General Betting
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23 Feb 15
18:02
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10 Jan 02
If there is a 60% chance of A being higher than B and a 60% chance of B being higher than C what chance should it be that A is higher than C assuming a random distribution?
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Imagine A is a country with land area 100 km2 - 60km2 is at altitude 5000m ( Andean plateau ) and the rest at altitude 1000m.
Imagine B is a country with land area 100 km2 - all of it at altitude 3500m. Pick a random spot in both countries - it's 60% likely that the spot in country A is higher. Imagine C is a country with land area 100 km2 - 60km2 is at altitude 2000m and the rest at altitude 8000m ( Himalayan plateau ) Pick a random spot in countries B and C - it's 60% likely that the spot in country B is higher. There are some steep cliffs in countries A and C, climbers love it. Then pick a random spot in countries A and C - it's 36% likely that the spot in country A is higher. To summarise: p(a>b) = 0.6, p(b>c) = 0.6, and p(a>c) =0.36 And that also explains why a political party can be more popular than two less worthy rivals but finish behind them. |
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What I meant was imagine if it was like a random draw of 1 billion numbers.
A gets 60 picks to B's 40 B gets 60 picks to C's 40 so A gets 36 picks to C's 16 so A is 4/6 to beat B, B is 4/6 to bt C, and A is 4/9 to bt C |
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Imagine in a numbers game like darts. If Van Gerwen was 4/6 to bt Taylor, Taylor was 4/6 to bt Anderson, How close would Van Gerwen be to being 4/9 to bt Anderson? Given that the distribution is not random, should he be bigger or shorter than 4/9 and why?
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My model would be A is a random number between 1 and 50000 so average 25000, B is a random number between 1 and 20000 so av 10000, and C is a random number between 1 and 8000 so av 4000.
So the average number drawn fits the criteria and there is a 84% chance A is greater than C |
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Another model:
A chucks a dart at a board of size 100 - 20% of the board he wins, the other 80% of the board it's 50/50 between him and B. B chucks a dart at board size 80 - 20% of the remaining board (area=16) he wins, the other 80% ( area=64) it's 50/50 between him and C. Extend the rule to A chucking against C - area of board is back to 100 - but 36% he wins, and the other 64% splits 50/50 between him and C. Outcome: A is 68% likely to beat C. |
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not sure mine is right will think some more
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LOL, I don't seem to have much luck with posting to these forums
 .Here are the URLs of the images that should have been in the post above: http://oi57.tinypic.com/xc1at.jpg http://oi59.tinypic.com/307z4g5.jpg http://oi61.tinypic.com/2a4rd3a.jpg http://oi61.tinypic.com/2hdv191.jpg |
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I should add that the above assumes that the distribution of performances for A, B and C are all independent of each other.
For something like darts this is probably fine, but for certain other games it may well not be. For example: * A could have a certain skill "X" that gives him an competitive edge over B. * B could have a certain skill "Y" that gives him an competitive edge over C. * C is impervious to A's superior skill "X" ability. Another (extreme) example to better illustrate this, would be if players A, B and C were to play rock-paper-scissors with the following strategies: * A plays rock 100% of the time. * B plays scissors 100% of the time. * C plays paper 100% of the time. A would beat B 100% of the time, B would beat C 100% of the time, BUT A would lose to C 100% of the time! |
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Another model
A is a random number between 11 and 60, B is a random number between 6 and 55 and C is a random number between 1 and 50. So for example 10% of the time A is 56-60 and is always higher than B, 10%* B is 6-10 and always lower than A, the remaining 80% it is 40% A higher, 40% B higher. so overall 60/40. Using that logic for A vs C. 20% 51-60, 20% 1-10 are A higher. Of the remaining 60% 30% A higher, 30% C higher. So overall 70% chance A greater than C. * I think this perhaps should be 9% as some of the time as when B is 6-10, A is 56-60 and already counted. |
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Here is the solution assuming the distribution of performances is normal:
http://oi58.tinypic.com/5mh2dv.jpg http://oi59.tinypic.com/mvj3us.jpg P(A>C) = ~69.4% To better visualize what's going on, here is a plot of the distribution of performances for A, B and C: http://oi57.tinypic.com/34y9s1v.jpg Whether any of these are an acceptable model for the distribution of performances is questionable, but hopefully this has gone some way toward answering your question. |
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Cheers for your input juk. It was just a random thought that crept into my head loosely related to something I was betting on. If I thought it was going to be much use I wouldn't have posted it on here. I just picked darts because it seemed the best example. They don't get injured or dq'd and the performances of each player are almost independent of each other.
Surprising (to me anyway) how close the answer was to 4/9. |
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For what it's worth and going off on a tangent, I think the athletic performance of men in sport is much more consistent than that of women. This view came about when thinking about performances that were well below what I expected. It dawned on me that a significantly disproportionate amount of the time it came from women. So I did some digging, not my own research, rather looking for other research on the subject, but it came to the conclusion that instinctively I felt was true. Could be confirmation bias of course, but anyway, it concluded that some women's performances were affected by the time of their menstrual cycle.
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