Probablity of winning versus probability of losing.
March 11th, 2013 at 5:32:19 PM
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I'm not sure how to word this question, but here goes. Is there a formula or calculation for determining the risk of your bet on the next toss? I understand the probability, but is here a calculation to rate the probability of winning versus the probability of losing?I should give the following example. If I place a $6 bet on the 6 and 8 I have $12 at risk. I understand the probabilty of a win is 10 combinations to 36 combinations or 28%. What is a good way to evaluate the chances the the next toss will be winning 6 or 8 versus the losing 7. 28% chance to win versus 17% chance of losing.can you divide this by that to get to one calculaion?
March 11th, 2013 at 5:39:20 PM
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It's very simple. Just divide.
Placing the 6 and 8:
Chance of any win on the next roll:
10/36
Chance of loss on the next roll:
6/36
Chance of no result on next roll:
20/36
Now if you want to disregard non-decisive rolls, you can say that you have a 10/16 chance of winning on the next decisive roll, and a 6/16 chance of losing on the next decisive roll.
So you are 10/6 times more likely to win on the next decisive roll than to lose.
Placing the 6 and 8:
Chance of any win on the next roll:
10/36
Chance of loss on the next roll:
6/36
Chance of no result on next roll:
20/36
Now if you want to disregard non-decisive rolls, you can say that you have a 10/16 chance of winning on the next decisive roll, and a 6/16 chance of losing on the next decisive roll.
So you are 10/6 times more likely to win on the next decisive roll than to lose.
March 11th, 2013 at 5:40:06 PM
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just use ratios. you have 10 winning combinations (6 or 8) and 6 losing combinations (any 7). and youre risking $12 ($6 each on 6 and 8) to win $7 (the payout if a 6 or 8 hits). so youre 5:3 to win your bet when youre laying $12 to win $7.
