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Football Gambling Question???

alexsanderngaongo

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I'm betting on a 12 team parlay where you can pick from 2 choices; the favorite or the underdog. What are all of the different combinations of teams without repeating a bet?
 
The result is 2704156. More good is to play 1 game max 2 on a tiket for sure profit. I invite you at youcanpayless com is free to join and you will have acces to sure tips.
 
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No offense to previous poster, but there are 4096 combinations for 12 team parlay. You get the answer from 2^12=4096.
 
2 choices and 12 teams result 24 choices. And we need 12 diferent choices from 24 choices. This means are combinations of 12 from 24. And formula of cominations will give this nice number 2704156. Srry but this is the reality
You are invited as well to EDIT from moderator: Don't spam.
 
Last edited by a moderator:
2 choices and 12 teams result 24 choices. And we need 12 diferent choices from 24 choices. This means are combinations of 12 from 24. And formula of cominations will give this nice number 2704156. Srry but this is the reality
You are invited as well to EDIT from moderator: Don't spam.




With all due respect, the only reality is that your logic and methodology are faulty. I will give a hypothetical example for a 2 team parlay (to save time and space). As in original post, only 2 choices, the favourite or underdog allowed.

Game 1 Miami +7.5 at New England -7.5
Game 2 Green Bay -6 at Detroit +6

Combination 1 Miami and Green Bay
Combination 2 Miami and Detroit
Combination 3 New England and Green Bay
Combination 4 New England and Detroit

There are only 4 different combinations possible. 2^2=4

Now using your methodology...2 choices and 2 teams result 4 choices. And we need 2 different choices from 4 choices Using nCr combinations formula (n=4, r=2)....4C2=6.... But there are not 6 different combinations possible, there are only 4.

So going back to original question- 12 team parlay- How many different combinations of teams without repeating a bet.....The answer is 4096 not 2704156.
 
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