A casino offers the following game a fair 6sided die numbere
Solution
Here as the game is about throwing a fair 6 sided die, each trial is independent.
Let A - your chance of winning
B - your friend win
B\' - your friend loses
i. if you are told that your friend lost.
Then friend must have thrown less than house.
Then possibilities for friend and house score are
(1,2) (1,3)...(1,6)
(2,3)....(2,6)
(3,4)...(3,6)
(4,5)(4,6)
(5,6)
There are 21 outcomes for friend losing to house.
If you have to win you have to throw more than the house.
Possibilities for winning are in case your friend loses,
(friend score, house score, your score) would be
(1,2,3) (1,2,4)...(1,2,6)
(1,3,4)(1,3,5)(1,3,6)
(1,4,5) (1,4,6)
(1,5,6) -- 10 ways
(2,3,4) (2,3,5)(2,3,6) (2,4,6) (2,5,6) (2,4,5) -- 6 ways
(3,4,5) (3,4,6) (3,5,6) -- 3 ways
(4,5,6) - 1 way
Hence prob = 20/216 =0.09259
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ii. In case your friend won the possibilities are
House, friend
=
(1,2) (1,3)...(1,6)
(2,3)....(2,6)
(3,4)...(3,6)
(4,5)(4,6)
(5,6)
Your chance of winning would be
same as that of your friend
Hence 21/216 = 0.09722
....

