree that the probability of each individual door contining the prize is 1/3.
Let’s say the prize is behind Curtain A. Then, Monty can choose either to open Curtain B or Curtain C, as neither contains the prize. For simplicity’s sake, we’ll assume that there’s an equal chance of each (I suppose Monty might open B more often than C but it really doesn’t matter, though it does make the math messier).
If the prize is behind Curtain B, then Monty will always open Curtain C, and if the prize is behind Curtain C, then Monty will always open Curtain B.
Let’s say that Monty opens Curtain B. He will do this ALL of the time when the prize is behind Curtain C, and HALF of the time that the prize is behind Curtain A.
That’s why it is now twice as likely that the prize is behind Curtain C. Hence, Probablity of A containing the prize = 1/3, and Probability of C contaning the prize = 2/3.
Baseball isn’t about probabilities, Joe. It’s about improbabilities.
Brute Force, each outcome equally likely:
If the prize is behind A:
If you pick A, Monty shows B or C, switch and lose.
If you pick B, Monty shows C, switch and win.
If you pick C, Monty shows B, switch and win.
If the prize is behind B:
If you pick A, Monty shows C, switch and win.
If you pick B, Monty shows A or C, switch and lose.
If you pick C, Monty shows A, switch and win.
If the prize is behind C:
If you pick A, Monty shows B, switch and win.
If you pick B, Monty shows A, switch and win.
If you pick C, Monty shows A or B, switch and lose.
6 out of 9 times, if you switch you win.
3 out of 9 times, if you switch you lose.
Voice of Reason @13…what’s the girl problem?
“It’s a 1-in-3 chance to start, but once Curtain 3 is out of the loop, the odds have to redistribute. No?”
No, not the way that you mean. Unless we are using quantum goats, or something.
“All these explanations that there’s still a two-thirds chance it’s not behind Curtain 1 seem to stress that Monty’s going to show you…”
Because the key element in the puzzle is that, when you have initially chosen correctly, Monty is given a choice; when you have chosen incorrectly, his actions are completely constrained.
Three games – let’s give Monty a coin with an A side and a B side to flip to determine which curtain he reveals.
Game 1 – Monty always reveals the curtain indicated by the coin.
1. Monty flips A, A is wrong, B is also wrong
2. Monty flips B, A is wrong, B is also wrong
3. Monty flips A, A is right, B is wrong
4. Monty flips B, A is right, B is wrong
5. Monty flips A, A is wrong, B is right
6. Monty flips B, A is wrong, B is right
When Monty reveals a wrong curtain, two choices are eliminated. Your chances of having chosen the correct one are now 50/50.
Game 2 – Monty always reveals the curtain indicated by the coin, unless that curtain has a prize. You get to see the coin!
1. Monty flips A, A is wrong, B is also wrong
2. Monty flips B, A is wrong, B is also wrong
3. Monty flips A, A is right, B is wrong
4. Monty flips B, A is right, B is wrong
5. Monty flips A, A is wrong, B is right
6. Monty flips B, A is wrong, B is right
Here, you have a huge advantage – the two cases that were eliminated in Game 1 remain in game two, but these are now guaranteed wins for you. You are still 50/50 on the other cases.
Game 3 – Monty always reveals the curtain indicated by the coin, unless that curtain has a prize. You don’t get to see the coin.
1. Monty flips A, A is wrong, B is also wrong
2. Monty flips B, A is wrong, B is also wrong
3. Monty flips A, A is right, B is wrong
4. Monty flips B, A is right, B is wrong
5. Monty flips A, A is wrong, B is right
6. Monty flips B, A is wrong, B is right
Here’s the catch: because in this game you don’t see the coin, the two cases which were eliminated in game 1, and sure wins in game 2, have now been merged with the other cases. The last two cases look exactly like the first, and the remaining cases look exactly like the second. In essence, Monty is leaving you to chose from 1,5,6 or from 2,3,4.
Red @ 35 – taken very loosely: a woman, known two have two children, reveals that one is a girl. What are the odds that that both children are girls?
There are two things that make this problem especially fun. One is that it is very sensitive to the phrasing of the question. The other is that there are incorrect solutions that also provide the correct numerical result.
As a result, you get four different groups talking past each other.
Group A gets to the correct answer in a completely naive fashion, and thinks everybody else is making the problem unnecessarily hard.
Group B correctly calculates the effect of the first nuance of the problem, but miss a second more subtle part of the problem, and get an incorrect answer.
Group C correctly calculates the effect of both nuances, arriving at the same solution as group A.
Group D bypasses the calculation completely, using a brilliant insight to achieve the correct answer directly.
Now you’ve got a prime mess: you really need to be paying attention to distinguish groups A and D, group B is trying to simultaneously accuse A (and D) of not getting it, while also defending themselves against the accusations of not getting it from group C.
For a typical demonstration of what this mess turns into, see Jeff Atwood’s blog (and the roughly 2500 comments over the course of a week).
http://www.codinghorror.com/blog/archives/001203.html
http://www.codinghorror.com/blog/archives/001204.html
This is best summed up by Damon Rutherford (#18) – it is conditional probability.
We are comparing the probability of it being behind A given that we know it is not behind C and the probability it is behind B given that we know it is not behind C.
This IS different to the probability it is behind A or B (with no option C).
This is explained in great detail here
http://en.wikipedia.org/wiki/Monty_Hall_problem
If after reading this you are still convinced it is 50-50 then I give up.
Voiceofunreason,
I’ve actually heard of that one. Isn’t the probability 1/3? If you list the children by age, she can have GG, GB, or BG.
If she says the oldest is a girl, then the odds are 1/2. GG, GB.
AMR #38:
That should convince everyone.
#38 is a very good way of explaining.
I think the people claiming it is 50-50 are focusing on the assumption of Monty showing you whats behind the wrong door. This stems from misreading the original problem.
“If you try this for a while, I don’t think it will take long to convince you that switching is better.”
I did once try this for a while because I also couldn’t get my head around the answer, and you are right it doesn’t take long to see that switching is correct.
If you do run into anyone who refuses to accept that switching is better, offer to play them a thousand times for money.
I love looking at statistics and diving into saber metrics while watching baseball, and often times I will during a game out of curiosity, but it’s a bit frustrating watching a team that doesn’t care for the stuff and thus goes the route of your friend.
By the way, this would NOT apply to Deal or No Deal – because the contestant is picking the cases, not the host, who would presumably know where the million dollar case was.
Sorry, still can’t make the leap. I’m sure the shortcoming is mine. But it seems statisticosemantic to me. I mean, I understand what the argument for two-thirds is, but not why the argument against one-half is wrong.
If Joe e-mailed you and me and said he was going to give a million bucks to one of us, he just had to figure out who, we’d each have a 50-percent chance. Then he decides to give it to me. The odds have changed; there aren’t the same number of choices (two) anymore. You don’t still have a 50% chance. And neither do I. The odds changed for both of us.
I know it’s not completely analogous, and probably egregiously so. But I still need my point answered in such a way that even I understand it. You don’t have a 50% chance. You *had* a 50% chance. So if I pick Curtain 1, and goats are behind Curtain 3, I might have had a 1-in-3 chance before the goats were revealed, but now the odds have changed.
Probably I just don’t get statistics. Because I keep wanting to go to the one-door-in-a-thousand scenario where 998 doors are revealed to be losers, and to say See? I don’t have a 0.1% chance now. I’m sure I do, and you’re all right, I just can’t get it.
Joe, As a person who holds 2 degrees in Statistics and who often teaches Stats classes within industry and in colleges, I have to say that this was the best explanation of the Monty Hall problem I have ever seen. Well done.
Heh, despite the stats, I’d have a hard time switched my initial pick & probably would stick with it…
Mike:
I think the best explanation is that the revealing isn’t random. In the one-door-in-a-thousand scenario each of the 998 doors is revealed, but not at random. Whoever is doing the revealing knows what doors are empty and thus picks those ones to show you. This makes the person *feel* like the odds are shifting in their favor (because there are fewer doors left), but they are not. But the game doesn’t work if the revealing is random. This would lead to the possibility of Monty Hall revealing the prize, at which point yes, of course, you would switch. And if the revealing was random and somehow Monty Hall didn’t reveal the prize in those 998 doors, and you are left with only your original choice and one other door then it *is* a 50/50 chance. But the odds of him not revealing the prize when opening doors at random is 2/1000 = 0.2%.
How bout thinking of this. I’m thinking of player who is currently in the hall of fame. You get one guess to who it is and write it down. I then reveal all the names of who I am *not* thinking of except for one. Wouldn’t you be pretty certain that the one i didn’t reveal is most likely the name i was originally thinking of, and that the one you originally picked is very unlikely to be the correct answer? You can at least be sure it isn’t a 50/50 choice.
Wow! I have always loved baseball, but I never realized how boring it was until JUST NOW! Joe, I love your writing, but I quit on this one shortly after your friend did.
Red @ 43
“I’ve actually heard of that one. Isn’t the probability 1/3? ”
No, it isn’t – unless you are very careful about how you phrase the question. Which Jeff wasn’t – so the second more subtle nuance “cancels out” the first. Basically, the missing piece is that in your three cases ( GG, GB, BG ), the probability that the mother will volunteer that information is not the same.
Not sure you’ll ever read this Sir Joe, but your post reminds me of a quote by physicist and Nobel Laureate Richard Feynman:
“I have a friend who’s an artist and he’s sometimes taken a view which I don’t agree with very well. He’ll hold up a flower and say, “Look how beautiful it is,” and I’ll agree, I think. And he says – “you see, I as an artist can see how beautiful this is, but you as a scientist, oh, take this all apart and it becomes a dull thing.” And I think that he’s kind of nutty.
First of all, the beauty that he sees is available to other people and to me, too, I believe, although I might not be quite as refined aesthetically as he is; but I can appreciate the beauty of a flower.
At the same time, I can see much more about the flower than he sees. I can imagine the cells in there, the complicated actions inside which also have a beauty. I mean it’s not just beauty at this dimension of one centimeter, there is also beauty at a smaller dimension, the inner structure.
Also the processes, the fact that the colors in the flower evolved in order to attract insects to pollinate it is interesting – it means that insects can see the color. It adds a question: Does this aesthetic sense also exist in the lower forms? Why it is aesthetic? All kinds of interesting questions which shows that a science knowledge only adds to the mystery and awe of a flower. It only adds; I don’t understand how it subtracts.”
Mike @ 49
“Sorry, still can’t make the leap.”
OK, if you didn’t follow the difference between game #2 and game #3 in post 40, this experiment might help.
Get yourself three index cards – preferably the same on both sides. Mark one card with a letter A on each side, one card with a letter B on each side, and one card with the letter A on one side, but B on the other.
Now, flip and shuffle the cards where you can’t see them, then place one card face up on the table in front of you. Make a note of which letter is facing you, then make note of the letter on the bottom. Return the card to the deck, shuffle, and repeat.
After 100 or so trials (more would be better), you’ll probably find that the letter on the front is the same as the letter on the back 2/3 of the time – that shouldn’t be a surprise; after all, there are two cards with the same letter on boths sides.
If you have been keeping track of the letters, you’ll also see that when the letter A is face up, the letter A is also face down about 2/3 of the time, and likewise for letter B.
This is our analog computer for simulating the game. In each case, you are understood to have chosen curtain C. The top of the card tells you which curtain is revealed. Given that extra information, should you switch? Well, curtain C has the prize is represented by the A/B card, curtain A has the prize is represented by the B/B card, and curtain B has the prize is represented by the A/A card. In other words, if you have dealt one of the cards where the front and the back is the same, then you want to switch….
Yes, you are right that the letter A face up tells you that the B/B card is impossible. You are also right that the A/B card and the A/A card are equally likely. However, the A/A card always lands A side up, and the A/B card does not. That’s a huge difference.
Dave/#52, I love the analogy. It illustrates why I feel like the issue is at least partly semantic. To answer: I would pick, say, Yaz. And I agree, absolutely, I’ve got a 1-in-289 chance of getting it right. And that doesn’t change, even after you list 287 of the 288 other guys. I always had a 1-in-289 shot. It was highly unlikely I’d guess right, but it was always possible. If it turns out to be Yaz, I hit on a 1-in-289 shot; it doesn’t change the odds.
But now, you give me the chance to switch. Do I still want to go with Yaz, or do I want to switch to Musial? This seems like a totally new game, with totally new odds. If I had guessed Musial instead of Yaz, I still would have had a 1-in-289 chance, not a 288-in-289 chance.
Of course, I bow to the people who understand this stuff and can do the proofs, as well as the people who did the flip-a-coin proofs.
Castlerook wins for being the first to provide an explanation that worked for me. Definitely prefer a mathy approach to all these explanations that just provide analogous ways to look at the exact same problem. Glad someone laid it out there so well.
AMR,
For whatever reason, after reading a dozen explanations, my monkey brain would not bend enough to accept this truth until I read your explanation.
You should become a teacher immediately. A teacher of monkeys.
Quickly skimmed through the comments but didn’t see anyone link to a simulation yet. Anyone who’s still doubting, should check out a site like this:
http://www.math.ucsd.edu/~crypto/Monty/monty.html
where you can run the simulation and see that switching gives you a 2 in 3 chance of winning.
Argh. Why does this problem keep coming up?
The entire premise of the dilemma is faulty. It pretends that you first make a choice on 1 in 3, and then make a second choice on 1 in 2, and whether switching would increase your chance of winning.
The problem is that the game is rigged, because someone who knows the winning door will always show you one of the two faulty doors. ALWAYS. No matter what you pick, you are shown one of the two faulty doors. This eliminates a choice. But this selection is not random, it is arbitrary. Introducing an arbitrary (and failproof) element in any probability scenario completely skews your percentages.
You are NOT playing a 1 in 3 plus a 1 in 2 game. The ONLY thing you are playing is a 1 in 2 game. The first choice you make is completely irrelevant. Whether you pick the right door or the wrong door or no door, doesn’t matter. Afterwards, you will be left with a choice of two doors, one of which you know contains the prize. This is a 50-50 chance, whether you choose to stick with your original door, or whether you choose to switch.
Any statistician who wants you to believe that switching gives you a higher chance of winning is conveniently ignoring the fact that an arbitrary element messed up his probabilities.
If there was no arbitrary element, but a RANDOM element to the probabilities, the idea would make more sense. But it’s not random. It’s arbitrary. You are ALWAYS shown a faulty door, and then asked to make another choice.
You only make one choice that matters. That is a choice between two doors. And that is a 50% chance. No matter WHAT happened before the arbitrary element was introduced.
Read through all the comments, and I have to say some of you do an excellent job of making the case for the switch theory.
I’m now confused whether the arbitrary element skews the percentages or whether it actually reinforces them.
I think I’m just not wired for probabilities.
“But now, you give me the chance to switch. Do I still want to go with Yaz, or do I want to switch to Musial? This seems like a totally new game, with totally new odds.”
And for a person just now wandering in, told to guess Yaz or Musial, it would be a new game. The difference is that you have the extra information that Musial is a stand in for “not Yaz”, and he doesn’t.
“I’m now confused whether the arbitrary element skews the percentages or whether it actually reinforces them.”
If the curtain revealed by the host were selected randomly (ie, he is just as likely to show a curtain with the prize as one without), then at the end when you are left with two choices, the switch strategy would be equivalent to the stay strategy.
The probabilities change in favor of switching when the curtains revealed by the host are constrained.
Whether that’s skewing or reinforcing depends on which answer you think is the natural one.
@Creston:
I think we should do an experiment. You can be Monty Hall, and put $20 under one of three cups. I’ll pay $11 to play the game, and we’ll play 10,000 times or so. If the odds are 50-50 like you say, you’ll make out like a bandit.
——-
An explanation I’ve always liked:
Let’s say my strategy is “I always switch”. In this case I only win if I guess wrong at the start, and the odds I guess wrong at the start is 2/3.
Let’s say my strategy is “I always stand pat”. In this case I only win if I guess right at the start, and the odds I guess right at the start is 1/3.
alas. though it SEEMS like the odds readjust, and as much as your intuition would like to think that it became a 50-50 problem, they don’t and it didn’t. keep reading the wonderful explanations and try out some of the experiments, and soon the solution really will just pop into place in your brain and you’ll wonder how you ever doubted. always change your choice when monty asks! loved this post, joe, and love these explanations/comments/confusions. truly wonderful!
Humans overanalyze, because our brains are wired to perceive patterns, even when there are none.
In a famous study, rats were placed in a maze that had two end points, A and B. A reward was placed at one of the end points. If the distribution of the rewards was 50-50, the rats proceeded to points A and B in a 50-50 manner. If, on the other hand, the reward was at A 60% of the time, and at B 40% of the time, the rats eventually learned to ALWAYS go to point A, thus receiving a reward 60% of the time. When a group of Yale undergraduates was given the same study, they continued to look for a pattern, even when the odds were 60-40; thus, they were still right only 50% of the time.
As for the Monty Hall problem, clearly the correct strategy is to switch, as has been pointed out by numerous other commenters. Being given information about one of the three choices, specifically that that is a bad choice, MUST influence the probability of the remaining two doors. I think roarke #26 has the best simple explanation of this.
And much thanks to Dave #31 for delineating one of my pet peeves: just because there are X possible outcomes does NOT mean that the probability of each outcome is 1/X. My first encounter with this fallacy was Woody Hayes’ observation that passing was bad (or at least not as good as running) because “there are only 3 outcomes [complete, incomplete, interception], and 2 are bad.” Obviously, if the probability of the “good” outcome is at least 51%, what difference does it make how many bad outcomes there might be?
(More on Woody Hayes, “genius”: he was fond of describing his run-only offense (see above) as “3 yards and a cloud of dust.” This was supposedly a good thing. But even as a 10-year-old, this prompted me to think, “So what? All that means is a lot of punts on 4th-and-1.”)
And as for Marilyn vos Savant: she does say a whole bunch of really dumb — and wrong — things, especially for someone who claims to be the world’s smartest human. Given that, I can see why many would assume that she was wrong about the Monty Hall problem. My favorite “puzzler” of hers was about walking from 1st Street to 20th Street in Manhattan. The so-called “problem” was which would take longer: getting from start to 10th Street, or 10th Street to end? She had a very long, pedantic answer/explanation for this; obviously, the answer is the latter because it’s a longer trip (1st to 10th equals 9 blocks; 10th to 20th is 10 blocks). Duh!
The answer to the “two girls” problem is obviously 50%. Assuming normal, random distribution of live-birth gender, we are simply asking “What are the odds of having a girl?” The gender of the older child has absolutely no bearing on the gender of the second child.
It’s a variation of one of my favorite puzzlers (that I used to win some small bets in high school): if I flip a coin 99 times, and I get 99 heads, what are the odds of getting heads (or tails, however you prefer to phrase it) on the 100th flip? The answer, of course, is 50% (assuming a fair coin); the previous outcomes of a random event have absolutely nothing to do with any future outcome. Most people confuse this with “What are the odds of getting 100 heads in a row?”; that is not the relevant question.
Thanks for this column, Joe — it was a lot of fun, and the comments have really been interesting, to say the least.
Greg/#58, Thanks!
Dave in NYC:
Sorry, but you are wrong about the two girls problem. The correct answer is 33%.
Here’s the problem again: “a woman, known to have two children, reveals that one is a girl. What are the odds that that both children are girls?”
Here’s the explanation.
There are four possible combinations for having two kids. You can either have BG, GB, GG, or BB. Each has an equal 25% chance of happening. All the problem told you is that one of the children is a girl. Not necessarily EXACTLY one– just at least one. This eliminates the BB possibility. There are three remaining solutions, only one of which has two girls. Therefore, the odds are 1/3.
The Wall Street Journal did a probability quiz on this. 43% of respondents got the right answer of 1/3. See here for more details: http://blogs.wsj.com/numbersguy/probability-quiz-results-and-winners-321/
If that doesn’t convince you, well, Google it.
This problem and discourse tormented me to the extent that I wondered if I could go on living. Alas, clarity came to me and I finally see the solution. The moment was borderline sexual.
Anyway, since everyone is offering their own way of thinking through this, perhaps my explanation will help just one person see the light (my blood-debt is due to VoiceOfUnreason above, thank you sir).
Here goes:
Imagine a map of the United States. Your dream girl is somewhere on that map. You get to pick one State. So you go up in the corner and pick Washington. Monty gets the rest of the map to find your dream girl. So he goes through the rest of the 49 States and gets rid of 48 of them. Now do you stick with Washington or switch over to the one State that Monty has left?
Don’t you assume that Monty probably better luck looking at the rest of the map and finding your girl? If you want any action tonight, you’d better switch, because Monty had the luxury of looking at 49 states, while you only had 1. Odds are he found that girl out in Mississippi or Maine. Dump Washington and take advantage of Monty’s work by going with the better odds.
Scale it back and it applies to the curtains. Whatever.
Nice work, Joe.
“my blood-debt is due to VoiceOfUnreason above, thank you sir”
No worries – there’s entertainment in discovering which explanation will turn on the light.
Dave @ 69
“Sorry, but you are wrong about the two girls problem. The correct answer is 33%.”
The analysis you offer is correct, as far as it goes. However, there’s another layer to this onion, an extra consideration you have not included. Whether this additional piece is important depends very much on the precise phrasing of the question.
See the links for the “advanced” version of the problem.
However, if the contestant is Matt Wieters, he’ll just use his x-ray vision and pick the right curtain first up, no switching required!
The best way I could explain it is to imagine you had the option of taking 1 door or 2. Obviously everyone would choose 2 doors even though you know 1 of them definately has goats, you just don’t know which one. All Monty Hall is doing is telling you which door has the goats, because he knows whats behind each door. The fact that he is confirming that one of your doors has goats does not decrease the odds that you have the prize.
I guess I am late to the party, but I have tried to explain the Monty Hall thing many times and I usually have the most success when I break it down very simply:
If you play the game 100 times and never switch, you will only win when you picked right initially. Since that is only going to be the case around 33 times, you would have won around 67 times had you switched (the only other choice.)
Not sure if this link has been provided, but it’s a nice simulator of the problem stated.
http://nlvm.usu.edu/en/nav/frames_asid_117_g_3_t_5.html?from=category_g_3_t_5.html
You can do up to 100 trials at a time and tell it to stick, switch, or alternate.
The Monty Hall problem, Hall of Fame voting, and Derek Jeter’s defense? Very few combinations of topics will get you as many comments.
Anyway, a similarly fun “paradox” is the boy/girl problem: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
First, the painful stuff. Joe, this is far from the best paragraph you ever wrote: I have no great for it. But I want to know how it works. For instance, while reading The Drunkard’s Walk, I became fascinated by this formula the author used to explain probability in a seven-game series. He used various statistical tools that I immediately that I immediately had to try myself. The first sentence should include a noun, e.g. “I have no great talent for it.” The last sentence repeats “that I immediately”.
As for the math, this is how bridge players might approach the Monty Hall problem. In fact, it’s analagous to “The Principle of Restricted Choice” which yields the same resulting odds (2:1 to switch) but I won’t detail here; you can go to: http://acbl-district13.org/artic003.htm or Google it if you’re interested. There are six possible outcomes before you pick:
Prize is behind A. Monty exposes B.
Prize is behind A. Monty exposes C.
Prize is behind B. Monty exposes A.
Prize is behind B. Monty exposes C.
Prize is behind C. Monty exposes A.
Prize is behind C. Monty exposes B.
Because Monty is never going to expose the winning prize, your choice did not matter. He is always going to pick a loser that you did not pick (unless you REALLY offended him, in which case he might not allow you to switch). There were always two possible curtain opens for any pick, and four other choices. Thus, whatever you picked (two outcomes) there were four other outcomes, even though he exposes the one you know to be wrong. Four to two is good odds; you should switch.
Is there any point in adding yet another comment? Probably not, but here goes.
The key is the assumption: “Monty ALWAYS opens a curtain that he knows has nothing there.”
Say you replace that with “Monty doesn’t know where the prize is. He’s just opening a curtain at random.” Then it’s 50/50- and the same applies even if there are a million curtains, although then he would be unlikely to find 999,998 empty curtains.
Or if you replace it with “Monty is trying to help you win- he will offer a switch only if you were wrong”, then you definitely switch.
Or, if you assume “Monty is trying to make you lose. He only offers a switch if he knows you had guessed right”, then you definitely don’t switch.
If someone offers you the chance to play this game, the tricky bit is working out which of the four assumptions (if any) is correct.
My challenge to those who think it’s 50-50 is to actually play the game. Get a friend, play this game with playing cards. Have your friend be Monty Hall – he must know where each card is and always choose a non-winning card to show you.
I guarantee it won’t take too many versions of the game before you decide it’s always in your best interest to switch.
“It’s a 1-in-3 chance to start, but once Curtain 3 is out of the loop, the odds have to redistribute. No?”
No. Curtain 3 isn’t out of the loop just because you know what’s behind it.
Suppose there were no reveal. That is, you choose Curtain 1, and then Monty — without opening any curtains — tells you you can switch your choice to take BOTH Curtain 2 and Curtain 3. Of course you would switch, because that gives you a 2/3 chance of winning instead of a 1/3 chance.
After Monty opens Curtain 3 and reveals nothing, now Curtain 1 (your choice) has a 1/3 chance, which Curtain 2 has a 2/3 chance. The odds do not “reset”.
All that math gave me a headache. I don’t understand it all but no matter how clueless I am I’d hesitate to call Marilyn Von Savant an idiot. I’d rather call David in NYC an idiot.
Glen, you’re right of course. The odds do not “reset.”. But they are also dependent on perspective. For instance, Monty knows what is behind each curtain, right? So to him, the odds of the curtains with duds behind them are precisely 0% that they will have season tickets. How can the odds of those curtains get “better” just because we change perspective to someone who knows less than Monty? Too much focus on the numbers distorts reality. It’s fun though!
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