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Re: Easy mathematics question
Posted:
Oct 8, 2007 8:10 PM
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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+g-1 attempts, one of them is sure
> to win. The probability that this is the striker is
> P(s,g) = sum_{j=s}^{s+g-1} (s+g-1 choose j) (2/3)^j (1/3)^(s+g-1-j)
> 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 -
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