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							<channel><title>New Posts For Thread: Place markets math</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math</link><description>Assume you know the exact probability for each runner to place in a place market. Is it possible to compute the probability of each combination of them placing?An example: In a "2 to be placed" market with 4 runners (A,B,C,D). The probability of each</description><item><title>YOU COULD HAVE LEARNT TO BUILD A NUCLEAR BOMB WITH LESS CALCULATIONS THAN YOU LOT HAVE COME UP WITH</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515209815#515209815</link><description>YOU COULD HAVE LEARNT TO BUILD A NUCLEAR BOMB WITH LESS CALCULATIONS THAN YOU LOT HAVE COME UP WITH</description><pubDate>Wed, 18 Jan 2012 18:56:59 -0600</pubDate></item><item><title>WHEN YOU HAVE FINISHED YOUR PROBLEM AND YOUVE GOT YOUR ANSWER ,YOU WILL HAVE ANOTHER PROBLEM WHEN YOU FIND OUT YOU STILL CANT WIN</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515209737#515209737</link><description>WHEN YOU HAVE FINISHED YOUR PROBLEM AND YOUVE GOT YOUR ANSWER ,YOU WILL HAVE ANOTHER PROBLEM WHEN YOU FIND OUT YOU STILL CANT WIN</description><pubDate>Wed, 18 Jan 2012 18:55:46 -0600</pubDate></item><item><title>Andriy,Please check you message box.</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515178229#515178229</link><description>Andriy,Please check you message box.</description><pubDate>Wed, 18 Jan 2012 05:15:55 -0600</pubDate></item><item><title>My gut feeling is that the chance of any runner finishing in the forecast is variable (not fixed)and changes according to whichever runner joins it in the forecast but then again what do I know!!!</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515092331#515092331</link><description>My gut feeling is that the chance of any runner finishing in the forecast is variable (not fixed)and changes according to whichever runner joins it in the forecast but then again what do I know!!!</description><pubDate>Mon, 16 Jan 2012 07:17:25 -0600</pubDate></item><item><title>I think dbze is spot on in his answer that we do not have enough information to determine this from the place probabilities. For instance would the joint probabilitesAB: 0.6AC: 0.1AD: 0BC: 0BD: 0CD: 0.3give probabilites for each runner to place as gi</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515088821#515088821</link><description>I think dbze is spot on in his answer that we do not have enough information to determine this from the place probabilities. For instance would the joint probabilitesAB: 0.6AC: 0.1AD: 0BC: 0BD: 0CD: 0.3give probabilites for each runner to place as gi</description><pubDate>Mon, 16 Jan 2012 05:09:25 -0600</pubDate></item><item><title>YOU CAN WRITE AS MANY NUMBERS AS YOU WANT ,BUT YOU STILL WONT WIN</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515066617#515066617</link><description>YOU CAN WRITE AS MANY NUMBERS AS YOU WANT ,BUT YOU STILL WONT WIN</description><pubDate>Sun, 15 Jan 2012 15:24:28 -0600</pubDate></item><item><title>You can make correlation/regression models for certain race type configurations (no. of runners, shape of odds of field), but even then you need to consider characteristics of some individual runners, eg. novice chasers which tend to win or fall ofte</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515061581#515061581</link><description>You can make correlation/regression models for certain race type configurations (no. of runners, shape of odds of field), but even then you need to consider characteristics of some individual runners, eg. novice chasers which tend to win or fall ofte</description><pubDate>Sun, 15 Jan 2012 13:29:00 -0600</pubDate></item><item><title>Agree Andriy - I wonder if one of the posters is fine-tuning their bot after trying to back VDV at 29.I think the assumption that win probability "exactly correlated" with place probability is too strong. For example, if there is a strong odds-on fav</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515056765#515056765</link><description>Agree Andriy - I wonder if one of the posters is fine-tuning their bot after trying to back VDV at 29.I think the assumption that win probability "exactly correlated" with place probability is too strong. For example, if there is a strong odds-on fav</description><pubDate>Sun, 15 Jan 2012 11:44:32 -0600</pubDate></item><item><title>^ And the above assumed also that win probability correlated exactly to place probability... but that's a different sory. It's interesting that there have been a few place/each way threads since the VDV debacle and place market issue, with people pos</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515045603#515045603</link><description>^ And the above assumed also that win probability correlated exactly to place probability... but that's a different sory. It's interesting that there have been a few place/each way threads since the VDV debacle and place market issue, with people pos</description><pubDate>Sun, 15 Jan 2012 06:09:26 -0600</pubDate></item><item><title>To get the 0.311 i quoted above i used the following method, considering that the relative probabilities would stay the same between 2nd/3rd/4th once winner was selected.Prob = P(A 1st, B 2nd)+ P(B 1st, A 2nd)= [0.35*(0.3/(0.2+0.3+0.15)] + [0.3*(0.35</title><link>https://community.betfair.com/general_betting/go/thread/view/94082/28841293/place-markets-math?post_id=515045099#515045099</link><description>To get the 0.311 i quoted above i used the following method, considering that the relative probabilities would stay the same between 2nd/3rd/4th once winner was selected.Prob = P(A 1st, B 2nd)+ P(B 1st, A 2nd)= [0.35*(0.3/(0.2+0.3+0.15)] + [0.3*(0.35</description><pubDate>Sun, 15 Jan 2012 05:47:02 -0600</pubDate></item></channel></rss>
