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[Fuseunderground]

[ringsting]

I think that where the card isn't clear is that it dosn't explain at what point in time the probability should be taken.

at point x in time, before any of the doors have been opened; the probability that A is correct is 1/3 and the probability that its not correct is 2/3. Hence when C is opened B becomes 2/3 as C=0 (according to the original calculation.)

however we move on in time and at point y in time, after C has been opened; we are simply left with a fresh choice of 2 doors, giving 50/50.

I think that you have to start each situation with a new calculation, otherwise you have to take into account all preceeding events such as: was the contestant chosen from a group of 100 people? if so you have the odds of 1/100 added onto the calculation ect

[mama_needs_a_new_car]

Ringsting:

[oliverkeers13]

No, similar to the theory of evolution, in that way. No matter how much sense it makes, some people will insist on believing in creationism.

[Ashin]

Re: Ringsting:

[GreenWindmill]

Look at it this way:

Let's assume the yacht is behind door 3 each time.

The gameshow host is always going to leave the door you pick and one other (which will be door 3 if you haven't picked it, and for argument's sake, door 1 if you did pick 3).

There are therefore 3 possible scenarios:

You pick door 1: The host eliminates door 2, you're left with 1 & 3. If you switch you win the yacht.

You pick door 2: The host eliminates door 1, you're left with 2 & 3. If you switch you win the yacht.

You pick door 3: The host eliminates door 1, you're left with 2 & 3. If you don't switch you win the yacht.

You win the yacht by switching in 2 of the 3 scenarios. In other words there's a 2/3 chance of winning if you switch and 1/3 if you don't.

As people have stated previously, this is more obvious with more doors - consider it with 100.

Not that this post is any real use, the correct answer has been confirmed looong ago, just trying to convert some non-believers!

[Factor41]

Aaaahhhh

[GreenWindmill]

[X9Tim]

[FairmontKing]

I studied this problem as an example in a graduate-level decision analysis class. I, too, struggled with the given solution at first. The reason the answer is not as straightforward as it should be is because we really don't know Jaunty's intentions and motives.

For the given answer to be correct, the following assumptions must be true:

1. Jaunty knows which door the prize is behind.
2. Jaunty decided ahead of time that regardless of which door you chose, he was going to show you a door without the prize.

Basically, our odds only improve if we switch because Jaunty is trying to help us.

[hobyrne]

Precisely what FairmontKing says. It depends on Jaunty's motivations and instructions. If the director of the show tells him, "You *must* open a goat door", that is one scenario. However, in the eerily coincidental [/sarcasm] parallel case of Monty Hall, Monty did _not_ *have* to open a door. It made for better TV, when he did, but he did not have to every time.

Therefore, if Monty was stupid, and wanted to save the network money, then every time someone chose a goat, he would end the game at that decision and the contestant is a loser, and every time someone chose the yacht, he would continue the game and give the contestant another chance to lose.

Of course, in this case, people would soon realise that the only time they are ever given a choice is when they have the winning door to begin with. So, what Monty really did, was to sometimes open a goat door when the contestant chose a goat. So, the best Monty could do to save the network money, is: If the contestant chooses the yacht, open a goat door. If the contestant chooses a goat, open the other goat door 1/2 of the time.

Conversely, if Monty wanted the contestants to win, he could end the game whenever someone chooses right, and give the contestant another chance to win if he chose wrong.

The card does not say that Jaunty *had* to open a goat door, just that he _did_. For that reason, the probabilities are 0.5 and 0.5.

[hobyrne]

Okay, since this _is_ a confusing issue, here is a worked-out example.

The three prizes are: Goat 1, Goat 2, Yacht. Making a distinction between the goats is important. If you flip two coins, there are three outcomes: two heads, one heads and one tails, two tails. If you flip a nickel and a dime, there are four outcomes: nickel heads & dime heads, nickel heads & dime tails, nikel tails & dime heads, nickel tails & dime tails.

Monty Hall has the following instructions: If the person chooses Goat 1, let him have it. If the persion chooses Goat 2, open the door with Goat 1. If the person chooses the yacht, open the door with Goat 1.

Chooses goat 1: Happens 1/3 of the time. Contestant loses.

Chooses goat 2: Happens 1/3 of the time. Contestant gets choice: wins if he switches.

Chooses yacht: Happens 1/3 of the time. Contestant gets choice: wins if he doesn't switch.

So, the contestant gets a choice 2/3 of the time. HALF of those times he gets a choice, he should switch, HALF of the times he gets a choice, he shouldn't switch.

BTW: Interesting end note. If the contestants switch half the time, the overall chance of winning the prize is 1/3... exactly the same as if this little game were done away with and he just always got his first choice. The extra game is just for show, to make it *seem* more exciting for TV.