[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.