Monty at it again
 

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 ?

Re: Monty at it again
 

[ 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.

Re: Monty at it again
 

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.

Re: Monty at it again
 

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.

Re: Monty at it again
 

Thanks .

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

Re: Monty at it again
 

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.

Re: Monty at it again
 

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 .

Re: Monty at it again
 

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

[/ QUOTE ]

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

Re: Monty at it again
 

First note that
P(ACuADuBCuBDuCD) =P(AC) +P(AD) +P(BC)+P(BD)+P(CD)

Since the events are mutually exclusive to each other . Clearly , they sum to 9/10 .

Also , P(AC)=P(AD)=P(BC)=P(BD) since there is no reason why one would be more likely than the other . For instance if A is one part to the car , then the probability c contains the matching part is no different than the probability d contains the matching part .

For P(AC) to occur , we need A to contain one of the two parts which is 2/5 . Also P(B|A) =1/4 since 2/5*1/4=2/20=1/10 . Moreover , P(C|A) +P(D|A) =3/4 which is the complement . Since P(C|A)=P(D|A) we get that

P(AC) = P(C|A)*P(A) but P(A)=2/5 and P(C|A) = 3/4*1/2 .
P(AC)=3/4*1/2*2/5 = 0.15

Re: Monty at it again
 

[ QUOTE ]
First note that
P(ACuADuBCuBDuCD) =P(AC) +P(AD) +P(BC)+P(BD)+P(CD)

Since the events are mutually exclusive to each other . Clearly , they sum to 9/10 .

Also , P(AC)=P(AD)=P(BC)=P(BD) since there is no reason why one would be more likely than the other . For instance if A is one part to the car , then the probability c contains the matching part is no different than the probability d contains the matching part .

For P(AC) to occur , we need A to contain one of the two parts which is 2/5 . Also P(B|A) =1/4 since 2/5*1/4=2/20=1/10 . Moreover , P(C|A) +P(D|A) =3/4 which is the complement . Since P(C|A)=P(D|A) we get that

P(AC) = P(C|A)*P(A) but P(A)=2/5 and P(C|A) = 3/4*1/2 .
P(AC)=3/4*1/2*2/5 = 0.15

[/ QUOTE ]


I had to take a day away from this because I went to the dentist and they gave me some "Goofy Juice"

I can see how you got the 3/4*1/2 by working the problem "in reverse"

I'm not sure I've got the whole thing though.

When I try to work through a problem I consider the same, or very similar, I get slightly different results.

Lets assume I have a group of 5 people, (A)lbert, (B)ill, (C)harley, (D)ennis, and (E)arl.

Each of these 5 people have purchased 2 tickets in a drawing.

The rules of the drawing are:
1) There are two equal prizes.
2) No individual can win more than one prize.
3) No person named Earl can win a prize.


Once Earl notices rule #3 he decides to give one of his tickets to Charley, and the other to Dennis.

What we want to find is the probability of each pair of people winning the prizes.

I'm not sure of the terminology so I'll assume P(A|) means probability of A being drawn on the first draw.
P(A|B) means probability of A being drawn on the second draw after B has been drawn on the first draw.
P(AB) means probability of both A and B being drawn with no regard to order

Given that

Albert has 2 tickets therefore P(A|) = 2/10
Bill has 2 tickets therefore P(B|) = 2/10
Charley has 3 tickets therefore P(C|) = 3/10
Dennis has 3 tickets therefore P(D|) = 3/10

P(B|A) = 2/8
P(C|A) = 3/8
P(D|A) = 3/8
P(A|B) = 2/8
P(C|B) = 3/8
P(D|B) = 3/8
P(A|C) = 2/7
P(B|C) = 2/7
P(D|C) = 3/7
P(A|D) = 2/7
P(B|D) = 2/7
P(C|D) = 3/7

P(AB) = P(A|)*P(B|A) + P(B|)*P(A|B) = 2/10*2/8 + 2/10*2/8 = 1/10
P(AC) = P(A|)*P(C|A) + P(C|)*P(A|C) = 2/10*3/8 + 3/10*2/7 = 9/56
P(AD) = P(A|)*P(D|A) + P(D|)*P(A|D) = 2/10*3/8 + 3/10*2/7 = 9/56
P(BC) = P(B|)*P(C|B) + P(C|)*P(B|C) = 2/10*3/8 + 3/10*2/7 = 9/56
P(BD) = P(B|)*P(D|B) + P(D|)*P(B|D) = 2/10*3/8 + 3/10*2/7 = 9/56
P(CD) = P(C|)*P(D|C) + P(D|)*P(C|D) = 3/10*3/7 + 3/10*3/7 = 9/35

Is there a mistake in how I worked the problem, or is my similar problem that much different than the 5 doors, two parts of the car problem?

Finally I resort to the counting method

column represents first ticket drawn, row represents second ticket drawn
<font class="small">Code:</font><hr /><pre>
A1 A2 B1 B2 C1 C2 C3 D1 D2 D3
-- -- -- -- -- -- -- -- -- --
A1 | . . AB AB AC AC AC AD AD AD
A2 | . . AB AB AC AC AC AD AD AD
B1 | AB AB . . BC BC BC BD BD BD
B2 | AB AB . . BC BC BC BD BD BD
C1 | AC AC BC BC . . . CD CD CD
C2 | AC AC BC BC . . . CD CD CD
C3 | AC AC BC BC . . . CD CD CD
D1 | AD AD BD BD CD CD CD . . .
D2 | AD AD BD BD CD CD CD . . .
D3 | AD AD BD BD CD CD CD . . .
</pre><hr />
AB = 8/74 = 4/37
AC = 12/74 = 6/37
AD = 12/74 = 6/37
BC = 12/74 = 6/37
BD = 12/74 = 6/37
CD = 18/74 = 9/37

So clearly I'm confused somewhere.