Monty at it again

[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 .

[mykey1961]

[ QUOTE ]
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 .

[/ QUOTE ]


Are you assuming that if the second draw is invalid, it invalidates the first draw as well?

[jay_shark]

Yes , my answer assumes that there is a re-deal if both or all three events occur . In fact , it isn't even necessary because when both events occur , with probability 1 , you will be given a non 1 card on your second deal , when your first card is a 1 .

Just verify that
1/10*9/10 +1/10^2*9.10+1.10^3*9/10 +....+ =1/10 .

[mykey1961]

[ QUOTE ]
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 .

[/ QUOTE ]


Ok I think I've read this enough to determine were are on completely different pages.

Albert's tickets are #1 and #2
Bill's tickets are #3 and #4
Charley's tickets are #5 and #6
Dennis's tickets are #7 and #8
Earl's tickets are #9 and #10

Earl gives ticket #9 to Charley
Earl gives ticket #10 to Dennis

An example of the draw:

First ticket drawn is #6 so Charley wins the first prize.
Second ticket drawn is #9, since this ticket also belongs to Charley, it's considered an invalid draw.
A second ticket is redrawn and is #3 so Bill wins the second prize.

[mykey1961]

I can also see how this differs from the 5 doors, and 2 car parts game.

In the car game we wait until after Monty opens a door to assign the A, B, C, and D.

If we pick door #1 and #2, and Monty opens door #4 then A = #1, B = #2, C = #3 and D = #5

[jay_shark]

Glad you cleared this up .

A has 1 and 2
B has 3 and 4
C has 5,6,9
D has 7,8,10 .

Now a number from 1-10 has been selected at random from the vault .With probability 1 there will be two distinct winners .

Lets compute the probability that A and B are the winners .

P(AB)=2/10*2/8 + 2/10*2/8 = 0.1
P(AC) =2/10*3/8 +3/10*2/7
P(AD)=P(BC)=P(BD)
P(CD) = 3/10*3/7*2 = 18/70

Notice that everything adds up to 1 .
P(AB)+P(AC)+P(AD)+P(BC)+P(BD)+P(CD)=1