Double monty - action at a distance?

[Simon]

This may mystify some.

Imagine two Monty Hall games taking place simultaneously on different sides of the world. The two games are completely independent.
Imagine you're a contestant in one of them.

There are three doors - two of them empty, one has a prize.
The host knows what's behind each door.
You pick a door.
The host eliminates an empty door from the remaining two.
You are asked whether you'd like to stick or swap.
You're aware of the familiar solution - that there is a 2/3 chance that the door you chose is empty and that the other has the prize.

Before you make your final decision, a phone call is made.
You are about to learn whether the contestant on the other side of the world chose the same* as you or the opposite**.

Depending on what you learn, does this knowledge in any way affect the odds of your game?

* same = you both picked an empty door or prize
** opposite = not the same

[CraigD]

What you learn from the phone call does change your expectation. It doesn’t, however, change the way you should play.

If you call reveals you chose the same, you know your probability of winning if you switch is 2/9, if you don’t switch, 1/9. If the call reveals you chose the opposite, your probability of winning if you switch is 4/9, if you don’t, 2/9

Regardless of what the call reveals, you should switch you guess after the host removes one of the doors you didn’t chose, as it multiplies you probability of winning by 2.

However, if the host always offers an incentive of, say, half the value of the prize to quit the game without choosing, you should accept the offer if the call revealed you chose the same, and reject it if it reveals you chose the opposite.

I determined the above empirically by modeling the different strategies for a large number of trials, not using analysis, because I am lazy, and computers are easy.

The MUMPS code, with results in comments (after ;s):

K W f C=1:1:100000 s R1=(3)+1,R2=(3)+1 s W=R1=R2,W(W)=W(W)+1 ;W(0)=66703 W(1)=33297
K W f C=1:1:100000 s R1=(3)+1,R2=(3)+1,R3=((R1=R2:1,1:2)),A=1,R3=R3-1 S:A=R1 A=A+1 S:A=R2 A=A+1 S:A=R1 A=A+1 s R1=R1#3+1 S:R1=A R1=R1#3+1 s W=R1=R2,W(W)=W(W)+1 ;W(0)=33257 W(1)=66743
K W f C=1:1:100000 s R1=(3)+1,R2=(3)+1,R3=((R1=R2:1,1:2)),A=1,R3=R3-1 S:A=R1 A=A+1 S:A=R2 A=A+1 S:A=R1 A=A+1 s R1B=(3)+1,R2B=(3)+1 s:(R1B=R2B)=(R1=R2) R1=R1#3+1 S:R1=A R1=R1#3+1 s W=R1=R2,W(W)=W(W)+1 ;W(0)=33363 W(1)=66637
K W f C=1:1:100000 s R1=(3)+1,R2=(3)+1,R3=((R1=R2:1,1:2)),A=1,R3=R3-1 S:A=R1 A=A+1 S:A=R2 A=A+1 S:A=R1 A=A+1 s R1B=(3)+1,R2B=(3)+1 s:(R1B=R2B)'=(R1=R2) R1=R1#3+1 S:R1=A R1=R1#3+1 s W=R1=R2,W(W)=W(W)+1 ;W(0)=66612 W(1)=33388
K W f C=1:1:100000 s R1=(3)+1,R2=(3)+1,R3=((R1=R2:1,1:2)),A=1,R3=R3-1 S:A=R1 A=A+1 S:A=R2 A=A+1 S:A=R1 A=A+1 s R1B=(3)+1,R2B=(3)+1 s R1=R1#3+1 S:R1=A R1=R1#3+1 s W=R1=R2,W(W)=W(W)+1 ;W(0)=33363 W(1)=66637
K W f C=1:1:100000 s R1=(3)+1,R2=(3)+1,R3=((R1=R2:1,1:2)),A=1,R3=R3-1 S:A=R1 A=A+1 S:A=R2 A=A+1 S:A=R1 A=A+1 s R1B=(3)+1,R2B=(3)+1 s R1=R1#3+1 S:R1=A R1=R1#3+1 s W=R1=R2,W((R1B=R2B),W)=W((R1B=R2B),W)+1 ;W(0,0)=22148 W(0,1)=44484 W(1,0)=11146 W(1,1)=22222
K W f C=1:1:100000 s R1=(3)+1,R2=(3)+1,R3=((R1=R2:1,1:2)),A=1,R3=R3-1 S:A=R1 A=A+1 S:A=R2 A=A+1 S:A=R1 A=A+1 s R1B=(3)+1,R2B=(3)+1 s W=R1=R2,W((R1B=R2B),W)=W((R1B=R2B),W)+1 ;                             W(0,0)=44291 W(0,1)=22433 W(1,0)=22245 W(1,1)=11031

[Simon]

Originally Posted by CraigD

What you learn from the phone call does change your expectation. It doesn’t, however, change the way you should play.

If you call reveals you chose the same, you know your probability of winning if you switch is 2/9, if you don’t switch, 1/9. If the call reveals you chose the opposite, your probability of winning if you switch is 4/9, if you don’t, 2/9[/code]

A curious and contradictory answer.

Firstly, what is your expectation before the phone call is made? Can I assume that we agree that it is 2/3 in favour of swapping? That's a crucial premise.

If so, then your probability figures in the second paragraph amount to no change. They translate to 2/3 in favour of swapping, regardless of what you learn.

However I contend that these figures are wrong and that it does make a difference what you learn about the other game.

Here is my solution:

Before the phone call is made, the odds are 2/3 you will get the prize by swapping and 1/3 you will get it by sticking

If you discover the other contestant chose the same as you, the probability of getting the prize goes up to 80% if you swap and down to 20% if you stick.

But if you find out the other contestant chose the opposite, the odds are 50/50.

Anyone disagree?

[CraigD]

Originally Posted by Simon

A curious and contradictory answer.

My result was wrong, because my hastily written code incorrectly defined the “same” condition as “other player’s first choice the prize door”. When fixed, it agrees with Simon’s answer.

For the 3 door game, knowing if you’re the same or opposite doesn’t require a change from the usual “always change” strategy to have the best strategy, winning with probability about 67%. For games with more than 3 doors, the “change only if opposite” strategy is best. For example, for 4 doors, “always change” wins with about probability 38%, “change if opposite” 47%. For 10 doors, AC has about 11%, CIO 19%. For 100, AC about 1%, CIO 2%.