6 Consider the following game of chance A fair coin is tosse
Solution
Probability of getting tail when a coin is tossed = 0.5
Initial bet is of $ k
If tail appears on the nth toss you recieve $ 2n
probability that 1st tail appears on nth toss = (probability that head appears on 1st toss) X (probability that head appears on 2nd?toss) ... X (probability that head appears on (n-1)th?toss) X (probability that tail appears on nth toss) = 0.5n-1 X 0.5 = 0.5n
Expected Winning = [ (2n- k) X (probability that 1st tail appears on nth toss)] + [ (-k) X (1 - probability that 1st tail appears on nth toss)]
Expected winnings = [ (2n- k) X (0.5n)] + [ (-k) X ( 1 - 0.5n)]
= 2n X 0.5n - k X 0.5n - k + k X 0.5n
= (2 X 0.5)n - k
= 1 - k ... (I)
Now Sometimes expected winnings of 0 is called a fair game
Suppose Expected Winnings = 0
From (I) we get,
1 - k =0
that is k = 1
So k should be 1 in order to have expected winnings equal to 0.
