Help in this Q dear experts......plz
20. In a game A throws two ordinary dice. If he throws 7 or 11 he wins. If he throws 2, 3 or 12 he loses. If he throws
any other number, he throws again and continues to throw until either the number he threw first or 7 turns up. In the first
case he wins and in the second he loses. Show that the odds against his winnings is 251 : 244.
Dear student,
Regards
P(4)=3/36 = 1/12 | P(5)=4/36 = 1/9 | P(6)= 5/36 | P(8)= 5/36 | P(9)= 1/9 | P(10)=1/12 |
3,1 | 4,1 | 5,1 | 6,2 | 6,3 | 6,4 |
2,2 | 3,2 | 4,2 | 5,3 | 5,4 | 5,5 |
1,3 | 2,3 | 3,3 | 4,4 | 4,5 | 4,6 |
1,4 | 1,5 | 2,6 | 3,6 | ||
2,4 | 3,5 | ||||
Regards