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Topic: Simple Football Pool (Read 1592 times) |
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Joe Fendel
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Simple Football Pool
« on: Oct 31st, 2005, 3:57pm » |
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You're in a Monday Night football pool with N participants. The rules are simple.
1. Every participant contributes $1 to the pot before the game on Monday Night.
2. The probability, p, of a home team win is public information. To simplify the situation, there are no ties possible.
3. Every participant submits a pick of either the home team or the visiting team in a sealed envelope.
4. After the game is over, the envelopes are opened, and the winners split the pot.
5. In the event that all participants picked the loser, everyone gets their $1 back.
When N=1, the game is trivial (and boring). When N=2, the optimal strategy is to always pick the favorite (i.e. if p > 0.5 pick the home team, otherwise pick the visitors). When N>2, a randomized strategy is called for when p is between 1/3 and 2/3 (because, for example, if p = 0.4, and both of your opponents have picked the visitors, then you do well to pick the home team, even though they're underdogs).
So let H be the chances you will pick the home team, given N opponents and a probability p of a home team win.
What is H as a function of N and p?
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towr
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Re: Simple Football Pool
« Reply #2 on: Nov 1st, 2005, 2:28am » |
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If everyone follows the same strategy, it doesn't actually matter what strategy you follow, since it's a zero sum game.
At least, it doesn't matter with respect to expected outcome. You could still try to maximize risk I suppose. (To make the bet as interesting as possible)
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| « Last Edit: Nov 1st, 2005, 2:29am by towr » |
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JocK
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Re: Simple Football Pool
« Reply #3 on: Nov 1st, 2005, 9:32am » |
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on Nov 1st, 2005, 12:08am, towr wrote:
That is the right strategy when N, the number of participants, is very large:
Suppose one of the participants selects the home team with a probability H < p or H > p. If the other participants all stick to H = p, they will have a (small) expected gain at the expense of that single participant.
Assuming all participants behav rationally, they will all stick to H = p.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Joe Fendel
Junior Member
Posts: 68
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Re: Simple Football Pool
« Reply #4 on: Nov 1st, 2005, 9:54am » |
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I agree that
limN -> infty H(N, p) = p.
Also, for p < 1/N, H(N, p) = 0, and for p > (1 - 1/N), H(N, p) = 1.
I'm also pretty confident that for p < 0.5, H(N, p) < p, and for p > 0.5, H(N, p) > p. (And H(N, 0.5) = 0.5).
But is there an explicit formula for H(N, p)? What about just H(3, p)? (I only play this with two other friends, after all.)
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JocK
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Re: Simple Football Pool
« Reply #5 on: Nov 2nd, 2005, 2:49am » |
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on Nov 1st, 2005, 9:54am, Joe Fendel wrote:
A stronger condition applies:
H(N, p) + H(N, 1-p) = 1
on Nov 1st, 2005, 9:54am, Joe Fendel wrote:
[...] But is there an explicit formula for H(N, p)? |
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If I'm not mistaken the Nash equilibrium for the N-player game can be written:
H(N, p) = 0, p < 1/N
H(N, p) = (N p - 1)/(N - 2), 1/N <= p <= 1 - 1/N
H(N, p) = 1, p > 1 - 1/N
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| « Last Edit: Nov 3rd, 2005, 2:13am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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