[singlewm35]

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.

[drfunkmaster]

WOW ... too much math

[BGboothA]

http://www.sbrforum.com/Betting+Tool...alculator.aspx

Math done for you...I believe this is what you are looking for.

[singlewm35]

After taking a break, and then resuming my search, I found this Calc. that will give me the answer. It'd be nice to know the formula used though, so calculations can be done when away from keyboard.

http://www.wagering-sports.com/gambl...in_parlays.asp

Put in the amount you wish to bet (remember, this # is multiplied by the # of possibilities).
Then put in the total # of teams you wish to play, and put in the # of parlays you wish it broke up into.
The calc. will then give you the payout based on the # of teams that actually won.