The Monty Hall Trap

[Antrax]

https://sites.google.com/site/psmartinsite/Home/bridge-articles/the-monty-hall-trap

This feels wrong to me. Luckily there are plenty of probability experts around - does this argument have any merit?

[hautbois]

I'd want to know the conditions of the simulation being run to come up with the statistics.

 

The correct answer depends on your assumptions about Monty Hall’s actions. As I state in my article, if you assume he is simply opening a door at random, the probability you are holding the grand prize is indeed 50-50. If you assume he will always show you a booby prize, then his action is irrelevant; the probability remains one third.

 

If Monty is opening doors at random, there is a 33% chance he will open the jackpot rather than a booby prize. That you are still in the game means your odds have changed to 50-50.

 

If he's specifically opening non-jackpot doors, your odds have changed to 66%.

 

The simplest way to understand. Imagine there are 1,000,000 doors. Pick one. Monty will open 999,998 doors with knowledge of where the jackpot is and specifically avoiding it. How confident are you that you knew where the jackpot was when you initially chose a door or would you like to rely on the extra, biased information Monty gave you?

[GreenMan]

The Monty Hall problem contains an unstated assumption that I've only ever seen pointed out once: Monty Hall will ALWAYS show you one of the booby prizes.

 

If that is his strategy, then the probabilities are as stated. However, he has no reason to follow a course whose counter-strategy is obvious.

 

If instead, say, he shows you a booby prize 100% of the time you chose the grand prize and 50% of the time when you don't, then switching and not switching are equally likely to be correct.

 

This doesn't change the bridge calculations, in which the defenders don't have the option of playing some of the time and not others.

[billw55]

The Monty Hall problem contains an unstated assumption that I've only ever seen pointed out once: Monty Hall will ALWAYS show you one of the booby prizes.

 

If that is his strategy, then the probabilities are as stated. However, he has no reason to follow a course whose counter-strategy is obvious.

Yes, that is the standard form of the Monty Hall problem.

 

And, the reason he follows this strategy is that it is not at all obvious to most people. I have posed this problem to quite a few people over the years, and every one has said that the odds are now 50-50 and switching doors doesn't matter.

 

IMO the simplest explanation is as follows: if your original choice is wrong, and you switch doors, what happens? (you win). How often is your original choice wrong? (two thirds of the time).

 

Like other problems in probability, it is somewhat counter-intuitive. I think the stumbling block for most people is not realizing that since he can always open a losing door, doing so contains no information about the original choice.

[GreenMan]

Yes, that is the standard form of the Monty Hall problem.

 

And as I said, the standard approach assumes facts not in evidence. MH's own strategy is unstated; everyone posing this problem assumes he opens a door 100 times out of 100, but if that's the case it should be stated explicitly.

[jjbrr]

Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have two children,” she replied.

 

What is the probability that both of her children are girls?

[billw55]

Walking down the street one day, I met a woman strolling with her daughter. "What a lovely child," I remarked. "In fact, I have two children," she replied.

 

What is the probability that both of her children are girls?

I'm going to say one third, hoping to not be wrong and look silly.

 

Then there is the birthday problem: 30 random people are gathered in a room. What is the probability that at least two have the same birthday?

[DrMunk]

Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have two children,” she replied.

 

What is the probability that both of her children are girls?

 

Don't answer this before you have taken this into account:

 

http://en.wikipedia.org/wiki/Sex_ratio

 

:)

[GreenMan]

BTW I griped about how the MH problem was posed in the link in the OP, but I found the application to the bridge problems fascinating.

[PhilKing]

Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have two children,” she replied.

 

What is the probability that both of her children are girls?

 

Very low. Boys don't go out strolling with their mother and sister, but a sister would have been there if she could.

 

Anyway, she was probably lying and was not even the mother.

[billw55]

Very low. Boys don't go out strolling with their mother and sister. The other sister would have been there if she could.

A true bridge player's answer :) don't rely solely on raw numbers when an inference is available.

[hrothgar]

There was a really good discussion of restricted choice on the forums a few years back.

 

As I recall, Phil and Justin had some really good posts...

[jjbrr]

I'm going to say one third, hoping to not be wrong and look silly.

 

Then there is the birthday problem: 30 random people are gathered in a room. What is the probability that at least two have the same birthday?

 

I apologize; I was trolling a bit, though I find this problem somewhat interesting notwithstanding. I think my problem boils down to semantics and interpretation of the problem.

 

Here's a quote from a guy who probably knows his stuff:

 

My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbor has one boy and one girl, with probability 1/2. The other possibilities---two boys or two girls---have probabilities 1/4 and 1/4.

 

Suppose I ask him whether he has any boys, and he says yes. What is the probability that one child is a girl? By the above reasoning, it is twice as likely for him to have one boy and one girl than two boys, so the odds are 2:1 which means the probability is 2/3. Bayes' rule will give the same result.

 

Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl? Observing the outcome of one coin has no affect on the other, so the answer should be 1/2. In fact that is what Bayes' rule says in this case. If you don't believe this, draw a tree describing the possible states of the world and the possible observations, along with the probabilities of each leaf. Condition on the event observed by setting all contradictory leaf probabilities to zero and renormalizing the nonzero leaves. The two cases have two different trees and thus two different answers.

 

This seems like a paradox because it seems that in both cases we could condition on the fact that "at least one child is a boy." But that is not correct; you must condition on the event actually observed, not its logical implications. In the first case, the event was "He said yes to my question." In the second case, the event was "One child appeared in front of me." The generating distribution is different for the two events. Probabilities reflect the number of possible ways an event can happen, like the number of roads to a town. Logical implications are further down the road and may be reached in more ways, through different towns. The different number of ways changes the probability.

 

click

[gwnn]

I think we could stop talking about the Monty Hall problem as we all know the right answer. Could we actually talk about the bridge hand (yes there is a bridge hand at the bottom of the page)? I have a feeling the author is drawing his conclusions a bit too hastily.. Is it really that simple?

[ArtK78]

I'm going to say one third, hoping to not be wrong and look silly.

 

Then there is the birthday problem: 30 random people are gathered in a room. What is the probability that at least two have the same birthday?

I don't remember the exact probability for 30 random people, I just know that if you have 23 random people in a room, the chances are greater than 50% that two of them have the same birthday.

 

This was one of the first problems posed in my basic probability and statistics course in my freshman year in college. It was easy to set up an algorithm for this and solve it on a computer (yes, we had computers when I attended college, but they were BIG computers). And I always thought that the solution was surprising. But it is correct.

[nate_m]

A lot of people have negative knee-jerk reactions to applications of Bayes' Theorem. However, even skeptics use reasoning similar to that used in the article. For example, when somebody leads from Jxxx or underleads an ace vs. a suit contract, there is an inference that they had no attractive lead. When somebody leads a shorter suit against a NT contract, the same reasoning makes it less likely they have a longer one on the side. Of course you have to be careful with your assumptions, but the general idea is sound.

 

Justin blogged an article on leads that uses similar reasoning to guess a missing queen.

[gnasher]

Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl?

Sounds like a trivial restricted-choice question. Most things can be simplified by converting them into bridge problems.

[gnasher]

I think we could stop talking about the Monty Hall problem as we all know the right answer. Could we actually talk about the bridge hand (yes there is a bridge hand at the bottom of the page)? I have a feeling the author is drawing his conclusions a bit too hastily.. Is it really that simple?

There are two hands. Which one are you worried about? This is one of my favourite bridge articles, so I'd prefer not to find out that it's wrong.

 

Having said that, there is one bit of analysis in the second deal that is incorrect:

 

If East had not opened, you would still know that clubs were three-five (He would not have played the nine at trick one from king-queen-nine.) and it would be clear to finesse West for the spade queen.

If a strong East played 9-Q-K, you should be wondering why he so helpfully told you the club distribution. If you also know that he knows you're good enough to wonder, you should be wondering if he wants you to wonder.

[Finch]

I don't remember the exact probability for 30 random people, I just know that if you have 23 random people in a room, the chances are greater than 50% that two of them have the same birthday.

 

This was one of the first problems posed in my basic probability and statistics course in my freshman year in college. It was easy to set up an algorithm for this and solve it on a computer (yes, we had computers when I attended college, but they were BIG computers). And I always thought that the solution was surprising. But it is correct.

 

I can't decide if that anecdote makes you older or younger than I am.

I was first introduced to that problem at school, and we found the probability for 23 people by using some handy approximations and a calculator.

[kenberg]

I will give it more thought but my first impression is that his reasoning is off. In these probability problems it is always essential to get a very precise fix on what is being assumed. I have only looked at what he calls problem 1 but here is what I get out of his assumptions.

 

By ducking the first trick, it becomes apparent (let's say it becomes certain for the purpose of analysis) that W has led from a five card suit. In addition we assume that W would always lead from a five card suit when he has one, and I think he supposes that the trick 2 play would always disclose whether the lead was from four or from five. If I am correctly understanding these assumptions, then I see no relationship to Monty Hall or to Restricted Choice. In the Monty Hall problem Restricted Choice applied. If the prize was behind the door originally selected, Monty had a choice of doors to open, if the prize was not behind the originally selected door, Monty's choice was restricted. But in the current situation, if I understand it correctly, W willl lead from a five card suit anytime he has one and the trick 2 play will always disclose it, and he will lead from his four card suit anytime he does not have a five card suit, and the trick 2 play will disclose that it was from only four. If so, W is basically a robot. He agrees to tell declarer that he has a five card suit whenever he has it, and he agrees to tell declarer that he does not have it when he does not have it. I see no issue of Restricted Choice.

 

Of course play at the table is not so simple and it may well be unclear whether the lead was from four or five. Or maybe I misunderstand the stated assumptions.

 

About MH. As noted, the statement has to be precise. For example suppose the doors are numbered 1,2,3 and suppose the contestant first picks door 1. Suppose MH will then always opens door 2 unless the prize is not behind door 2. This means that if MH opens door 3 then surely the prize is behind door 2 (a really Restricted Choice) while if MH opens door 2 it is 50-50 whether the prize is behind door 1 or door 2, because MH robotically opens door 2 in either case. There are many such variants.

 

Anyway, I look forward to learning what it is that I misunderstood, but for conditional probability to apply there must be some non-robotic choices somewhere.

 

Added: Now as to the calculation of probabilities, I have not done it. Maybe his numbers come out. We want the probability that W was dealt exactly two clubs, given that he was dealt at least one five card suit and, as will be known at crunch time, either one club spot or Qx. This requires some thought. Here is where I think he is right that it is easy to be simplistic. The fact that the suit is spades is not relevant. He will lead whatever five card suit he has, and knowing that it is spades rather than hearts or diamonds is irrelevant. So I need to think about the probabilities. The fact that two five card suits is possible (not likely but possible) confuses the issue a bit.

 

All I am saying so far is that I don't see that this involves Restricted Choice. Well, I suppose the fact that he led spades leads to thoughts that if he had two five card suits he might have led the other. A non-robotic play. Maybe that's it.

 

I will think a little more.

[Vampyr]

Back to the article:

 

Since you already knew that at least one of the other two doors held a booby prize, you have learned nothing. You still have the same one chance in three that you started with.

This is very obviously wrong, since there are only two doors left. Even if the author doesn't understand that switching changes your odds to 2/3, he ought to be able to see that you can't have a one in three chance if there are only two doors.

[Antrax]

I may have misstated my question. I'm actually not that interested in the original Monty Hall, in the birthday paradox and in the 99% accurate test.

Solution 1's reasoning seems off to me, for the exact reason kenberg gives. It's irrelevant that west has more *spades* than his partner, just that he has more *known cards not containing the ♣Q* than his partner.

Suppose your opponents at the other table somehow reach three notrump from the North hand. East leads a red suit and (surprise!) he has more cards in that suit than West. Is your opponent supposed to finesse against West for the club queen while you finesse against East?

Yes, if E shows ♦ are 3-5, then the odds favor finessing against W. If E shows diamonds are 3-5 and W shows spades are 5-3 on the same deal, then it's a toss.

Given that spades is West's longest suit, the expected spade break is roughly four and a half-three and a half. West rates to have one more spade than East. So the actual five-three break is only one card away from expectation. It is equivalent to a random suit's breaking four-three.

 

That means you can counter the bias by pretending that East has only one extra unknown card instead of two. You cash the club king and lead toward the ace. East follows. Now he has zero extra unknown cards. So it's a toss-up.

This just seems wrong to me. The first sentence is mumbo-jumbo, and everything else hinges on that. After cashing the ♣K and leading a club from N, you've eliminated the possibility E has Q doubleton or any singleton. How that swings the pendulum back towards the drop is beyond me.

[GreenMan]

But in the current situation, if I understand it correctly, W willl lead from a five card suit anytime he has one and the trick 2 play will always disclose it, and he will lead from his four card suit anytime he does not have a five card suit, and the trick 2 play will disclose that it was from only four. If so, W is basically a robot. He agrees to tell declarer that he has a five card suit whenever he has it, and he agrees to tell declarer that he does not have it when he does not have it. I see no issue of Restricted Choice.

 

The article addresses those concerns. In the initial condition:

 

"West leads a low spade. You duck in both hands; East continues spades. It appears from the carding that spades are five-three."

 

Similarly for the other two cases.

 

Also at the end:

 

"These arguments assume that West can be relied upon to have led his longest suit. If the auction makes certain leads unattractive, or if West has led from a sequence, or if West is simply known to be perverse, then he might have a longer suit and none of this applies."

 

Obviously the opps may be able to fool you, but sometimes you do have a good indication, even against humans, and then the article's arguments will apply.

[awm]

I think the basic line of thought is right, but would disagree with the actual odds here.

 

The basic theory is that on certain hand patterns, the person on lead might've had a choice of equally appealing leads. These should be discounted in the probability distribution because "if he had that hand, he might've lead something else." If we assume that LHO would always lead his longest suit or pick randomly from suits of equal length, the the odds worked out at the end of "solution one" are likely correct.

 

However, in reality I think LHO is virtually always leading a spade from 5-5 in the pointed suits, because the auction given calls for a major suit lead. And if he was 5-5 in the majors he might've bid (and in any case, 5-5 majors is much less likely because we can see a lot of hearts between declarer and dummy). This swings the odds to the point where finessing looks better than the drop.

 

The inference is stronger if LHO leads from a four-card spade suit. Presuming that he'd prefer a five-card suit, the only way he can have singleton club is if he's precisely 4441. Further, he would have a choice of equally appealing leads from that hand. Playing LHO for a singleton is really quite poor in this case! If he has doubleton club, he must be 4432 or 4342, whereas with three clubs he can be 4333 or 4243 or 4423. Note that 4432 and 4423 have equally appealing heart leads, whereas 4333 is presumably always leading a spade. It follows that we should finesse LHO for the queen of clubs (4333 or 4243 or half 4423 more likely than half 4432 or 4342).

[GreenMan]

This is very obviously wrong, since there are only two doors left. Even if the author doesn't understand that switching changes your odds to 2/3, he ought to be able to see that you can't have a one in three chance if there are only two doors.

 

You started with a 1/3 chance. You knew that at least one of the other doors had no prize behind it. Nothing has changed. The chance your door was the right one is still 1/3.