Probabilities from movie 21

justsomeguy

What gives the advantage is a combination of two factors.
1. The contestants initial chance of picking the right door is 1/3. Nobody will argue this.
2. The host then opens one of the other two doors, reducing that doors chance of holding the prize from 1/3 to 0/3, and shifting that probability to the remaining unpicked door -- as combined they had a 2/3 chance of holding the winner.

If a 2nd player comes in who knows only which door the host opened, he has a 1/2 chance of picking the right door from the remaining two. If he knows only what door the 1st contestant picked, his odds of picking the right door are also 1/2. If he knows both things, then he can pick the door that the 1st contestant should switch to, and get the same 2/3 odds.

I don't know what you mean here, but I described all possible scenarios above.

In this scenario, sure, short everything. However stock markets are not nearly so unpredictable, and news that comes out on monday is going to affect the market on monday as soon as the news breaks -- not the next day.

scalpmaster

Wrong.Only applies to the first contestant.2nd player is shown 2 doors only.

Correct, the critical question is does it matter whether the host had opened the SAME empty door knowingly or randomly(independently guess) from the other two doors?
If the host knows everything, it is guarranteed that the game will proceed to the 3rd stage(to switch or not) since he will always pick an empty door.
If the host just guess one of the other 2 doors, the game may stop at 2nd stage(host reveal door).
The question is whether the overall situation is affected if the host guess the same empty door and the game still proceeds to stage 3 as per normal? Mathematicians seem to think it is...but why in layman terms?
http://math.stackexchange.com/questi...l-problem?lq=1
2nd player probability depends on whether he knows how many other doors existed before he is shown the 2 doors game he is playing and 1st contestant's choice. If he does not know anything, from his perspective, it is always 50-50.

justsomeguy

Yes, "from his perspective." If door A is the 1st contestants choice, B is the hidden door with the goat, and door C is the remaining door, there is still a 2/3 chance that door C holds the prize.

The fact that the 2nd contestant doesn't know which door (A or C) the 1st contestant picked does not change the odds of a door holding the prize -- which is what the puzzle is all about -- it just changes the odds of the 2nd contestant picking the door with better odds.

Do you see the difference?

scalpmaster

All the above is self explanatory when I said "from his perspective", but thats not my question which is "does the overall mathematical condition change if the host guess instead of knowing the same empty door to reveal?"

justsomeguy

Your question needs more rules specified as I said when you first asked it.

If the host opens a random door there are two special cases you need to address in the rules. Specifically, what happens when the host opens a random door and reveals the prize. Can the contestant switch to that door? What if it's the contestants door?

scalpmaster

Then there is simply no stage 3 for player 1 and no 2nd player.Game over.

justsomeguy

If it's an instant game over if the host opens your door, win or lose, then the fact that he's opening a random door doesn't matter.

1/3 times the host will open your door.
1/3 of those times (1/9 total) you instantly win.
2/3 of those times (2/9 total) you instantly lose.

The rest of the time (2/3) you have a chance to switch.

1/3 of those times (2/9), you picked the right door to begin with.
2/3 of those times (4/9), you picked the wrong door, and switching will let you win.

So under your new rules:
1/9 times you win instantly.
2/9 times you lose instantly.
2/9 times you win if you don't switch
4/9 times you win if you switch.

So overall your chances of winning are 5/9 if you switch when given the opportunity, and 2/9 of winning if you don't.

There is a variant of this that asks if the host opens a door at random that is not your door and reveals a goat, what are your odds if you switch?

This changes things up a lot and results in the 1/2 chance of winning that you seek. It's the same as this problem except you have to factor out the cases where the host revealed the prize or opened your door, and their impact on the outcome.

Sorry for the delay, forgot about the post when I was busy and just saw the email notification and realized I hadn't replied.

Please feel free to double check the math, I've been having a long week!

broccoli7

This is also known as the Monty Hall problem. Discussed here (http://bayesianthink.blogspot.com/20...inference.html)
Your calculation is correct. Basically in light of new evidence, it makes sense to change the choice of doors