virchis said:
Yes, there is. I just divide the dollars returned on one side by the fractional odds on the other. For example:
1120 to win 1000 at -112, would be roughly 2120/1.952=1086 and change
1086 to win 1034 at -105
Lose roughly 86 either way or win 2034 if it hits. 2034/86 = 23.65:1
Looks like neilm beat me to it...same example too. :shocked:
Just seems like there is a better way to do this, let me know if i am wrong.
Instead of buying the half point on -7 for the 10 cents do this.
Bet -7 -102
and +7.5 -105
Bet 1020 to win 1000 -102
and bet 1035 to win 986 -105
Then it seems like you are risking 35 units to win 986.
Which seems even better than previous calculations.
Is this right?
Instead of buying the half point on -7 for the 10 cents do this.
Bet -7 -102
and +7.5 -105
Bet 1020 to win 1000 -102
and bet 1035 to win 986 -105
Then it seems like you are risking 35 units to win 986.
Which seems even better than previous calculations.
Is this right?
virchis said:
Yes, but this is interesting, and now we're going to get into advanced middling. Yes, that example is better odds. The question is going be whether you are going to be bumping against the max at the books you're playing at or the most you're willing to put into play. I still think the 7 is worth moe than 10 cents so the buy is good, but if you can simply bet double on your second example, then go ahead and do that. Either way is fine, depends on your tastes, the maximums, your bankroll, etc.
