Pot Odds in Omaha Hi/Lo

[Max_Speed]

Hi,
I have posted in the probability forum. I hope that some of you may be able to help me out. Many thanks, Max

http://forumserver.twoplustwo.com/sh...14&fpart=1

[Buzz]

Max - In the article to which you refer in your post on the probability forum, Fred Remzey observes that when you split the pot, the amount you actually win is less than half of what you would get if you scooped.

You can easily demonstrate this phenomenon to yourself by stacking up chips of two different colors, one color to represent the chips you, yourself, contribute to the pot and another color to represent the chips your opponents contribute to the pot.

Suppose everyone involved in a hand contributes ten chips to the pot.

When you have two opponents and everyone contributes ten chips to the pot, then there are thirty chips in the pot. Take twenty chips of one color and make two stacks to represent ten chips contributed to the pot by each of two opponents. Then make a stack of ten chips of a different color to represent the ten chips you have contributed to the pot. When you scoop, you get your own ten chips back and win all twenty chips contributed by your two opponents. When you win half the pot, you get your own ten chips back but only win five of the chips contributed by your two opponents. Thus with two opponents the amount you win is four times as much when you scoop as when you split.

When you have three opponents and everyone contributes ten chips to the pot, making a total of forty chips in the pot, when you scoop the pot, you get all forty chips in the pot. In other words you get your own ten chips back and win the thirty chips contributed by your three opponents. When you split a forty chip pot with one opponent, you get twenty chips. Ten of these chips are the ten chips you have contributed yourself. In other words you get your own ten chips back but only win ten chips. Thus with three opponents, you win three times as much when you scoop as when you split.

Fred states that you can halve your scoop odds and then subtract one to get your split odds. That’s not quite correct, but it’s in the ball park.

The simplicity of Fred’s method is appealing. However, I'm not interested in calculating my split odds because I'm rarely, if ever, drawing for exactly half of the pot. Let me explain.

More often than not, you’ll have some outs for either the low half the pot or the high half of the pot while you’ll have other outs for the whole pot. For example, when you hold Ah-2h-4c-Kc, and the board after the turn is 4s-8h-Tc-Jd, you have 16 outs for the low half of the pot (3c, 3d, 3h, 3s, 5c, 5d, 5h, 5s, 6c, 6d, 6h, 6s, 7c, 7d, 7h, and 7s) - and another 4 outs (Qc, Qd, Qh, and Qs) for a scoop.

Further complicating the issue is the quartering factor - one or two of your opponents may be drawing for the same hand - especially when you are drawing for low or for a wheel with ace-deuce or ace-trey, or drawing for a high straight with two honor cards. (Honor cards are ace, king, queen, jack, and ten).

And there are some additional complications I’m not going to address.

I approach the problem a bit differently. Rather than calculating “split pot odds,” I sometimes estimate what I call “scoop equivalent” hand odds. Then I compare these “scoop equivalent” hand odds to implied pot odds. (Thus I would never need to use “split pot odds.”)

I find I don’t need to make a “scoop equivalent” hand odds calculation very often - maybe one time - or at most a few times during a playing session.

In brief, I multiply my outs to half (or possibly a quarter) of the pot by a factor or 0.3 or 0.4, depending on the situation. In the example given above, I’d multiply the 16 outs for the low half of the pot by 0.3 to get a product of 4.8. Then I’d round that up to 5 and add it to the 4 outs for the queens to get a total of 9 “scoop equivalent” outs. (I’d use 0.3 rather than 0.4 here because of the danger of being quartered or sixthed by an opponent or two who might also be drawing to the acey deucy nut low).

(Note that if I had used 0.4 instead of 0.3 as the multiplicand it wouldn’t make much difference. I’d get 6.4 as the product, round down to 6 and end up with a total of ten “scoop equivalent” outs instead of nine).

Then, on the turn, there are 44 unknown cards. The equivalent of 9 cards are favorable while the equivalent of the other 35 cards are unfavorable. Thus the hand odds would be 35 to 9 against, or slightly less than 4 to 1 against. Accordingly, the pot would have to contain at least four times as much as it would cost me to call on the third betting round in order to justify a call from me. That four times as much would include chips my opponents would contribute on the final round if I made my hand.

My method seems crude to me, but I can do the simple math in my head very quickly and easily while seated at the table, usually without calling time out. Here are the steps for the example given:
16*0.3=~5.
5+4=9.
44-9=35.
35/9=~4.
Voila! Four quick steps to tie together outs for the whole pot with outs for half or a quarter of the pot and come up with hand odds of about 4 to 1 against, thus needing implied pot odds of at least 4 to 1 if I want favorable (drawing) odds.

Simple as it is, my method is more complicated than Fred Remzey’s method. However, merely dividing the pot size in two and then subtracting one (Fred Remzey’s approach) doesn’t get me anything that is useful to me.

But in any event, Fred and I are in complete agreement that when you split the pot you win less than half of what you would win if you scooped. I hope you follow my explanation above and see it too.

Just my opinion.

Buzz