Why are shorthanded games swingier than full table?

[Eihli]

I thought more hands = less variance.

[squiffy]

I guess, you are forced to play a lot more hands to stay even. The more hands you play, the more money you risk -- either win or lose.

Because you have to pay more blinds per hour, twice every 6 hands instead of twice every 8 hands, you are losing more money per hour. You have to win more to make this up.

Since there are only 6 players at the table, when you do win a pot, you don't necessarily win as much. So you can't just sit back with a flush or straight draw and let the fishies pay you off. You have to go in with high cards and slug it out with more marginal hands. If you are a newbie, you are going to make more mistakes.

It's a faster-moving, faster-paced game, I guess with more bluffing and aggression.

For some reason, you cannot afford to just sit back and wait for the nuts, I suppose because they blinds will be eating you alive. And talented players can just fold on those occasions when you do have the nuts.

Just a guess. Based on what little I have seen or read.

[Ulysses]

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I thought more hands = less variance.

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They are way more aggressive, so you'll often find yourself putting in many more bets w/ small edges than you might in a full game.

You also rarely get the huge overlays you sometimes get in full games. For example, you might flop the nut flush in a full game against 4 opponents where 3 of them are drawing dead. If one has a smaller flush and another has a set, you'll probably cap it on the flop and turn 5 ways with only 1 guy drawing live against you on the turn.

Those are two factor that really increase the swings.

[Nate tha' Great]

In addition to the reasons cited by squiffy and Ulysses, another consideration is that the play at shorthanded games is more opponent-dependent, so sitting down at a particularly fishy or tough table can have a big impact on your bankroll.

[nykenny]

no. more information = less variance

the more acurate you can make your decision the less variance i believe. if you never have any clue whether your starting hand is best, whether you are ahead on flop, turn or river, and constantly have to pay off then you will suffer high variance.

less aggression = less variance.

looser opponents = less variance.

more opponents = less variance.

heads up with aggressive opponent = HIGHEST variance

hope this makes sense.

Kenny

[Dynasty]

[ QUOTE ]
I thought more hands = less variance.

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More hands dealt will lower your variance. After an infinite # of hands, variance is 0. In short-handed games, you are often playing more marginal hands such as J9o or A7o for a raise. It may be correct to play those hands but doing so isn't going to lower your variance.

There are other factors as well.

[Nottom]

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More hands dealt will lower your variance. After an infinite # of hands, variance is 0.

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This doesn't seem right. I might not know the proper definition of variance, because it sure seems to me that more hands would increase total variance but would decrease your pre-hand variance.

[Dynasty]

Your variance after an infinite # of hands is 0 because if your expected winrate is 1.2 big bets per hour, then you will achieve exactly 1.2 big bets per hour.

Perhaps I'm stating it wrong?

My point is that after you play for 100 hours, 1,000 hours, 10,000 hours, 100,000 hours, etc., your actual winrate gets closer and closer to your expected winrate. So, variance has less of an impact on what your actual winrate is. After an infinite # of hands, it has no impact on your winrate.

[Robk]

What you say is correct with "variance" replaced by "win rate confidence". Variance is an average of the squared difference between your results and your expected results. This won't converge to zero as n -> infinity.

[Nottom]

I can see that. Maybe one of the math gurus can step up and prove something, but I would imagine it would be possible for your "win rate variance" to converge on 0, while your actual Variance continues to grow.

For example, if we could create a clone of a 1BB/hour player who played exactly the same and had them both play indefinately. Their win rates would begin to converge, but I would imagine that the difference in their total bankrolls would be quite large.

[rigoletto]

Variance is an average of the squared difference between your results and your expected results. This won't converge to zero as n -> infinity.

I'm no math guru, but at n -> infinity I would expect my results and expected results to be the same and the square of 0 is 0 (give and take). At least that's what I understood from Dynasys post.

[Robk]

OK maybe I didn't phrase that carefully enough. The key phrase was "is an average". In other words I played 4 hours. My expected result is 80 dollars, since my winrate is 20 per hour. But actually I won 200. So the difference between them is (200-80) = 120. Then I square this = 120^2. Then I average these numbers for all my sessions. Now does it make intuitive sense that this average won't converge to zero if you play infinite sessions? (Of course the tricky point which I'm entirely glossing over is how to properly weight these in the average, given that sessions are different lengths. That's why there's no easy way to calculate a poker players' variance, and you have to use an approximation like the one Mason gives in the Essays section.)

[Tommy Angelo]

Short-handed games are not inherently any swingier than full games. If the shorthanded games you play in are swingier than the full games you play in, it is completely and only because the players, maybe you included, choose to make it that way.

[J.A.Sucker]

Dynasty is exactly correct. Your AVERAGE will converge to you winrate, in actuality as your # of trials goes to infinity. This is the exact basis of uncertainty (which std deviation and/or variance characterizes for a sample of a given population). Think of it this way, as you have experienced infinite trials, you have sampled the entire universe of populations, and therefore there is no uncertainty left - you are right at the true average.

[MMMMMM]

In Stud, higher ante structures tend to produce swingier games.

Since the ante structure in Hold'em is a function of the blinds vis-a-vis the number of players, 4-handed Hold'em has a higher ante structure than 9-handed Hold'em. You are forced to get involved on marginal values more often.

Doesn't poker naturally (and correctly) get swingier the higher the ante structure?

Of course what you do can influence it too, as you point out.

[Robk]

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Your AVERAGE will converge to you winrate, in actuality as your # of trials goes to infinity.

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I agree with this 100%.

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This is the exact basis of uncertainty (which std deviation and/or variance characterizes for a sample of a given population).

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I don't know what you mean by "the exact basis of uncertainty". What I mean by uncertainty is the term Z in this explanation by Bruce:

"In particular, you will say something like "I have X% confidence that my win rate is W +/- Z". But with a low number of hours, you can make X high, but then Z will be very high, or you can make Z low, and then X will be very low. You would like to make X high and Z low, and that usually takes a lot of hours, many more than most people realize. To calculate this tradeoff, use this equation:

x*(SD/EV)/sqrt(N)*100 = Z"

Note that we agree that for any fixed X, as N -> infinity, z -> 0, which is obvious from the equation. But you seem to be saying that because Z -> 0, SD (or equivalently variance) -> 0, which is not the case.

[squiffy]

I am not sure I follow your analysis. But I will give it a stab.

The ante amount is the same in a 6 person 1-2 limit game and a 10 person 1-2 limit game. So if I am playing against 5 opponents I have one chance in six, on the average, of winning the $3 ante.

If I am playing in the 10 person game, I have one chance in ten of winning the $3 ante on average.

So I guess, I would be willing to risk only .30 in the 10 person game and perhaps .50 in the 6 person game.

Plus, in the six person game, I am losing more money to the antes every round, so I need to play more hands more frequently to make up for that increased lost. And the more hands I play the more I lose in rake.

So this increased drain means I have to play more hands, and each of these hands is more marginal and more risky than I would play in a 10 person game, so the variance or swings in my bankroll are much higher?

But I think variance in the bankroll is more than just the ante structure. But the number of players of are having to support that ante structure.

I guess I am a little confused about this. The ante structure is exactly the same in a 1-2 6 handed and a 1-2 10 handed.

So the difference in ante structure cannot account for the increased variance. It is the lower number of players who are now required to pay for and support the SAME ANTE structure.

And with a lower number of players, perhaps AGGRESSIVE play is more highly rewarded and more statistically correct in a short-handed game than in a 10 person game.

Oh, I just re-read your post. Maybe you mean that it has a higher RELATIVE ante-structure, meaning per player????

So I probably agree with you, but am not sure about the semantics.