### An Ugly Example of Mathematical Variance

Yes, Poker.com has implemented a major new software upgrade, enhancing in particular their multi-table functionality. I've hit three other sites this week that have also done major upgrades, so it's obviously Summer Upgrade time in the online world.

Ahem. Dear Poker.com, now that you've upgraded your software, could you give the ol' random-number generator a swift kick in the capacitor? God, but it's been

*ugly*there in recent days. Fresh off winning four hands out of 101 in a ring game where the flop-seen percentage for the table was in the 40's, I followed that up earlier today with this gem of a session:

That's how you drop 30 big bets in a big hurry. What a shame to be so totally card-dead in a game that had a 51% flops-seen percentage when I first sat down. (Of course, I drag that down by about 1% just by taking a seat, but that's another matter.)

But rather than go off on some ridiculous and untrue "Online Poker is Rigged" rant, let's use the ugliness to illustrate the statistical probabilities inherent in such a bad variance run. I can't count the number of times I've encountered a post or a comment by someone who has a run like this, then says something inane like, "The odds against this are like 10 million to one!! It's fixed!!"

Well, no, it's not. And the odds aren't that hard to calculate, either. Winning

*x*number of hands (such as 4 of 101) goes into an area of probability called combinations and permutations, nerdy stuff that was among my favorite subjects, and more than I feel like writing right now. But an oh-fer run like today's doesn't even require math that complex.

In posting an oh-fer, one loses every hand. And to calculate the odds for a long winless run, all we need to do is take the probability of losing a single hand and raise it to a certain power, where that power equals the number of hands played.

Fortunately, I know my game, and I can tell you that in a full ring game I'll win about 9% of all hands dealt. I don't win 10% because my game is naturally tighter than most of the opponents at the stakes that I play; I'm participating in several percent less hands than the average, but while I normally win a greater percentage of the hands I do enter, it still doesn't balance out when I look at the count of overall hands. Nor should it. I do loosen up my game as the tables themselves get looser and looser, but I still maintain tightness relative to the table.

If my odds of winning a given hand are about 9%, then the odds of me losing that same hand are the remainder, or 91%. And the odds of me losing 72 consecutive hands is then an easy calculation: it's that 91% raised to the seventy-second power, or 0.91 ^ 72.

What do you think the odds are of an 0-for-72 run for a middlin' tight player? One in a million? One in a hundred thousand?

Way too high. The calculation shows that 0.91 ^ 72 = 0.0011, or 00.11%. That's a hair less than 1,000 to 1. Ugly, really ugly, but far from a statistical impossibility. It's in the same range as your opponent catching the perfect two cards on both the turn and the river to overcome your all-but-insurmountable lead after the flop. If my odds of winning at this particular table had dropped down, to say, 8.5% on a given hand, then the overall odds on such a run would drop to .00167, or about 600 to 1.

One of the hidden benefits of a quickie statistical analysis is that it allows a better understanding of some of the short-run variations that occur. Just to cite one example, it's easy to see that if I'm winning hands at a 9% rate, then I should win roughly 1 out of every 11 hands I play. So how often will I sit down at a table and lose my first 11 hands?

Easy math again, using the same principle as before: 0.91 ^ 11 (the number of consecutive hands played) = 0.3543. I'll lose my first 11 hands at a table a hair over 35% of the time, a bit more than one time in three. It doesn't feel right, but it's what occurs. The act of winning hands tends to cluster to a greater degree than we realize.

That's all for this Poker Math 102 lesson. Now, to wait out that variance...

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