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wracket
wracket
1426 posts

Re: Let's Make a Deal

c in dc wrote:
Mu Mu wrote:
cyberpainter wrote:
c in dc wrote:


Wracket, you are a man after my own heart.
Here is a probability puzzle for you to ponder:

Say you are on a Let's Make a Deal type game show. There are 3 doors to choose from; one has the grand prize, the other 2 nothing. You choose one door. The host then reveals one of the 2 remaining doors, and it has nothing behind it. The host then offers to trade you the other unchosen door for your chosen door.
Should you accept his offer?


Wouldn't you have a 50/50 shot either way at that point?



Hmmm, each door has a 1/3 chance of containing the prize. When one door is shown to have no prize, the two remaining doors have a 1/2 chance of containing the prize. I agree with CP, your chances remain 1/2 no matter which door you choose from the remaining 2 doors. Correct?


Nope. You wanna splain this one, Wracket?

Sure. You should take the host up on changing, and here's the "short" answer why. Mu is right that there is a 1/3 chance of each door containing the prize, so that means that a third of the time your initial choice will be the one with the prize and 2/3 of the time it won't be. Therefore, when the host shows one of the two remaining doors, since he has to open a door without a prize, he has effectively doubled your chances of winning by changing...which can be demonstrated with a basic three-part binomial diagram (which I can't be bothered to try to recreate here). So here's a quick-and-dirty written version:

Assume for simplicity's sake that you've chosen door 1. There is still an equal 1/3 chance of the prize being behind any door, so bring out your diagram to show the three possible locations of the prize. Now, from each of those three possibilities, let's look at what can happen next. If you've chosen door 1 and the prize is indeed behind door 1, the host can either open door 2 or door 3. As there is an equal probability of either of these outcomes, each individual scenario has the probability of 1/3 x 1/2 = 1/6. Either way you lose by switching here, so that means you have a 1/6 + 1/6 = 1/3 contribution to your overall chance of losing by choosing to switch from door 1 if the prize is behind door 1.

If, however, the prize is behind door 2, the host has no choice as to which door he will open--he can only open door 3. Therefore, you don't need to divide this probability by two...it is simply a 1/3 contribution to your overall chance of winning by choosing to switch from door 1 if the prize is behind door 2.

Likewise, if the prize is behind door 3, the host can only open door 2...once again, 1/3 contribution to your overall chance of winning by choosing to switch from door 1 if the prize is behind door 3.

So now that we have covered all the possible scenarios, we see that we have a total of 1/3 chance of losing by switching and a 1/3 + 1/3 = 2/3 chance of winning.

I know it sounds weird at first, but do the binomial tree and see!

Maybe C can do a better explanation if mine was too convoluted...

Jul 11, 2011, 20:15



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