SOLUTION: This problem i tried using probability distributions but i couldnot get please help Thanks in advance A game consist of a sequence of plays on each play either you or your opp

Question 38018: This problem i tried using probability distributions but i couldnot get
please help
Thanks in advance
A game consist of a sequence of plays on each play either you or your opponent scores a point, you with the probability p(less than �), he with probability q=1-p. The number of plays is to be even 2, 4, 6, etc� To win the game you must get more than half the points. You know p, say 0.45, and as you get a prize if you win. You get to choose the number of plays in advance How many should you choose?
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
Seems to me, as anyone who has gambled in Las Vegas knows, in the long they will get you and take your money. The same is true here. In the long run your opponent will certainly win. BUT, in the short run anything is possible, as I saw eleven come up on a Vegas craps table four times in a row (about 95,000 to 1). Thus it would be my conclusion that in order to afford yourself the best chances of winning, you should make as short a run at this game as possible, thus 2 plays.
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