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ATC, While both of us seem to find this stuff not only fascinating but also wealth-optimising, apparently we are in the vast minority. The problems are relevant, not all that hard, yet reasonably challenging. And whether you get them right or wrong, you'll probably end up learning something useful. But since there appear to be no other takers, you may undisqualify yourself. Once these are resolved, I'll pose a follow up that comes even closer to real AFL life. Restating Problem 2: Under same terms as before, 4 Rounds left B is 2 games behind A What is the probability of B overtaking A? 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. |
Problem 2:
What is the probability of B overtaking A? 8.9844% Will leave the general solution for someone else. |
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8.9844% = Agreed Problem 3 can keep for now. Round to 9% as we move to Problem 4: This should be quicker and easier than the others, so therefore there should be an avalanche of responses. And the answer seems counter-intuitive to me. Problem 4: 4 Rounds remaining. Team G is 2 games out of the 8. It is directly below 5 Teams all equal on wins. A, B, C, D, & E. G has an equal 9% chance of finishing ahead of any one of the other 5. The only 2 possibilities are: - G fails to make the 8 - G makes the 8, kicking out 1 of A - E So what is the probability of G making the 8? In case I'm being too subtle, this hypothetical problem is moving quite close to the current real AFL one. |
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. ---- 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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