[Guest]

Reading the arguments in Xamdam's Marble Puzzle has inspired me to pose this variation of the Monty Hall Problem (which I didn't find in a quick search):

You are on a game show and you are presented with three closed doors. One of them hides 1000$ in cash while the other two hide nothing.

"Pick the door that you think hides the cash!" instructs the host.

You pick door number #2.

"Now go and open one of the remaining 2 doors." the host says. "But be warned, if the door you open has the cash behind it, you lose!"

You cautiously walk to door #1 and open it slowly. You peek inside and see... nothing! Sighing with relief you return to your place beside the host.

"Now you have a choice. You can either switch your choice to door #3 and keep whatever is behind it. Or you can stay with door #2 and keep whatever it hides. And to make the choice a little easier, I'm going to give you 300$ cash that you can keep no matter what if you decide not to switch doors!"

From a statistical point of view, which choice nets the most money? Why?

[Guest]

Stick to your choice of Door#2. Probability of finding $1000 is 50% on each door. Sticking to the choice will also reap in additional $300. No matter where the booty is - you will win $300 (min) in either case

[Guest]

Ill sitck with my choice. as long as u dont lose the 300$ cash it doesnt matter cause its a win-or nothing situation the probability of getting a 1000$ extra is 50%

Edited January 22, 2010 by ashsk8rkrish

[Guest]

I think stciking to the same option is correct because at the start each door had a probablity of 1/3 to have the cash behind it but since one of the door which the person had not selected does not has money behind it so the other door which he had not select had a probablity of 1/3 to ahve the money behind it now it means that the option that he has selected has a probablity of 1-1/3 = 2/3 So a higher probablity of winning

[Guest]

Not sure about this but....

..you might call this "cheating" though I don't claim this as my own answer (but if this offends any purists on here, I apologise now!).... This quiestion is posed in the film 21 and the answer is that you have the best chance of winning the $1000 if you switch doors. The film attributes this to variable change. But then I don't know if this still applies to this puzzle...

And as i'm not a gambler, I would stick with my original choice and take the guaranteed $300!!!

[Guest]

I really like this puzzle because I had to argue with my friends over the original Monty Hall puzzle where the host opens the door showing no money behind it. But I think this is an entirely different situation...

So in the original 3 doors puzzle, the answer is to switch your door. You have only a 1/3 chance of winning sticking with your door but a 2/3 chance of winning if you switch doors. This is all based on the fact that the host knows where the money is. Therefore, he will always open a door with no money behind it. See the link above for a more detailed explanation.

You, on the other hand, have no idea where the money is. The door you opened with no money could just as easily have had the $1000 behind it. So by having you choose the door to open, the probability of the money being behind the other 2 doors is unaffected. This is the same way that Deal or No Deal works. If you get down to 2 cases in the end and the million dollars is still left, the 2 remaining cases both have an equal chance of holding the money. Switching or keeping your case is a toss up. However, if Howie had picked the other 34 cases to open himself, (assuming he knows where the money is), I would switch cases. The odds of it being my case would only be 1/36, but the odds that the other case has the money is 35/36!

Long story short, you should stick with the door you got. That way you're guaranteed the $300 and you have a 50/50 chance of getting the $1000.

[Guest]

I just saw found a site where you can actually play the game yourself. It even keeps track of the stats so you can see if they follow the logic presented.

Play the game yourself!

[Guest]

YOU SWITCH. If you stick with your original choice, your odds are 1/3. If you switch, your odds increase to 1/2. When you originally chose door #2, you had a 1/3 shot of guessing correctly. Now that you've eliminated one door, door #2's odds of being correct is still 1/3 because that is what they were when you chose it. But now if you switch, you are switching to a door that has 1/2 chance of being correct. The average person is going to stick to their original choice out of pride which obviously benefits the gameshow.

[Guest]

YOU SWITCH. If you stick with your original choice, your odds are 1/3. If you switch, your odds increase to 1/2. When you originally chose door #2, you had a 1/3 shot of guessing correctly. Now that you've eliminated one door, door #2's odds of being correct is still 1/3 because that is what they were when you chose it. But now if you switch, you are switching to a door that has 1/2 chance of being correct. The average person is going to stick to their original choice out of pride which obviously benefits the gameshow.

Why would door #2's odds stay at 1/3? It's an entirely new choice between the two remaining doors. The previous choice of three has no bearing on the second decision. It's 50/50. Therefore you should stay. Door #2 offers a 50% chance at $1300 and 50% at $300 netting $800... while Door #3 offers a 50% chance at $1000 and a 50% at $0 netting $500

Now... if the host had opened door #1 (and known it was empty as in the classic scenario), it changes things. Door #2 would then offer a 1/3 chance at $1300 and 2/3 at $300, netting $633.33... Door #3 would offer a 2/3 chance at $1000 and a 1/3 at $0, netting $666.66. So in this case, switch doors for an additional $33.33

Edited January 22, 2010 by Jet Alone

[Guest]

You still have a 2/3 chance of winning if you switch. The fact that you get to choose the empty door is irrelevant since we are discarding the scenario where there is money behind it - it just becomes the Monty hall problem.

If you stick: $300 + 1/3 * $1000 = $633

If you switch: $1000 * 2/3 = $667

So you should switch. However, I don't like gambling and I'm guaranteed $300 if I stick so I would stick, and if I'm lucky i might get $1300.

[Guest]

Louie and Jet Alone got the right answer and demonstrated the knowledge I was looking for.

Interestingly enough, unlike the original Monty Hall problem, the probability is 50/50 after opening door #1. Monty's knowledge that the door he will open is incorrect makes a big difference. Look at it this way. In the original problem, 2/3 of the time you started with a wrong door and switching wins. In this problem, 1/3 of the time your original choice is correct. 1/3 of the time your original choice is incorrect and you open the other incorrect door. And the other 1/3 of the time your original choice is incorrect and you open the door with the money and don't make it any further. Given we didn't open the door with the money in this case we discard the last scenario and end up with a 50/50 chance we were right in the first place.

To be honest I was kind of hoping there would be some debate and argument about this one. I even threw in the 300$ as a distraction for those familiar with (but not fluent in) the original problem. Ah well.