Monty at it again

[jay_shark]

Here is a variation to the problem that I just thought of .

You're on a game show and you're given 2 choices of 5 doors . Behind one door is a car and behind the others are goats .For simplicity , say you pick doors 1 and 2 . The host knows what's behind each door and he quickly reveals to you two goats , say door 3 and door 4 . He informs you that you may keep whatever is behind doors 1 and 2 or you may opt to relinquish that offer and select door 5 .

What do you do ?

[rufus]

[ QUOTE ]
Here is a variation to the problem that I just thought of .

You're on a game show and you're given 2 choices of 5 doors . Behind one door is a car and behind the others are goats .For simplicity , say you pick doors 1 and 2 . The host knows what's behind each door and he quickly reveals to you two goats , say door 3 and door 4 . He informs you that you may keep whatever is behind doors 1 and 2 or you may opt to relinquish that offer and select door 5 .

What do you do ?

[/ QUOTE ]

Without knowing the behavior of the host, this is unclear. Now, with the assumption that the host always shows two empties, the chances from the first two doors are 2/5 while the chance from the last is 1/3, so the 2/5 gives you a better chance of getting the car.

[Silent A]

It's 3/5 to switch assuming Monty always reveals 2 goats.

Whenever your initial guess is wrong, the car is behind the 5th door. Since your initial guess is wrong 3/5 of the time, you should switch.

[PantsOnFire]

The chance of a car being behind your two doors is 2/5. The chance of the car being behind the other three doors is 3/5. God bless Monty for getting rid of two of those useless doors. I would switch.

However, I like this variation you came up with. For people who don't know the solution to the problem, it seems even more confounding to give up two doors for just one.

Nice work.

[jay_shark]

Thanks .

I came up with it so that you would respond to this post :P

[mykey1961]

How about a twist on the twist.

5 doors, 3 goats and 2 parts of the car. You must choose both parts of the car to be a winner.

After you make your selection, Monty opens 1 of the goats and gives you the option to change 1 or both of your doors.

What was the original probability a part of the car was behind each door? and your probability of selecting the correct 2 doors.

After Monty opens a goat door, what are the new probabilities for each door? and which combination of remaining doors gives you the best probability of winning, and what is that probability.

Assume the doors are A, B, C, D, and E. You selected A, and B, and Monty opened E.

[jay_shark]

There is a 1/10 chance that you have both parts belonging to doors A and B .

P(AC+AD+BC+BD+CD) =9/10
we know P(AC) =P(AD)=P(BC)=P(BD) but we have to check P(CD) .
Also P(AC) = 2/5*3/4*1/2 = 0.15

So we have 0.15*4+P(CD)=9/10
P(CD) = 0.3 .

You should select Doors C and D .

[mykey1961]

[ QUOTE ]
Also P(AC) = 2/5*3/4*1/2 = 0.15

[/ QUOTE ]

Can you explain the thought process that gives you this line?