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Old 10-23-2002, 04:57 PM
PseudoPserious PseudoPserious is offline
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Join Date: Oct 2002
Posts: 151
Default Re: Proposition bet based on a hold \'em situation

Okay, here goes:
(irchans, prepare to correct errors)

Laying odds of 130-100 means you have to win the money 57% of the time to break even. You win the money whenever your straight holds up 26 or more times out of 50. Assuming that (1) ties are redealt (i.e., don't count towards the 50 hands) and (2) if you're tied at 25-25 at the end of 50 you deal one tie-breaker hand, you don't need that much of an edge in any one particular hand to have a 57% win rate...your per-hand win only needs to be around 51.25% or so.

Here are some of his possible holdings, along with your win percentage on one hand and your win percentage of winning 26+ out of 50:

Ks Qs - 57% - 84%
9s 9c - 60% - 93%
2s 4s - 63% - 97%
Jc Jh - 65% - 97%
As Ac - 92% - very large
Ad Ac - 96% - very very large

If your friend picks any of these, bully for you. Lay him more action. However, there's also:

Qs Ts - 47% - 33%

So you'd have to outlaw that hand too.


Cheers,
PP
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