Does anyone know the formula to calculate how many wins you would get for XXX number of losses in a round robin?
Example...
If I play 10 teams, and break it into X number of 6 team parlays, if two out of the 10 picks are losses, how many of the 6 team parlays win?
I've looked for online calculators but they all seem to be either limited to max of 5 picks or only calculate the payout of parlay bets.
I use this formula to calc. the total possibles:
N!
---- = Tot. combinations
r!(N-r)!
that gives me the # of combinations.
If N=10 (number of picks)
r=6 (size of parlay)
10!
----
6!(10-6)!
10x9x8x7 5040
-------- = ------ = 210
4x3x2x1 24
Now out of the 210 combinations, how many would not contain one or both of the 2 losers? Thus giving me the # of winning combinations.
thanks in adv.
Example...
If I play 10 teams, and break it into X number of 6 team parlays, if two out of the 10 picks are losses, how many of the 6 team parlays win?
I've looked for online calculators but they all seem to be either limited to max of 5 picks or only calculate the payout of parlay bets.
I use this formula to calc. the total possibles:
N!
---- = Tot. combinations
r!(N-r)!
that gives me the # of combinations.
If N=10 (number of picks)
r=6 (size of parlay)
10!
----
6!(10-6)!
10x9x8x7 5040
-------- = ------ = 210
4x3x2x1 24
Now out of the 210 combinations, how many would not contain one or both of the 2 losers? Thus giving me the # of winning combinations.
thanks in adv.
