Question

Karl and Patrick are playing with a fair coin. The one who obtains more tails than heads in 4 tosses is declared the winner. What is the probability that Patrick wins?

Answer 1

these are tricky counting problems where one has to be careful ...
if they each toss the coin 4 times, each guy will have 16 possible outcomes (2^4) and the prob. of a any specific outcome is 1/16. there is 1 way of getting all (4) tails, 4 ways of getting 3 tails, 6 ways of 2 tails, 4 ways of 1 tail, and 1 way of no tails (ie all heads). so the probs are:
P(0 tails)=1/16
P(1 tail)=4/16
P(2 tails)=6/16 (not 1/2 !)
P(3 tails)=4/16
P(4 tails)=1/16
we need P(# of Patrick tails > # of Karl tails) =
P(1T and 0T)+P(2T and (0 or 1) T)+P(3T and (0 or 1 or 2) T)+P(4T and (0 or 1 or 2 or 3) T) =
(4/16*1/16)+(6/16*5/16)+(4/16*11/16)+(1/16*15/16) = 93/256 = 0.36328


another way of lookin at it:
The prob that Patrick wins = prob that Karl wins = (1 - P(they tie)) / 2
prob they tie = P(# of Patrick tails = # of Karl tails) =
(1/16*1/16)+(4/16*4/16)+(6/16*6/16)+(4/16*4/16)+(1/16*1/16) =
1/256 + 16/256 + 36/256 + 16/256 + 1/256 = 70/256 = 0.2734

therefore, to check... P(Patrick wins) or P(Karl wins) or P(they tie) =
93/256 + 93/256 + 70/256 = 1 as the probability of all possible outcomes must total 1