11th August 2006, 05:52 AM
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Member
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Join Date: Jan 1970
Location: Sydney
Posts: 402
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Quote:
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Originally Posted by AssumeTheCrown
No sign of the avalanche so i will fire in my answer.
If G has a 9% chance of finishing ahead of each of A,B,C,D & E and assuming independance then the probability of G NOT making the 8 is 0.91^5= 0.624 or 62.4%. Therefore the probability of G finishing ahead of at least one of A,B,C,D & E and hence making the top 8 will be 1-0.624 = 0.376 = 37.6%.
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Agree on 37.6% = $2.66 Implied Odds
So for A, B, C, D, E
Prob - Odds
92.5% $1.08
Before I tried this hypothetical exercise I'd thought Geelong's chances were as hopeless as Anthony Mundine's of getting elected to parliament, but fortunately this has now stopped me making really dumb bets on the 8.
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Problem 3:
The general solution.
# Rounds Remaining = Column A
# Games B is behind A = Column B
In Column C give the formula for B overtaking A.
My Answer
=BINOMDIST(A1-1-B1,2*A1,0.5,TRUE) + BINOMDIST(A1-B1,2*A1,0.5,FALSE)/2
I wonder whether anyone will trouble to paste that formula in and try it.
This (or adaptations) should have a number of applications, presumably particularly for in-play Golf.
But judging from the lack of interest about such a powerful yet simple technique,"Most Punters Must Enjoy Losing Money".
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