Can you help me with a problem?

There are 2 people that are playing a game in which a coin is tossed 11 times. The first player gets a point for a toss of heads. The 2nd player gets a point for a toss of tails. whoever gets 6 points wins. Suppose that so far the first player has 2 points and the second player has 4 points. What is the probability that the first player wins the game? I know the first player needs 4 more points to win. I think that is 1/2*1/2*1/2*1/2 but I'm not sure of anything else. Can you help

Thanks,

Hi,

The coin has been tossd 6 time and hence is to be tossed 5 more times. One way to solve this problem is to list the possible outcomes from tossing a coin 5 times and the count how many result in the first player winning.

 HHHHH HTHHH THHHH TTHHH HHHHT HTHHT THHHT TTHHT HHHTH HTHTH THHTH TTHTH HHHTT HTHTT THHTT TTHTT HHTHH HTTHH THTHH TTTHH HHTHT HTTHT THTHT TTTHT HHTTH HTTTH THTTH TTTTH HHTTT HTTTT THTTT TTTTT

The 6 results in yellow have 4 heads before two tails and hence these are the winning outcomes for the first player. Each of the outcomes listed is a result of tossing the coin 5 times and hence each of the outcomes has probability 1/2*1/2*1/2*1/2*1/2 = 1/32. Thus the probability that the first player wins is 6/32 = 3/16.

A second way of looking at this problem is to say that the game stops as soon as one of the players wins. Hence the game stops with player one as the winner if the next 4 tosses are heads or the next 5 tosses are THHHH, HTHHH, HHTHH or HHHTH. The probability of 4 tosses in a row being heads is 1/2*1/2*1/2*1/2 = 1/16. Each of the outcomes from tossing a coin 5 times has probability 1/2*1/2*1/2*1/2*1/2 = 1/32 hence the probability of one of the 4 outcomes listed is 4/32. Thus the probability that player on wins is

1/16 + 4/32 = 3/16

Cheers,
Andrei and Penny

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