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Re: Easy mathematics question
Posted:
Oct 8, 2007 8:10 PM


On Oct 8, 5:46 pm, Robert Israel <isr...@math.MyUniversitysInitials.ca> wrote: > Anthony writes: > > Say in a game of penalty shootouts, the striker had a 2/3 chance of > > scoring and the goalkeeper had a 1/3 chance of saving. I'm just making > > up these figures. So that basically would mean that you could score on > > either the right, left or centre of the goals. > > > Now say if I made a penalty shootout game first to 10 points. Well. > > That would mean that the striker would have to score 10x and the > > goalkeeper would have to save 10x. Just remember it's only the player > > vs. the goalkeeper. Now, it would be harder for the goalkeeper to get > > 10 points because he only has a 1/3 chance of saving, whereas the > > player has a 2/3 chance of scoring. How could I make it that the > > Goalkeeper and player have an equal chance of winning, for example in > > points (e.g. Player has to score 5 points and goalkeeper has to score > > 8 points  I know these figures aren't probably correct), however > > could someone help me with this question. > > Suppose you made it so the striker must score s points and the goalkeeper > must save g points to win. Thus in s+g1 attempts, one of them is sure > to win. The probability that this is the striker is > P(s,g) = sum_{j=s}^{s+g1} (s+g1 choose j) (2/3)^j (1/3)^(s+g1j) > For any given g, the closest P(s,g) gets to 1/2 is with > s = 2 g. For example, if the striker must score 10 times to win > and the goalkeeper must save 5 times to win, the probability of the > striker winning is approximately 0.4755, and this is the best you can > do with s + g <= 15.
That's a relief. I thought my random number generator wasn't working.
>  > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Hide quoted text  > >  Show quoted text 



