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The monty hall problem
| Musings on the Monty Hall problem
|
| Connor Osborn
| 06/20/15
The problem:
Suppose you're on a game show, and you're given the choice of three doors:
Behind one door is a car; behind the others, goats. You pick a door, say No. 1,
and the host, who knows what's behind the doors, opens another door, say No. 3,
which has a goat. He then says to you, "Do you want to pick door No. 2?"
Is it to your advantage to switch your choice?
My intuition:
Every game boils down to two mystery doors. The door you originally chose and
the door the host didn't reveal. Every game has a goat door that can be
removed. Again you will face two mystery doors, neither of which transparently
reveal a car. It doesn't matter if you switch because your odds are 50/50.
I'm wrong and most people are. You should always switch.
The error in my intuition:
There are two goats and three doors. Each door is twice as likely to have a
goat than a car, i.e each door has 2/3 probability of hosting a goat. The host
changes this probability in a subtle way. The host will never reveal a goat
behind your door. The probability of your door having a goat doesn't change.
However, the host always reveals a goat from the other two. The probability of
those doors does change!
Initial odds(goat):
*------------------------------------*
| 6/3, 2 goats |
*---------* *---------------------*
| 2/3 | | 4/3 |
*-------* *-------* *-------*
| | | | | |
| 2/3 | | 2/3 | | 2/3 |
|o | |o | |o |
| | | | | |
| | | | | |
*-------* *-------* *-------*
Door 1 Door 2 Door 3
Host reveals goat!
Final odds:
*------------------------------------*
| 6/3, 2 goats |
*---------* *---------------------*
| 2/3 | | 4/3 |
*-------* *-------* *-------*
| | | | | |
| 2/3 | | 1/3 | | 3/3 |
|o | |o | | (__) |
| | | | | o o\ |
| | | | | ('') |
*-------* *-------* *-------*
Door 1 Door 2 Door 3
Door 1 is twice as likely to have a goat!
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