General Betting
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10 Oct 10
00:13
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25 Oct 03
In a ficticious competition, A v B, your research shows that 'A' has won 44 out of 100 and 'B' has won 5 out of 10 against other competitors. They are facing each other for the 1st time.
On the face of it 'B' has a better strike rate (50% v 44%)
but it's only over 10 games so it's less reliable than 'A's stats.
So how would you approach pricing this up if this is the only info you have?
On the face of it 'B' has a better strike rate (50% v 44%)
but it's only over 10 games so it's less reliable than 'A's stats.
So how would you approach pricing this up if this is the only info you have?
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A- 2.27
B- 1.78 As it's the only info we have the reliabilty of B's stats cannot be quantified and so can't be considered as a factor in the pricing. |
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It depends on the situation and the game, I'd price B at 2.1 or maybe evens, not because I don't think 5/10 is a large enough sample (as Tober says this doesn't really affect the price), but because a lower number of games may indicate being newer to a higher level, and therefore there might be an element of 'selection bias', in other words he's facing player A because he's happened to win some good games recently.
Otoh, if he's facing A because he's recently had poor form and 'dropped down' then chances he's been running under EV and will perform better than stats suggest. Note, I'm not saying someone is ever 'due', but that if someone is running under EV then their new stats should be EV, not higher than that to 'correct'. I think when you have such a small sample size on someone, it's probably best to step away from the stats. If B happened to lose 1 more game then they'd be 4/10 and tober's prices would reverse. |
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I'm pretty sure that this is solvable using the probabilities of each event happening. Not sure that I have the maths to do that, certainly not at this time of morning!
However, the chances (using binomial) of the "true" chance of winning (assuming for some reason all other opponents are equal. Maybe they've played team C that many times) against the other teams are: To win at least 5/10 50% of the time requires a 45.2% chance (or skill rating as I'm going to call it) To win 44/100 50% of the time requires a 43.5% skill rating Changing those to 75% of the time gives, 5/10 = 55.5% Skill rating 44/100 = 46.9% Somewhere if you plot all these chances I suspect you get an area of overlap in the graph that tells you the right number. I need coffee However, off the top of my head, I'd think that 55.5/46.9 = 2.18 will be too high and 45.2/43.5 = 2.04 is too low. |
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Get me a Drink i would price up evens both of em , evens take your pick ..why its a bit like tossing a coin it dont matter if heads (A) has won 44 out of a 100 and tails (B) has won 5 out of ten .....its still a true even money chance next toss spin ...... I got a Drink thanks .....
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But zipper that is a very naive estimator.
If as Lori says we assign each a "skill rating", p1 and p2, then we can calculate Prob(p2>p1). If that's significantly less than 0.5 then it's bad value to price this as evens. I can't work out right now how to calculate Pr(p2>p1) but my gut feeling says this should be possible. |
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ror when the margin is 0.01 to 0.05 there aint no skill ....only a guess .
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I think when you have such a small sample size on someone, it's probably best to step away from the stats. If B happened to lose 1 more game then they'd be 4/10 and tober's prices would reverse.
On the other hand, if B won one more game then it would be 6/10... |
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tobermory is disregarding B's stats because of the sample size, so whether B wins or loses the next match is irrelevent. I think its a good place to start where he has, assuming B to be of average ability and therefore pricing purely on A's stats. If it was me I would maybe be tempted to bring the prices slightly closer together...something like 2.22 - 1.82
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