(originally posted 10:08 AM, Feb 11, 2006)A couple days ago (yes, in a bar) I became intrigued with a fairly well known game problem, here's my version:
- There are three doors, one of which is a winner.
- You pick one door, but it isn't opened yet.
- At this point at least one of the unpicked doors (maybe both) are losers.
- A losing unpicked door is opened.
- You now have the option of either sticking with your original choice, or switching to the remaining unopened door. Which has a better chance of being the winner?
Here is a Perl script I wrote which will play the game using both the "switch" and "stay" strategies:
http://eigenstate.net/misc_scripts/make_a_deal.pl What do you think?
Tags:
game theory, perl, lets make a deal
12 Feb 2006 - addition:
Here is one way of explaining what is going on:
When the losing choice is revealed, you gain some new information.
At the start, you have equal knowledge of all the choices, so at first every door has 1/n chance (1/3 if there are three doors).
You then divide doors into two groups: one door in group A (your first pick) and the rest of the doors in group B. With three doors group A carries 1/3 of the chances and group B carries 2/3.
When one of the doors in group B is revealed to be a loser you know something new:
You know that the group of chances that was in group B at the start must now be redistributed among the remaining doors in group B.
So if group B had 2/3 of the chances at the start, it *still has* 2/3 of the chances, but those all ride on the one remaining door in group B. So, the remaining door in group B has a 2/3 chance fo winning, whilst the door you originally picked, in group A still has only a 1/3 chance.
What has happend is that the revelation of the loser in group B has told you that all the other doors in group are more likely to be a winner.
If you try it with 10 doors it is still better to switch - to pick any remaining door in group B - than to stay with the door in group A.
Where N is the number of doors, the chance of the first pick winning is 1/N, and the chance of a door in group B after eliminating a loser from group B is: (N-1)/N
It is the -1 that makes swicthing worth it.
We can generalize the problem further by saying that T is the Total number of doors, A is the number of doors in group A, and B is the number in group B before a loser is revealed. Then the chance of a door in group A is A/T (for example, 1/3) and the chance of a door in group B, after the elimination is B - 1 / T.
So if there are 10 door total, and you put 3 doors in group A at the start:
T = 10
A = 3
B = 7
Each group A door has a 1/10 chance (3/10 for the group as a whole.) After eliminating a loser from group B, the group as a whole still has 7/10 chances and thus each remaining group B door has a 6/10 11 2/3% chance of being the winner. (thanks to keith for the correction)