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StephenW
02-15-2005, 11:08 AM
hi, I am completely stuck as to an odds calculation, I bet on 2 in play sporting events, one with odds 1/16 and one with odds 1/6, both needless to say lost. Going by the odds I make it that the 1/16 was 94% and 1/6 was 86%,what is the probality that neither bet wins?
( I realize the bookmaker has got an edge on these odds but I am interested never the less)

Thanks
gaming_mouse
02-15-2005, 11:38 AM
(15/16)*(5/6)=.79

About 79%
StephenW
02-15-2005, 12:30 PM
[ QUOTE ]
(15/16)*(5/6)=.79

About 79%

[/ QUOTE ]

Thanks, Correct me if I'm wrong, but is that the % chance that both will win, I still can't figure the math for niether to win, for 1/16 I think there is a 6% chance it will lose, so I figure it should be even less than 6% for both to lose
gaming_mouse
02-15-2005, 02:34 PM
[ QUOTE ]
Thanks, Correct me if I'm wrong, but is that the % chance that both will win, I still can't figure the math for niether to win, for 1/16 I think there is a 6% chance it will lose, so I figure it should be even less than 6% for both to lose

[/ QUOTE ]

I calculated the chance that neither will win.
olavfo
02-15-2005, 02:46 PM
Odds expressed as A/B means you will win B out of A+B times (if we assume that the odds reflect the true probabilities and that there's no bookie overround).

Odds 1/16 = you win 16 out of 17 = 1/17 chance to lose

Odds 1/6 = you win 6 out of 7 = 1/7 chance to lose

The probability of both teams losing is then

(1/17)*(1/7) = 1/119 = 0.84%

olavfo