scenario:
I wish to place a bet on the results of 3 games. each game has three possible results, home win, visitor win or tie.
How many individual bets do I need to place to ensure I cover every possible outcome?
What if I wish to cover 4, 5 and 6 games?
2007-10-22 06:45:38 · 4 answers · asked by Petero 6 in Games & Recreation ➔ Gambling
Sorry William
But I know your calculations are incorrect. I have tried covering all the scenarios and It is more the 9 (I gave up trying to calculate myself).
2007-10-22 07:56:13 · update #1
William was on the right tra... well, no he wasn't.
It's actually 3x3x3 = 27
{3 outcomes from game 1} x {3 outcomes from game 2} x {3 outcomes from game 3} = 27 possible outcome sets.
For more games, it's 3x3x3x3, or 3x3x3x3x3x3x3x.... You get the idea.
3 = 27
4 = 81
5 = 243
6 = 729
7 = 2,187
8 = 6,563
9 = 19,689
Like I said, you get the idea...
2007-10-22 09:43:33 · answer #1 · answered by Tank 64 3 · 0⤊ 0⤋
You need to place nine bets to cover every possibility on 3 games with three different outcomes (3X3).
You need to place 12 bets to cover every possibility on 4 games (3X4).
etc.
The bookmakers account for this and you will not be able to make money placing these bets.
2007-10-22 06:59:43 · answer #2 · answered by William H 5 · 1⤊ 1⤋
1 game = 3 possibilities
2 games=9
3 games=27
4 games=81
5 games=243
6 games=729
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2007-10-22 12:25:00 · answer #3 · answered by Elizabeth Aragon 3 · 0⤊ 0⤋
why even bet if your gona hedge all your bets??
2007-10-22 08:23:34 · answer #4 · answered by pit boss 2 · 0⤊ 0⤋