### Monty Hall - The Last Straw

Jul. 14th, 2010 05:44 pm**lenv1**

I'm used to seeing people butcher the Monty Hall problem. In fact, the only person who gets it right (besides me) is Monty Hall. He did a great interview in the New York Times a few years ago. Now a Canadian, bridge playing mathematician is bridge-blogging about it and I just can't stand it any more. So, here is a breakdown of the usual problem statement, the most common answers, and the correct answer.

BTW, this is almost entirely not about bridge. The common analogy of MH to restricted choice works well because, in a bridge setting, all the usual assumptions are met. In short, the usual answer there (finesse when you see an honor, it's about 2-1 in your favor) is correct. If that's all you're interested in, you don't need to read any further.

You go on Let's Make a Deal and Monty Hall shows you three doors. He says there is a goat behind each of the two losing options, and some prize you want (e.g. a car) behind the other door. He tells you to pick a door. You pick a door and he opens a different door, revealing a goat. Then he asks whether you'd prefer to switch to the third door, or stick with your original choice. What do you do, and why?

People make excuses for why certain answers are right, even after you demonstrate they are completely wrong. In particular, when you point out a direct contradiction, they claim it's a linguistic issue and you know what they meant. Fine. Except if they said something other than what they meant, and what they said was wrong, it's fair game. The real problem I have with this sort of imprecision is that they say one thing meaning another, then add a third thing that makes perfect, intuitive sense, given what they said, but not given what they "meant". That leads to fuzzy thinking. I expect better from the smartest woman in the world. And I demand better from mathematicians.

In a way, Marilyn was right, given certain (false) assumptions. The mathematicians were entirely correct, given different (false) assumptions. Monty is awesome, obviously. Let's look at various assumptions people have made, but not stated:

Marilyn assumed, without stating, numbers 1, 2, 3, 4, 5 and 6. The mathematicians (note: these are the ones who disagreed with her, not every single mathematician in the world) assumed, without stating, 7. When pressed, they might admit that they had also assumed 1, 3, 4 and 5, but those were not necessary for their argument. FWIW, Monty seemed to believe 2 and 3. The fact that Marilyn made SO MANY more untested assumptions makes me lean towards the mathematicians (or, really, their argument. I'm doing it again...), but their one assumption is so grossly unreasonable they come off looking extremely foolish. I don't mind pointing this out, since I don't know any of their names.

Should you switch doors? Of course not! If you want a car, and don't know what's behind each door, and believe that Monty doesn't want you to have a car, and believe Monty has free will, the bulk of the time he offers a swap its not in your best interest. If, however, you grant all the assumptions of the mathematicians (or don't grant assumption #4), what you do doesn't matter. If you grant Marilyn's reasonable (but incorrect) assumptions, you should swap.

There's a bridge blog about this type of math puzzle and how it applies to different bridge situations, by the Canadian, bridge playing mathematician I mentioned earlier. I didn't even make it to the bridge part of the blog. Here's the relevant part. BTW, it looks like he found this problem in a probability book (one I haven't read) so I hope this isn't too much like a game of telephone, where my version bears no resemblance to the original formulation. The blogger writes (with my interruptions):

I could pick nits here (boys are more likely to be born than girls, coin tosses are biased and serially correlated, and birth order is irrelevant (it's a proxy for conditional probability)) but the math and conclusion are sound, and the assumptions are even stated! Oh, the probability of two finesses succeeding is either 24% or 26% (depending on whether you finesse through the same hand both times).

Next we ask if one child is known to be a girl what are the chances the other is also a girl? It would be wrong to argue that given that one is a girl doesn’t affect the odds the other is a boy, so the chances of 2 girls should be 50%. The correct argument is that the birth order of boy-boy has been removed from consideration, so there are 3 possible sequences remaining leaving the chances of 2 girls at 1 in 3. Similarly, if we are assured that at least one finesse wins, the chance of the other also succeeding is 1 in 3.

Call me a nit picker, but this is COMPLETELY false. Grab the nearest five year old you can, and pose the question. You have two children. ONE is a girl. Then, what must the other one be? The kid will get it right, two kids, one girl, therefore one boy. Yes, he meant at least one is a girl, of course, but that's TOTALLY different a paragraph from now.

Next we ask, what are the chances of 2 girls given one of them is named Florida? There are those who would argue that whether the girl was named Florida, or Jane, or Laura should make no difference to the odds that their other child is a boy. Although the name Florida is unusual, there is no causal effect at work. Consider the problem statistically and imagine going through US census data looking for all parents with 2 children one of whom is named Florida. Can we expect to find that the other child is a more likely also to be a girl? It doesn’t make sense that we should.

Now my head explodes. Here are my issues:

Although there seems to be no causal link between the name and the probability of 2 girls, the argument doesn’t solve the Girl-Florida problem as posed. The correct solution is obtained by incorporating the information that a daughter is named Florida, condition FL, into the possible sequences of births. The possibilities are the following 4: boy- FL, FL-boy, FL-girl, and girl-FL, in half of which the other child is a girl. This corresponds to our intuitive feeling that it doesn’t matter whether the girl was named Florida, or Jane, or Laura, the chances are 50-50 the other child is also a girl. What does matter is that the naming of one child changes the odds for a second girl from 1 out of 3 to 1 out of 2.

AAAARGH! This is the exact same INCORRECT argument he disdainfully claimed was common in centuries past. Again, birth order has NOTHING to do with anything. He just stated by fiat that four possibilities are equally likely, when that's not true given the selection criteria. In practice, if you pick some normal name (e.g. Joan) there's a pretty good chance the probability of two girls will be about 50%. That's not because anyone named the child (note, it applies equally to second children named Joan) but because of the way you select your population. I just wish people would state their assumptions clearly.

And if you don't believe me, I'll write a simulation to prove it to you.

Bring money...

BTW, this is almost entirely not about bridge. The common analogy of MH to restricted choice works well because, in a bridge setting, all the usual assumptions are met. In short, the usual answer there (finesse when you see an honor, it's about 2-1 in your favor) is correct. If that's all you're interested in, you don't need to read any further.

**The Usual Setup**You go on Let's Make a Deal and Monty Hall shows you three doors. He says there is a goat behind each of the two losing options, and some prize you want (e.g. a car) behind the other door. He tells you to pick a door. You pick a door and he opens a different door, revealing a goat. Then he asks whether you'd prefer to switch to the third door, or stick with your original choice. What do you do, and why?

**The Usual Answers**- Marilyn Vos Savant, with the world's highest recorded IQ, wrote in her Sunday column that you should switch. She said you had one chance in three of picking correctly at first, and nothing has changed, since Monty could always show you a goat, regardless of which door you picked originally. She was wrong.
- A bunch of mathematicians (professors, PhD's, etc.) wrote in to say various mathy and nasty things (e.g. You haven't learned the math they dole out in eighth grade). They believe that there are two doors remaining, and each has a different prize, so it's a straight 50-50 chance whether to switch or not. Needless to say, they were hella wrong.

**Monty Hall's Answer**

- He said that if you picked a goat, he could just let you have the goat. If you picked a car, he could offer to let you swap. Therefore, you shouldn't swap. I'm not going to argue with Monty.

**Collaborators**People make excuses for why certain answers are right, even after you demonstrate they are completely wrong. In particular, when you point out a direct contradiction, they claim it's a linguistic issue and you know what they meant. Fine. Except if they said something other than what they meant, and what they said was wrong, it's fair game. The real problem I have with this sort of imprecision is that they say one thing meaning another, then add a third thing that makes perfect, intuitive sense, given what they said, but not given what they "meant". That leads to fuzzy thinking. I expect better from the smartest woman in the world. And I demand better from mathematicians.

**Unstated Assumptions**

In a way, Marilyn was right, given certain (false) assumptions. The mathematicians were entirely correct, given different (false) assumptions. Monty is awesome, obviously. Let's look at various assumptions people have made, but not stated:

- Each person playing the game is always shown the goody behind a door other than the one chosen, and is always offered the opportunity to swap the first door for the third door (Note: I didn't mean the first door for the third door, I meant the prizes behind those doors, but you "knew what I meant". Isn't that confusing? I'll try to avoid that from now on...)
- Monty knows where the goats are
- Monty doesn't want you to win a car
- You want to win a car (I have to admit, I like goats, though not in that way...)
- You can neither hear nor smell the goats specifically enough to locate them
- Monty picks a door, on purpose, that hides a goat (as opposed to using some random mechanism independent of his knowledge and goals)
- Monty picks a door via some random mechanism, independent of his knowledge and goals. In particular, that random mechanism is equally likely to pick either door you hadn't chosen

Marilyn assumed, without stating, numbers 1, 2, 3, 4, 5 and 6. The mathematicians (note: these are the ones who disagreed with her, not every single mathematician in the world) assumed, without stating, 7. When pressed, they might admit that they had also assumed 1, 3, 4 and 5, but those were not necessary for their argument. FWIW, Monty seemed to believe 2 and 3. The fact that Marilyn made SO MANY more untested assumptions makes me lean towards the mathematicians (or, really, their argument. I'm doing it again...), but their one assumption is so grossly unreasonable they come off looking extremely foolish. I don't mind pointing this out, since I don't know any of their names.

**The Answer**Should you switch doors? Of course not! If you want a car, and don't know what's behind each door, and believe that Monty doesn't want you to have a car, and believe Monty has free will, the bulk of the time he offers a swap its not in your best interest. If, however, you grant all the assumptions of the mathematicians (or don't grant assumption #4), what you do doesn't matter. If you grant Marilyn's reasonable (but incorrect) assumptions, you should swap.

**Caveats**- Some believe that the problem statement implies you will always be offered the option to swap. Even when I discuss this, and THEN ask the victim to state the problem in their own words, they never, ever state this assumption (assumption number 1)
- When anyone offers to prove you're wrong about, essentially, anything by showing you the results of a computer simulation, you should be extremely wary. In particular, if a mathematician offers to play a simulation game with you, and tells you to BRING MONEY, what they mean is that they'll write a computer program (or pull tickets out of a box, or flip coins, or whatever) to demonstrate the results of their unstated assumptions. It's easy to write a simulator, in good faith, that proves your point, as long as you don't question all your assumptions
- That's why restricted choice is 2:1, wihtout any further (non-bridge) assumptions. When you have ATxxxx opposite Kxx, and it goes K-x-x-quack, you always get to decide whether to finesse or play for the drop. Monty can't make your decision for you

**A Girl Named Florida**

There's a bridge blog about this type of math puzzle and how it applies to different bridge situations, by the Canadian, bridge playing mathematician I mentioned earlier. I didn't even make it to the bridge part of the blog. Here's the relevant part. BTW, it looks like he found this problem in a probability book (one I haven't read) so I hope this isn't too much like a game of telephone, where my version bears no resemblance to the original formulation. The blogger writes (with my interruptions):

Suppose that a couple have produced 2 naturally conceived children. What are the chances they are both girls? We assume that at the time of conception a boy is as likely to result as a girl. The event is mathematically equivalent to tossing a coin. First we present a false argument that was common in centuries past which goes as follows. There are 3 equally probable states: 2 boys, 2 girls and a boy and a girl. The chance of producing 2 girls is 1 in 3? That is wrong because the probability of a given outcome of a series of random events is proportional to the number of ways in which that outcome could have been produced. One must take into account the birth orders, of which there are 4: boy-boy, boy-girl, girl-boy, and girl-girl. The chance of producing 2 girls is 1 in 4.

Pascal would have got it right, as would most bridge players who are asked, given that 2 finesses are to be taken, what are the chances both succeed? There are 4 equally likely possible outcomes of the play, for two of which one finesse wins and the other loses. The chance of both finesses succeeding is 1 in 4.

Pascal would have got it right, as would most bridge players who are asked, given that 2 finesses are to be taken, what are the chances both succeed? There are 4 equally likely possible outcomes of the play, for two of which one finesse wins and the other loses. The chance of both finesses succeeding is 1 in 4.

I could pick nits here (boys are more likely to be born than girls, coin tosses are biased and serially correlated, and birth order is irrelevant (it's a proxy for conditional probability)) but the math and conclusion are sound, and the assumptions are even stated! Oh, the probability of two finesses succeeding is either 24% or 26% (depending on whether you finesse through the same hand both times).

Next we ask if one child is known to be a girl what are the chances the other is also a girl? It would be wrong to argue that given that one is a girl doesn’t affect the odds the other is a boy, so the chances of 2 girls should be 50%. The correct argument is that the birth order of boy-boy has been removed from consideration, so there are 3 possible sequences remaining leaving the chances of 2 girls at 1 in 3. Similarly, if we are assured that at least one finesse wins, the chance of the other also succeeding is 1 in 3.

Call me a nit picker, but this is COMPLETELY false. Grab the nearest five year old you can, and pose the question. You have two children. ONE is a girl. Then, what must the other one be? The kid will get it right, two kids, one girl, therefore one boy. Yes, he meant at least one is a girl, of course, but that's TOTALLY different a paragraph from now.

Next we ask, what are the chances of 2 girls given one of them is named Florida? There are those who would argue that whether the girl was named Florida, or Jane, or Laura should make no difference to the odds that their other child is a boy. Although the name Florida is unusual, there is no causal effect at work. Consider the problem statistically and imagine going through US census data looking for all parents with 2 children one of whom is named Florida. Can we expect to find that the other child is a more likely also to be a girl? It doesn’t make sense that we should.

Now my head explodes. Here are my issues:

- Those who argue the name makes no difference are correct.
- There might be no causal effect, but he hasn't demonstrated that. Kid named Florida gets teased, the parents don't want another girl, and so on. Ok, maybe that's a stretch
- Even granted there's no CAUSATION, that doesn't mean there's no CORRELATION. The self selected group of people who name a girl Florida might be less likely to want girls (e.g. the ratio of boys to girls among third children in China is almost 3-1, not that I assert they name any of them Florida)
- Remember when I complained that he said one of them was a girl? Now he looks for families where that girl (that one girl) is named Florida). Sound reasonable? Try restating it correctly: There are two children, at least one is a girl, and SHE is named Florida. That's doesn't parse. One is a girl and named Florida parses fine, but it isn't what he means.
- He assumes that two siblings can't have the same name. That's false (see Michael Jackson's two sons, or George Foreman's five sons) but irrelevant, as it happens

Although there seems to be no causal link between the name and the probability of 2 girls, the argument doesn’t solve the Girl-Florida problem as posed. The correct solution is obtained by incorporating the information that a daughter is named Florida, condition FL, into the possible sequences of births. The possibilities are the following 4: boy- FL, FL-boy, FL-girl, and girl-FL, in half of which the other child is a girl. This corresponds to our intuitive feeling that it doesn’t matter whether the girl was named Florida, or Jane, or Laura, the chances are 50-50 the other child is also a girl. What does matter is that the naming of one child changes the odds for a second girl from 1 out of 3 to 1 out of 2.

AAAARGH! This is the exact same INCORRECT argument he disdainfully claimed was common in centuries past. Again, birth order has NOTHING to do with anything. He just stated by fiat that four possibilities are equally likely, when that's not true given the selection criteria. In practice, if you pick some normal name (e.g. Joan) there's a pretty good chance the probability of two girls will be about 50%. That's not because anyone named the child (note, it applies equally to second children named Joan) but because of the way you select your population. I just wish people would state their assumptions clearly.

And if you don't believe me, I'll write a simulation to prove it to you.

Bring money...