#11
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Re: Monty at it again
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 ? |
#12
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Re: Monty at it again
[ 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. |
#13
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Re: Monty at it again
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...% |
#14
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Re: Monty at it again
[ 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. |
#15
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Re: Monty at it again
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 . |
#16
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Re: Monty at it again
[ 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? |
#17
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Re: Monty at it again
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 . |
#18
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Re: Monty at it again
[ 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. |
#19
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Re: Monty at it again
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 |
#20
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Re: Monty at it again
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 |
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