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 09-15-2017, 11:24 AM #1 Csaba veteran     Join Date: Jan 2009 Posts: 2,981 Combined odds of dependent events in an either/or situation I'm looking for a formula to calculate the combined odds of one of 3+ events occurring in an either/or situation. For example, what are the combined odds of either player A, B or C scoring the first goal in a football match if their different odds are 4/1, 10/1 and 15/1. I think the formula for 2 events is (x*y)/(x+y) but haven't been able to find anything for 3+ events. Thanks in advance for any responses.
 09-15-2017, 02:04 PM #2 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 4,202 Re: Combined odds of dependent events in an either/or situation I always convert odds to probabilities before doing any calculation. Assuming the odds are applicable, if the odds against an event are X to 1, the occurrence probability is 1/(X+1). For your example, we have P(A 1st) = 1/(4+1)= 0.20, P(B 1st)= 0.0909 and P(C 1st) = 0.0625. The total is 0.353 and it represents A or B or C score first. This implicitly assumes odds for all other players scoring first –or no goal – add up to a 100% probability. However, if this is a gambling situation, to give the bookie profit, that will not normally be the case. Given a probability of P, the corresponding odds against are (1-P)/P, so in this case it is 647/353=1.83 to 1. Again, this assumes all probabilities are accounted for such that the bookie has no expected profit. For 2 players, using the odds numbers directly, I get (XY-1)/(X+Y+2) which differs from what OP posted so this illustrates my lack of knowledge of how football odds are developed
 09-15-2017, 02:22 PM #3 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,837 Re: Combined odds of dependent events in an either/or situation Hopefully, we are using the same terminology. Ignore "vig" in what follows (so all the odds are "true"). Then the presumed underlying probability of an event offered at 4:1 odds is 1/(4+1) = 1/5 = 20%. Same for all other odds. Going the other way, if an event has a presumed underlying probability of 20% occurring, then the true odds would be given as (1/20%)-1 to 1. Or (1/.2)-1 to 1. Which is 5-1 to 1 or 4 to 1. Under a scenario in which all the events are mutually exclusive, such as the player who scores the first goal in a game, then there is a straightforward relationship between individual odds and the combined odds of the event of any of a group of several players scoring the first goal. If there are N players you wish to find the combined odds for: Combined Odds = [Product (1+Individual Odds) / Sum{Product of all the ways you can "choose" (N-1) of the players} (1+Individual Odds)] - 1 to 1 Example 1: Two players with individual odds of 3:1 and 4:1. Then the presumed underlying probabilities that each of these two players scores the first goal is 1/(3+1) and 1/(4+1), respectively. Which are 1/4 and 1/5, respectively. Or 25% and 20%. Of course, the presumed underlying probability that either scores the first goal is the sum or 25% + 20% = 45%. And the true odds for an event that is 45% likely is (1/.45)-1 to 1 or 1.2222222 to 1 or which is 11/9 to 1. Let's check the above formula to see if it gives that result. Combined Odds = [(3+1)*(4+1) / (3+1)+(4+1)] - 1 to 1 = [4*5/(4+5)] - 1 to 1 = (20/9) - 1 to 1 = 11/9 to 1 (confirmed) Example 2: Three players with individual odds of 4:1, 10:1, and 15:1. Then the presumed underlying probabilities that each of these three players scores the first goal is 1/(4+1), 1/(10+1), and 1/(15+1), respectively. Which are 1/5, 1/11, and 1/16, respectively. Or 20%, 9.09090909%, and 6.25%, respectively. Of course, the presumed underlying probability that any of the three scores the first goal is the sum or 20% + 9.09090909% +6.25% = 35.34090909%. And the true odds for an event that is 35.34090909% likely is (1/.3534090909)-1 to 1 or 1.829581994 to 1 (which is 569/311 to 1). Let's check the above formula to see if it gives that result. Here N = 3, so N-1 = 2. So we are going to do the sum over all the pairs of players {AB, AC, BC}. Adding 1 to each of the respective odds gives 5, 11, and 16. Then the formula becomes: Combined Odds = [(5*11*16) / {(5*11)+(5*16)+(11*16)}] - 1 to 1 = [880 / 311] - 1 to 1 = (569/311) to 1 = 1.829581994 to 1 (confirms above answer) Hope this helps and hope this makes sense. Last edited by whosnext; 09-15-2017 at 02:34 PM.
 09-27-2017, 01:32 PM #4 Csaba veteran     Join Date: Jan 2009 Posts: 2,981 Re: Combined odds of dependent events in an either/or situation Thank you both for the responses. Sorry it took me so long to have a look at this thread, I've just finished moving house.

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