Sports betting - round robins

[mulebennett]

This is a fairly elementary question, but I'm not a good math person and thought you guys would definitely be the people to turn to.

There's six teams I want to bet on (teasing, specifically) this weekend. I want to round-robin them so that every possible two-, three-, four-, five- and six-team bet is included.

Would that be 15 twos; 20 threes; 15 fours; 6 fives; 1 six for a total of 57 wagers?

If so, what happens when one of those teams loses? How many of those games are affected? Obviously four of the fives would be gone, the six would be gone, but how many of the twos and threes and fours do I lose?

[hitch1978]

6 teams -

ABCDEF (1 combo, 0 if 1 loser)

5 teams -

ABCDE
ABCDF
ABCEF
ABDEF
ACDEF
BCDEF (6 combos, 5 if 1 loser)

4 teams -

ABCD
ABCE
ABCF
ABDE
ABDF
ABEF
ACDE
ACDF
ACEF
ADEF
BCDE
BCDF
BCEF
BDEF
CDEF (15 combo's, 5 if 1 loser)

3 teams -

ABC
ABD
ABE
ABF
ACD
ACE
ACF
ADE
ADF
AEF
BCD
BCE
BCF
BDE
BDF
BEF
CDE
CDF
CEF
DEF (20 combos, 10 with one loser)

2 teams - Same no' of combos as 4 teams, just pick every combination of 4 teams from above and back the OTHER teams.

I think I may have screwed up, and I'm sure that it is much easier using the math but that might be of some use to you.

[TomCowley]

You know how to calculate the number of bets for each number of teams, like 6c3 = 20. To figure out how many bets are still live after one team loses, just do the calculation for n-1. So if you have 6 teams round-robined in 3s, for 20 games, once one loses, do 5c3=10. So you have 10 live bets and 20-10 lost bets. Once another team loses, do 4c3=4, so the second loss costs you 6 more bets. Repeat for each set (2-teamers, 3-teamers, etc.)

FWIW, you're better off putting all of your money into the 2/3 teamers (whichever has better odds, +100/+180 play the 2-teamers, -110/+180 play the 3-teamers) and ignoring the 4-5 teamers. An additional small play on the 6-teamer is fine if the 6-teamer has equivalent odds to the other group you're playing.