Math Question / Sports betting

The Prophet 22The Prophet 22
in Off Topic Lounge
If I am playing 8 games on one ticket and I want to play A - D on every ticket. How many tickets will I need to play to cover every combination on tickets with the other 4 games for vistors and home.

Therefore if A-D win, I should have one winning ticket no matter what happens with E - H. Correct?

Games
A
B
C
D
E
F
G
H

Eample for 5 games

One ticket
A
B
C
D

E vistor

Second ticket
A
B
C
D

E home

As long as A - D win, E game wont matter as I have one ticket for the vistor and one ticket for the home. My question is how do I do this for 8 games. You actually have to fill out every ticket.


Prophet22

Comments

  • moosemoose
    If zero is home and one is away then 1111 converted from binary to decimal is 15, add 0000 and you get 16 different combinations.

    0000
    0001
    0010
    0011
    0100
    0101
    0110
    0111
    1000
    1001
    1010
    1011
    1100
    1101
    1110
    1111

    Each column corresponds to game EFGH with 0 home and 1 away.

    Or 4 games with 2 outcomes is 4^2 = 16
  • woog30woog30
    I suggest not betting crazy parlays, you will not be profitable long term.
  • MickeyHoldemMickeyHoldem
    woog30 wrote: »
    I suggest not betting crazy parlays, you will not be profitable long term.
    Are you kidding.... I'm in the process of filling out 85,900,584 lotto max tickets... easy money!!!
  • The Prophet 22The Prophet 22
    woog30 wrote: »
    I suggest not betting crazy parlays, you will not be profitable long term.

    I understand the parlay betting "scam". Not many here on the forum like NFL spreads for betting purposes only fantesy stats. This is an exercise with a few friends. My above question will only result in 7 team parlays and one team of 8.

    Thanks for your help
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