Omaha High/Low:

I'D RATHER BE LUCKY THAN GOOD

BY: Russ Fox

"Good luck is a lazy man's estimate of a worker's success." -- Anonymous

I was playing in my usual $6/$12 Omaha game, with (mostly) the usual group of suspects, when Ida entered the game. Ida said she'd never played the game before (Paul the prop later told me he'd never seen her in the game), and muttered, after scooping a pot with 2 9 9 K (making nines full), "I'd rather be lucky than good." I held my tongue. However, I think it's worth examining that statement in depth.

Obviously, short-term luck can have a tremendous influence on whether you are a winner or a loser while playing poker. If on every hand you start with the best draw but you don't ever catch, you're unlucky. However, over time (or, statistically, over an infinite amount of hands) we will all get the same cards and make the same draws (and miss the same draws, too). That is, luck evens out.

Unfortunately, none of us live to see an infinite amount of hands. The late Hermine Baron, one of the best bridge players that ever lived, felt that she never held good cards. She thought that I was the only person who approached her in holding bad cards. Yet we were playing duplicate bridge, when half the people in the room held exactly the same cards as us! Superstitions and perceptions die very slowly.

Now let's look at an Omaha hand. Assume I hold A 2 5 7 , and the board is K Q 9 /3 . I have seven nut outs and two non-nut outs. Assume that I'm playing $6/$12, and the pot has $150 in it. One of my opponents bets, another calls, one folds, and I, the last player to act, must decide whether to call or not. Should I call?

This is a simple pot odds question. I have a 7/(52-4-4) = 7/44 = 15.9% chance of making your nut hand. This is equivalent to odds of 5.29 to 1. Does this make putting $12 into a $174 pot a good call? Well, it's an easy call, assuming we're dealing with independent events.

So, is each hand an independent event? If you're playing poker in a public casino or cardroom, certainly. In the above example, out of every 1000 hands you will win, on average, 159 hands and lose 841. Does that mean if we deal this hand 1000 times that will happen? No, it means that 159 wins is the expected value.

I'm sure you've all seen pictures of a bell curve. Let's assume the following experiment. I will deal this end-game situation 1000 times, and record the number of wins for my hand. I will then repeat this experiment an infinite number of times. The plot would be that of a bell curve (or a normal distribution), with the peak centered at 159. However, there will be cases where there will be fewer than 100 winning hands and cases with more than 220. That's the impact of randomness.

Now let's examine how randomness impacts an end-game situation in poker. On the hand in question most of the time (84.1%) I will lose. However, the amount of money in the pot compensates me for that loss. We look at that by determining the expected value (EV) of the result:

EV = [0.841 * (-$12)] + [0.159 * ($174 + x * $12 * n)], where x = the percent chance of your opponents calling you on the river with the winning hand and n = the number of opponents calling you on the river with the winning hand.

A reasonable assumption is that one opponent will call you on the river no matter what; solving for x=1 and n=1,

EV = [0.841 * (-$12)] + [0.159 * ($174 + x * $12 * n)]
= [-$10.092] + [0.159 * ($174+ 1*12*1)
= [-$10.092] + [0.159 * $186] = -$10.092 + $29.574 = $19.482

By calling on the turn you will, on average, net $19.48.

Of course, if you're a pessimist you remember the times you lose with your draws; if you're an optimist, you remember your wins. I try to think of myself as a pragmatist, and realize that as long as I've got the odds right, I'll win more times than I lose. I just hope I live long enough to experience an infinite number of hands.

Next month we'll take a look at a factor that is guaranteed to increase your wins (if you don't know it already) at Omaha.

© The Poker Forum.com, all rights reserved