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The generalised three-door problem

Posted May16-11 at 11:22 AM by CompuChip

Today I was having a discussion with a colleague about the three-door problem. I'm sure you've all heard of it in some form or another. Most commonly it is given as a game show concept, where a contestant has the choice between three doors. One conceals some prize (a car, or a large sum of money) and the other two don't (they contain a goat or nothing at all). After choosing one of the doors, the game show host opens one of the remaining two doors and shows that the big prize is not behind it. Then the contestant gets to say if (s)he wants to stick with his/her choice, or switch to the other closed door. The chosen door is then opened and (s)he goes home with the car or the goat :-)

Everybody knows (or should know!) that the winning strategy is to always switch doors, because it gives a 2/3 probability of going home with the money, instead of 1/2. My colleague claimed that this would remain the same, even if the option to switch doors was not given to all the contestants, but for example only to two thirds of them. I could easily show him that, although switching doors remains the best strategy, the probability of winning does decrease a little (but is still larger than 1/2 - it is 5/9 to be precise).

This led me to the following little generalisation of the problem:
Suppose that there are three doors. The probability of picking the winning door is pw, the probability of picking the losing door is pl = 1 - pw. If all the doors are identical, we will of course usually have pw = 1/3, pl = 2/3). To add in the new element, assume that the option to switch is offered with probability sw to a contestant who has initially chosen the winning door, and with probability sl to someone who has picked a losing door. Finally, the contestant will have a strategy, which we capture in a single parameter r: the probability with which (s)he switches doors, IF the option is given to him/her.

Now it is easy to find the three possibilities where the contestant goes home happy:
1) (s)he picks the winning door, is offered the option to switch and declines
2) (s)he picks the winning door, and is not offered the option to switch
3) (s)he picks a losing door, is offered the option to switch, and does so

The calculation is now simple statistics: the probability P for winning the prize is:
P = P1 + P2 + P3
where the probabilities corresponding to the above cases are
P1 = pw sw (1 - r)
P2 = pw (1 - sw)
P3 = pl sl r

The total probability is thus (after rearranging),
(*) P = pw + (pl sl - pw sw) r.

We can perform a simple check for the cases we already know, assuming that pw = 1/3, pl = 2/3:
*) If the contestant never switches (r = 0), P = pw = 1/3
*) If the contestant always switches (and is always offered the option to do so, of course, sw = sl = 1), P = 1/3 + (2/3 - 1/3) = 2/3
*) If 2/3 of the contestants is offered the option to switch (randomly) and they always do so in that case, P = 1/3 + (2/3 - 1/3) * 2/3 = 1/3 + 2/9 = 5/9

So far, so good.

Now, you may ask, what is the best strategy for a contestant? Well, formula (*) shows that P is simply linear in r, so the optimal solution lies at r = 1 (i.e. always switch) if the slope
(pl sl - pw sw) > 0
is positive. For pw = 1/3, pl = 2/3 this means
sl > sw / 2.
So if the option to switch is always given, r = 1 is the best strategy. If sl = sw / 2 then P = 1/3 independent of whether you switch or not - so this is the ideal strategy for the production company (that is, if THEY do know which door contains the prize, they let 2/3 of the winners choose to switch but only 1/3 of the losers, and people will go home with nothing 2/3 of the time). In all other cases, the slope is negative and the optimum value is P = 1/3 at r = 0.

It turns out that r = 1 is the best strategy in 75% of the cases (this can be easily seen by graphing sw vs sl, for example). So if the contestants have no idea how sw and sl are chosen, always switching remains the safest bet (even if sw and sl are not both 1). If the show is "fair" and the host does not know where the prize is, i.e. sw = sl, then r = 1 is definitely the best thing to do.

See you on tv!
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