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.

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

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.

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.