Probability question?

**u21c3f6** · 09-12-14, 08:14 PM

Based on your numbers, yes, the chance of both teams winning would be approx 68%.

Joe.

PS. The above does not account for juice, if any.

**donkeyshark** · 09-12-14, 08:55 PM

Originally posted by MustWinPlease

Ive been thinking about this for a while

The perceived probability of odds of 1.20 are 0.83 (83%)

So does that mean, if you place a bet on a team with odds of 1.20, then place a bet on another team (in another game) with the same odds, does that mean the chances of winning are 0.83 x 0.83 and therefore = 0.68 (68%?)

I'll address two things. First, it's unclear to me what you mean by the "perceived" probability of 1.20. One can derive the implied probability of a bet paying 1.20 to 1 as follows:

1.20/(1.20+1) = 1.20/2.20 ~= 0.54545 ~= 54.5%

Your question is with regard to independent events. The probability of winning two bets which are not correlated (eg - betting on two different games) is the product of the probabilities of winning each game. In this case, it would be 0.54545 * 0.54545 ~= 29.75%

**CidaliaRod** · 09-14-14, 10:31 PM

Originally posted by donkeyshark

I'll address two things. First, it's unclear to me what you mean by the "perceived" probability of 1.20. One can derive the implied probability of a bet paying 1.20 to 1 as follows:

1.20/(1.20+1) = 1.20/2.20 ~= 0.54545 ~= 54.5%

Your question is with regard to independent events. The probability of winning two bets which are not correlated (eg - betting on two different games) is the product of the probabilities of winning each game. In this case, it would be 0.54545 * 0.54545 ~= 29.75%

I believe that MustWinPlease was using decimal odds, so 1.20 odds would mean that a winner pays 0.2 to 1, not 1.2 to 1. If that is the case then MustWinPlease calculations are roughly correct.