Monty at it again

[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?

[jay_shark]

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

[mykey1961]

[ 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.

[jay_shark]

Does Earl give his tickets specifically to Charley and Dennis ?

What happens if one of them is given two prizes ? Does he give one of them away at random ?

[mykey1961]

[ QUOTE ]
Does Earl give his tickets specifically to Charley and Dennis ?

What happens if one of them is given two prizes ? Does he give one of them away at random ?

[/ QUOTE ]

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.

If ticket is drawn that would cause an individual to win a second time, the draw is considered invalid, and another ticket is drawn.

[jay_shark]

There are two prizes . Lets call them c1 and c2 .
There are also 8 non-prizes , I will call them g1, g2, ...,g8 .

There are also 4 players involved and Charley and Dennis get 3 tickets each .The number of ways of selecting 2 people from 4 is 4c2 . Anytime you select 2 players , there are an additional 2 ways to distribute the prizes to them .

The number of ways the prizes can be distributed is :

n(S)=2*8*7*6c3 + [2*8*7*6*5c3 + 2*8*7*6*5c3]*2 + 2*8*7*6*5*4c2

Let A be the event that Albert and Bill get the same ticket . We wish to find the number of favorable cases , or n(A) where P(A) =n(A)/n(S)

n(A) = 2*8*7*6c3
P(A) = 2*8*7*6c3/n(S) = 2240/49280=0.04545 or 4.5454...%

The probability Charley and Dennis get the prizes is :
2*8*7*6*5*4c2/49280 = 40.9090...%

The probability Albert and Charley or Albert and Dennis or Bill and Charley or Bill and Dennis get the prizes is:
6720/49280 = 13.6363...%

[mykey1961]

[ QUOTE ]
There are two prizes . Lets call them c1 and c2 .
There are also 8 non-prizes , I will call them g1, g2, ...,g8 .

There are also 4 players involved and Charley and Dennis get 3 tickets each .The number of ways of selecting 2 people from 4 is 4c2 . Anytime you select 2 players , there are an additional 2 ways to distribute the prizes to them .

The number of ways the prizes can be distributed is :

n(S)=2*8*7*6c3 + [2*8*7*6*5c3 + 2*8*7*6*5c3]*2 + 2*8*7*6*5*4c2

Let A be the event that Albert and Bill get the same ticket . We wish to find the number of favorable cases , or n(A) where P(A) =n(A)/n(S)

n(A) = 2*8*7*6c3
P(A) = 2*8*7*6c3/n(S) = 2240/49280=0.04545 or 4.5454...%

The probability Charley and Dennis get the prizes is :
2*8*7*6*5*4c2/49280 = 40.9090...%

The probability Albert and Charley or Albert and Dennis or Bill and Charley or Bill and Dennis get the prizes is:
6720/49280 = 13.6363...%

[/ QUOTE ]

When I first looked at this my only response was " Huh? "

If a person were allowed to win twice, there would be at most 100 possible payouts. 10 valid tickets on each of 2 drawings = 100 possible results. Since the same player can not win twice, 26 of those possible results are invalid.

[jay_shark]

First , let me make sure we're talking about the same question . I'm assuming now that the question is equivalent to picking numbers 1-10 and that only card 1 is the winner but a player cannot win twice . If by chance a player picks #1 twice , or possibly 3 times , then there is a re-deal and the players choose new cards .

The probability Albert wins one prize is, by the inclusion/exclusion principle : 1/10+1/10 -1/100 . Or , 1/10*9/10 +9/10*1/10 +1/10*1/10 =0.19

The probability Dennis wins one prize is : 3*1/10 - 3c2*(1/10*1/10) +3c3*1/10*1/10*1/10 = 0.271

Make sure you understand how I got these numbers before we move on .