"Monty Hall" Three Doors Problem

[Mike Strand]

Glad to see this new general topic in the forum!

Here's a problem that is quite simple in appearance, but requires close attention to fundamental ideas in mathematical reasoning:

The contest host Monty Hall shows the contestant Joe three closed doors. He tells Joe that behind one of the doors is a valuable prize. The prize goes to Joe if Joe chooses the door with the prize. Joe doesn't know which door holds the prize. He chooses one of the doors, maybe by some random process or perhaps by whim. Call his choice door A. Monty then opens up one of the remaining two doors -- the one not holding the prize (Monty knows which one -- call this door B). Replace the previous sentence by: "Monty, who knows where the prize is, then opens up one of the remaining two doors -- one not holding the prize -- call this door B". Then Monty asks Joe to stick with the original choice (A) or make a switch to the third door (C).

Question: Should Joe switch his choice to the third door (C)?

Use simulation or mathematical reasoning (e. g. principles of probability theory) to decide whether Joe should switch or not. It would be a service to the rest of us to explain your reasoning or your simulation.

This puzzle can be extended to 4, 5, ..., n doors, and the answer to the question will be the same.

[chaz wyman]

Glad to see this new general topic in the forum!

Here's a problem that is quite simple in appearance, but requires close attention to fundamental ideas in mathematical reasoning:

The contest host Monty Hall shows the contestant Joe three closed doors. He tells Joe that behind one of the doors is a valuable prize. The prize goes to Joe if Joe chooses the door with the prize. Joe doesn't know which door holds the prize. He chooses one of the doors, maybe by some random process or perhaps by whim. Call his choice door A. Monty then opens up one of the remaining two doors -- the one not holding the prize (Monty knows which one -- call this door B). Then Monty asks Joe to stick with the original choice (A) or make a switch to the third door (C).

Question: Should Joe switch his choice to the third door (C)?

Use simulation or mathematical reasoning (e. g. principles of probability theory) to decide whether Joe should switch or not. It would be a service to the rest of us to explain your reasoning or your simulation.

This puzzle can be extended to 4, 5, ..., n doors, and the answer to the question will be the same.

The answer is that he should ALWAYS change his choice as that consistently increases his chances of winning. After having a 33% chance, by changing his choice he now increases his chance to 50%. This at first seems counter intuitive as it seems that it should not make any difference. And yet you can demonstrate that it actually works in cases where the choice is repeated over several cases.

[Mike Strand]

Thanks, Chaz. I believe you are correct that Joe will have a better chance to win by switching. Check the probability, though -- I believe it increases from 1/3 to 2/3, not just to 50%.

If you need to see my argument, I'll post it. Remember, after Joe chooses a door, Monty always opens a door among the remaining two without the prize before asking Joe to decide whether to switch. So Joe naturally switches to the one door Monty doesn't open.

My statement of the problem was awkward. I should have said that "Monty, who knows where the prize is, then opens up one of the remaining two doors -- one not holding the prize -- call this door B" . It could be that neither of the remaining doors holds the prize, or one does. Monty always opens one that doesn't.

Of course, this problem rests on the "usual" simplifying assumptions: Joe has no "inside knowledge" where the prize is placed - the prize is assigned randomly to a door with equal probabilities, outside of Joe's knowledge. So the game is "fair". This means that even if Joe always picked the left-most door in repetitions of the game, the probability of winning is the same.

[chaz wyman]

Thanks, Chaz. I believe you are correct that Joe will have a better chance to win by switching. Check the probability, though -- I believe it increases from 1/3 to 2/3, not just to 50%.

If you need to see my argument, I'll post it. Remember, after Joe chooses a door, Monty always opens a door among the remaining two without the prize before asking Joe to decide whether to switch. So Joe naturally switches to the one door Monty doesn't open.

My statement of the problem was awkward. I should have said that "Monty, who knows where the prize is, then opens up one of the remaining two doors -- one not holding the prize -- call this door B" . It could be that neither of the remaining doors holds the prize, or one does. Monty always opens one that doesn't.

Of course, this problem rests on the "usual" simplifying assumptions: Joe has no "inside knowledge" where the prize is placed - the prize is assigned randomly to a door with equal probabilities, outside of Joe's knowledge. So the game is "fair". This means that even if Joe always picked the left-most door in repetitions of the game, the probability of winning is the same.

Okay - but the reason I said 50% is that he has a choice of two at the end. SO he has two choices the first is 1-3, the second 1-2. I'm not sure it is fair to call that 2/3. I'm no maths expert and still find this bizarre.

In the scenario I understand that the selection process is always random - no one knows what the doors hold. It might be possible for the right door to be chosen first time - otherwise you would be giving an active advantage by eliminating a bad choice - which would not make the test as remarkable.

[Richard Baron]

It's 2/3. The point is that the prize does not have any chance of moving after Monty opens a door (say one of B and C, the player having chosen A), behind which the prize is not to be found. So it is not 1/2 because it is not the case that after that opening, the prize is randomly allocated between A and and the other unopened door. It is already behind one of A and that other door, and it stays there.

Suppose that the player always chooses A, and plays the game several times.

Each time, the prize is first positioned at random.

One third of the time, it will be behind A, and the player will lose by switching.

Two thirds of the time, it will be behind B or C, and the player will win if he switches to B-C and knows which one of B and C to pick.

But he does know which one to pick, because Monty has opened the wrong one out of B and C.

[chaz wyman]

It's 2/3. The point is that the prize does not have any chance of moving after Monty opens a door (say one of B and C, the player having chosen A), behind which the prize is not to be found. So it is not 1/2 because it is not the case that after that opening, the prize is randomly allocated between A and and the other unopened door. It is already behind one of A and that other door, and it stays there.

Suppose that the player always chooses A, and plays the game several times.

Each time, the prize is first positioned at random.

One third of the time, it will be behind A, and the player will lose by switching.

Two thirds of the time, it will be behind B or C, and the player will win if he switches to B-C and knows which one of B and C to pick.

But he does know which one to pick, because Monty has opened the wrong one out of B and C.

Sorry are you refuting that switching is always an advantage?

[Richard Baron]

Sorry are you refuting that switching is always an advantage?

No, I am saying that switching is always the right thing to do, because it increases the probability that you will win from 1/3 to 2/3.

There are plays of the game on which you will lose by switching: those plays when the prize is behind the door you picked first. But that cannot be helped. It is still true that always switching will lead to more wins in the long run than never switching, or than sometimes switching and sometimes not switching.

If you play the game just once, switch, and then lose, you will kick yourself. But if you played just once, did not switch, and lost, you would also kick yourself. And you reduce the probability of having to kick yourself by switching.

[Mike Strand]

Thanks Chaz and Richard, for your interest.

It's important that Monty always opens one of the doors other than Joe's first pick, and it's a door without the prize. Joe switching forces him to choose the remaining unopened door. Then you can use the law of total probability to show Joe's winning chances are 2/3, given he always switches (decides to switch with probability 1). Now, if Joe flips a coin to decide whether to switch in the first place, the chance to win reverts to 1/3 (or does it? -- maybe greater than 1/3? Another puzzle).

Monty opening a prize-less door -- or even more to the point, Monty always leaving one unopened door other than the one Joe picked -- are the keys to giving the advantage to switching.

The general principle might emerge more clearly by thinking of 4 doors, which introduces two cases: (1) Monty opens 2 of the remaining 3 doors (two without the prize), leaving 1 unopened door, or (2) Monty opens only 1 door without the prize, leaving two doors unopened. If Joe is certain to switch, then in case (1) the winning probability is 3/4. In case (2), Joe now has to make a decision between two unopened doors other than the one he first picked. This still gives a better than 1/4 chance of winning.

But let's ignore case (2), and have Monty always leave one unopened door, like in the original 3-door problem. Now the solution becomes even clearer with a huge number of doors. If 100 doors, and Monty opens 98 of the remaining 99 doors not picked by Joe, and none of them have the prize, Joe would be crazy not to switch to the one remaining door, which now clearly has a probability of 99/100 of holding the prize!

[Richard Baron]

Now, if Joe flips a coin to decide whether to switch in the first place, the chance to win reverts to 1/3 (or does it? -- maybe greater than 1/3? Another puzzle).

The overall probability of winning becomes 1/2. Suppose that heads means switch. Half of the time he does not switch, and has a probability of winning of 1/3, and half of the time he does switch, and has a probability of winning of 2/3.

The general principle might emerge more clearly by thinking of 4 doors, which introduces two cases: (1) Monty opens 2 of the remaining 3 doors (two without the prize), leaving 1 unopened door, or (2) Monty opens only 1 door without the prize, leaving two doors unopened. If Joe is certain to switch, then in case (1) the winning probability is 3/4. In case (2), Joe now has to make a decision between two unopened doors other than the one he first picked. This still gives a better than 1/4 chance of winning.

Case (2) means that switching gives a probability of winning of 3/8. Assume you picked A initially. There is a 3/4 chance that the prize was originally placed behind one of B, C and D. One of these is eliminated by the opening, so if it is behind one of B, C and D, you are left with a 1/2 chance of picking the right one out of the two that have not been eliminated. 3/4 x 1/2 = 3/8.

[Mike Strand]

Right on, Richard! Even in the 100-door case, if Monty opens 98 doors and Joe flips a coin to decide whether to switch, the probability of winning is 1/2 : 1/2 times 1/100 plus 1/2 times 99/100 is 100/200. 1/2 times 1/n plus 1/2 times (n-1)/n is still 1/2.

Incidentally, due to possible lack of clarity in my original problem statement, Chaz may have been thinking of Joe flipping a coin to make the decision to switch, and thus Chaz's answer of a 50% chance of winning. It's notable that the 50-50 chance applies in all cases of Monty leaving one door unopened and Joe flipping a coin to decide whether to stick with his original choice or switch to the other unopened door, no matter how many doors are in the contest.

One generalization now would be to have Monty leave two doors unopened, and Joe either switching with certainty or flipping a coin to decide whether to switch. You showed the solution, Richard, in the 4-door case. I think it can be shown that any device that gives Joe a chance to switch improves his odds of winning, if Monty has opened some doors. And switching with certainty is the best strategy.

Leaving aside the detailed exercise of figuring out the probabilities, let me get to another point, maybe the main point of my posting this puzzle in the first place: In solving problems like this, involving probability concepts, without the use of a large number of simulations, it's interesting to me that mathematicians have developed models and theories to anticipate the outcomes of simulations. Like figuring the odds in poker without having a computer perform a billion simulations.

To me, this is, well, more than interesting. Consider the human ability to show with algebra that the infinite sum 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... (where in the denominator 2 is raised to successive integer powers) approaches the number 2 from below and never passes 2, but the infinite sum 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... gets bigger and bigger without bound.

One wonders how human ingenuity can solve these unimportant (?) matters, but bigger concerns go unsolved. Maybe it's because the factors influencing these math problems are well-defined and limited in number and known to the problem-solver. The factors influencing actions and outcomes in life at large are not all known, nor are their effects, even qualitatively, let alone quantitatively. Hence the development of models to try to approximate complex reality.

[Richard Baron]

I think it can be shown that any device that gives Joe a chance to switch improves his odds of winning, if Monty has opened some doors. And switching with certainty is the best strategy.

Yes. Assume there are n doors in total, and Monty opens m of them, where m < (n - 1).

If the player does not switch, the probability of winning is 1/n.

There is a probability of (n - 1)/n that the prize is behind a door other than the one that the player originally chose.

If the player switches, and the prize is behind one of those n - 1 doors, the player has to choose the right one out of (n - 1 - m) doors, and has a probability of doing so of 1/(n - 1- m).

So switching gives a probability of winning of [ (n - 1)/n ] x 1/(n - 1 - m) = (n - 1) / [ n (n - 1 - m) ]

The increase in the probability of winning from switching is (n - 1) / [ n (n - 1 - m) ] - 1/n

This equals [ n - 1 - (n - 1 - m) ] / [ n (n - 1 - m) ] = m / [ n (n - 1 - m) ]

The numerator, m, is positive. The denominator is also positive because n is positive and m < (n - 1).

So there is always an improvement in probability from switching.

The fraction m / [ n (n - 1 - m) ] becomes undefined if m = n - 1, but that is when all the doors apart from the one the player first chose are opened and are revealed to have nothing behind them. Then the player should stick to the original choice.

m = 0 gives no improvement from switching, but that is when Monty does not open any door, so the player gains no information and will gain or lose nothing by switching.

[Mike Strand]

Looks good to me, Richard! Thanks for carrying this through here. The general solution gives the solution in any particular case, given the rules of the contest and some of its variations.

I mentioned how interesting is was (at least to me) how human logic and imagination can solve puzzles like this without resorting to simulation. I believe that games of chance, and experience with them, were fundamental to developing probability theory. In effect, first people did the simulations, either by design or through circumstance (like the love and practice of gambling), and then thought about what was happening and developed the models.

Even the models for games like this are approximations, albeit excellent ones. For example, the chance of a physical die showing any particular face is not exactly 1/6. The die isn't perfectly balanced -- the spots themselves, which are indentations, cause some of the imbalance, as well as the way the die is manufactured and shaped. The face probabilities vary from die to die. But for practical purposes, given well-manufacture dice, the variations are tiny, and the theoretical probability models work well. Other games, like betting on horses, can't be modeled as completely and effectively.

My examples of adding up an infinite series exemplifies how algebra and the idea of a limit can solve seemingly intractable puzzles.

I'm not sure what bearing these musings have on a general treatment of "Logic and the Philosophy of Mathematics"; I only hope they may stimulate further thought and discussion.

[Richard Baron]

But for practical purposes, given well-manufacture dice, the variations are tiny, and the theoretical probability models work well. Other games, like betting on horses, can't be modeled as completely and effectively.

Another area that may be of interest is that of phenomena where we know perfectly well what we are modelling, and the equations to use, but the equations can sometimes be too difficult to solve exactly and we have to resort to numerical methods. The obvious example is fluid dynamics. We know the Navier-Stokes equations, but we often cannot solve them exactly, and we don't even know whether solutions always exist in three dimensions.

Yet another area lies within the epistemology of mathematics itself. At what point do you start to have confidence in an unproven hypothesis because all the results so far point to its being correct, eg having confidence in Goldbach's conjecture because it has been checked up to 10^17, or because the distribution of primes makes it likely (a more sophisticated argument than the argument that it has turned out correct so far), or for both reasons?

[Mike Strand]

I guess the "true" mathematician (whatever that means) would insist on a proof. It took about 360 years to come up with a proof of Fermat's Last Theorem (I believe it was proved in the year 1995), but it did get proved. Confidence isn't enough. Give me proof! (says this mathematician). Or refute it! (negative proof, which still settles the issue).

If it's a principle that has useful application (say has made life better or yielded profits), maybe confidence is enough for now, and epistemology has a practical interpretation. Perhaps not so for the purist.

In statistical estimation based on samples, it's desirable to obtain an interval estimate which covers the value being estimated with high probability. Certainty, no, but maybe still useful.

Thanks for your comments about numerical methods to approximate solutions to model equations. Numerical analysis and approximation theory are a fascinating part of math, and it's interesting that the techniques themselves are based on abstract mathematical theory. Not knowing whether a solution exists, for example in fluid dynamics -- non-linear dynamics and strange attractors? The big whirlpool at the end of the movie, "Dead Man's Chest"?