The Game Show problem. Is she right?

[pico]

http://www.marilynvossavant.com/articles/gameshow.html

is there a consensus on this problem?

[Ganchrow]

Yes.

If you don't switch you have a 1 3 probability of winning.

Switching yields a 2 3 probability of winning.

Switching is optimal.

[Ganchrow]

You might also be interested in these two slightly more complex variations on the so-called "Monty Hall problem":

4 Door Monty Hall Problem
Monty Hall-type problem

[pico]

just thought about something. people who does the shell game (1 pea and 3 caps) doesn't really have to cheat to win. because if they can distribute the peas randomly, that is 1/3 chance of winning and the payout is even money.

but cheating increase the house edge to 100%.

[rookie]

If player picked door 1 and switched:

[B]GAME #     DOOR 1   DOOR 2   DOOR 3   RESULT[/B]
GAME 1     AUTO     GOAT     GOAT     Switch and you lose.
GAME 2     GOAT     AUTO     GOAT     Switch and you win.
GAME 3     GOAT     GOAT     AUTO     Switch and you win.

If player picked door 1 and stayed:

[B]GAME #     DOOR 1   DOOR 2   DOOR 3   RESULT[/B]
GAME 4     AUTO     GOAT     GOAT     Stay and you win.
GAME 5     GOAT     AUTO     GOAT     Stay and you lose.
GAME 6     GOAT     GOAT     AUTO     Stay and you lose

Above scenarios are quoted in Savant's explanation. But, in Game 1 and Game 4, the host has two options - Door 2 and Door 3. So, you have an added scenario with same result - which makes probability 1/2.

Sorry about the formatting. Space bar / Tab isn't working.

Can anyone explain what I am missing ?

[Ganchrow]

Each of the above "games" are equally likely, all occurring with probability 1 3 .

Each of the host "options" (as you phrase it), are not equally likely. Specifically, the car being behind door 1 and the host revealing door 2 is exactly half as likely as the car being behind door 2 and the host revealing door 3.

If game 2 or 3 were to occur then the door revealed by the host would be predetermined. If game 1 were to occur, however, then the host would reveal door 2 half the time and door 3 half the time.

So we have the following probabilities:

Pr(car behind 1, host reveals 2) = 1 2 * 1 3 = 1 6
Pr(car behind 1, host reveals 3) = 1 2 * 1 3 = 1 6
Pr(car behind 2, host reveals 3) = 1 3
Pr(car behind 3, host reveals 2) = 1 3

[pico]

actually the optimal strategy is if you feel the game show host likes you, you always switch. if not, stick with your pick.

i.e. if you a hot blond who has not won anything on the show so far, and this is your last chance at winning something. the chances are, the host is trying to let you win by giving you a choice of switching

[rookie]

Cheers Ganchrow.

Pico, when you next meet a hot blonde, can you include this problem in your talks and let the forum know the details ?

[reno cool]

Many years ago this problem got me interested in probability. At first I also thought it wouldn't matter 50/50. Its amazing to hear people call themselves experts in the article and be wrong.
An easy way of looking at it is if you originally picked wrong and switched you would then be right 100% of the time. If you originally picked right and switched you would be wrong 100% of the time. Since you're more likely to be wrong to start--66.7% you swtch.

[13th Inning]

Assuming your original choice is door #1, and the host shows you one of the empty doors, there are six possible outcomes:

If the car is behind door #1:
KEEP DOOR #1 = WIN
CHANGE DOOR = LOSE

If the car is behind door #2;
KEEP DOOR #1 = LOSE
CHANGE DOOR = WIN

If the car is behind door #3:
KEEP DOOR #1 = LOSE
CHANGE DOOR = WIN

You're obviously not going to choose the door that he just showed you was empty - that's the key to this whole thing.

We can break down the 6 possible outcomes into 3 "keeps" and 3 "changes".

As you can see above, if you keep your door, you win 1 out of 3 times. If you change your door, you win 2 out of 3 times.

[mtneer1212]

These are independent events. Your chances of being correct are 1 out of 3.

[Dazzez]

These are independent events. Your chances of being correct are 1 out of 3.

What are independent events?

Your chances of being correct if you switch doors are 2 out of 3.
Your chances of being correct if you don't switch doors are 1 out of 3.

If you believe otherwise you're wrong.

[donjuan]

These are independent events. Your chances of being correct are 1 out of 3.

OK. Here's the deal. I put up 20k and you put up 20k. We meet and find a neutral party to run the trials just as stated in the problem. Each time, I risk $110 to win your $100 and I get to choose whether or not to switch. We repeat until one person goes broke. Sound good?