View Full Version : Proposition bet based on a hold 'em situation

10-23-2002, 04:00 PM
A poker acquaintance and I were considering making the following bet involving a hold 'em flop and 50 random turn/river cards.

The flop will come down 7s9dJs and he will give me 10-8o (no spades), in other words the flopped nuts. He will pick one hand to play for 50 turn and river plays and wants me to lay %30 on his action or, in other words, $130 to $100.

Let's say he picks his best possible drawing hand--I'm guessing he would pick KsQs or any set--who would be getting the better end of this arrangement?

The winner would be determined by whichever hand wins the majority of the 50 hands.


10-23-2002, 04:15 PM
I wouldn't take the bet. Without running any numbers, if I were your friend I'd choose 10s8s, thereby and winning whenever another spade appeared and tieing you every other time. Kind of cheeky of him to ask for YOU to lay the odds.

I'll look to see what happens if you don't let him take 10s8s though.


10-23-2002, 04:18 PM
Well we didn't iron out all the details, but the one exception would of course be his holding 10s8s...thereby sharing the nuts on the flop and giving him the redraw. He would have to pick some other hand to play....Most likely QsKs, giving him the best hand every time a spade or a 10 fell....It's probably a bad bet.

10-23-2002, 04:39 PM
his best drawing hand is QsTs that I can find, giving him 48.48% return of the pots.

Others of interest:
9s9h 39.39%
KsQs 43.33%

In the KQ scenario, there are no split pots, so if you give him 1.3-1 odds you will win 56.67 bets in 100 trials
he will win 43.33*1.3 = 56.33 bets.

If that IS the best hand, you have the edge.

10-23-2002, 04:57 PM
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.


10-23-2002, 05:05 PM

Two things:

1) Are you sure that the made straight is a favorite over Qs Ts? I think it's a slight dog. (~48% for the straight, ~52% for the draw)

2) The odds he's laying (130:100) is on who wins the most times out of 50 hands, not on each individual hand. For any draw his friend picks that's any sort of apprieciable underdog to the made straight, it's an overwhelmingly favorable to take. Conversely, even with the very slight edge that Qs Ts has, it's still a 2-1 favorite to be ahead over 50 hands. (of course, if I have it backwards with Qs Ts, then it's a great bet to make in all cases).


10-23-2002, 06:53 PM
I pick the Ts 8d. Pay me now or pay me later?


10-24-2002, 03:26 AM
You should not take the bet. According to the winning chart I have, a straight wins the pot 9% of the time and the flush wins 9% of the time. With you putting in $30 more, you will lose it! <font color="blue"> </font color>

10-28-2002, 02:32 PM
I would pick Ts8s and have a good laugh at you while free-rolling you out of your money.