Thanks - so it sounds like comparing the variance of one player vs another could be somewhat problematic unless you used players with similar winrates over the sample. Sound right?

Variance is a measure of data spread, both statistically and in everyday use. If you are comparing the variance of two players, their win rates may vary but their variances may be similar or vice versa. So, I would think the answer to your question, is another question – why are you making the comparison?

I think a more useful analysis is to compare your variance for different playing conditions, such as NLHE vs. Omaha or Live vs. On-line. You can use the variance measure to estimate confidence intervals and risk of ruin (ROR), so if going poker bankrupt is of concern you would use a site like PrimeDope, which has variance as a factor to see which game is better for you.

Not to get too technical, but here is a formula you can use to estimate risk of ruin:

ROR = exp(-2*B*W/S^2),

where B is your bankroll, W is your win rate and S is your standard deviation (Variance, V =S^2)

Example. You have a bankroll of 500 bb (B), a win rate of 6bb/hr (W) and a standard deviation of 70bb/hr (V=4900). Then

ROR= exp(-2*500*6)/4900) = 29%.

Risk of ruin reduces to 9% with a 1000 bb bankroll and to 1% with a 2000 bb bankroll.

There are more exact formulas which consider the number of hands or tournaments played.

Sure it does. As a simple example, consider a hand where you shove for $100 with 22 and get called by a player with AKo. You win 52% of the time, so your hand EV is 100 x 0.52 - 100 x 0.48 = $4. Variance for that hand is
(100-4)^2 x 0.52 + (-100-4)^2 x 0.48 = 9984. Standard deviation is the square root of this or $99.92.

Not all situations lend themselves to such easy calculation, but there certainly is a theoretical EV for each hand (the whole goal of all of us here is to maximize that EV - an impossible goal if it didn’t exist), and therefore a variance (or standard deviation) for each hand.

It is true that the variance or standard deviation over multiple hands is more important. This just is the result of the variances of each individual hand, though, just as your win rate is the result of the win rate on each individual hand. Your win rate, long term, is the sum of the EV of each hand you play (not the optimal EV, but the EV as you and your opponents played it). The variance of a multi-hand session is simply the sum of the variances of the individual hands. Standard deviation of multiple hands is only a bit more complex — it’s the square root of the sum of the squares of the individual hand sd’s.

If we take variance to be the standard deviation from your true win rate then speed of play makes a difference for sure. More hands you play the more likely you are to reach your true win rate.

Are you suggesting that you could realistically calculate variance on a hand by hand basis (for all hands, not just the ones where there’s all in prior to the river)? If so, how would you go about doing it? Also, why would you want to perform this calculation at such a low level? What questions can be answered by having this level of information?

Of course not. It’s not realistic to calculate your EV exactly while playing. You’d have to know your opponent’s hand to do so. The best you can do is to estimate your EV in a given hand. Since you need to know your EV to calculate variance, you cannot calculate variance while playing. The real point is that if you want to reduce variance (I really don’t think that’s always the best idea) you need to reduce the variance of each hand. You may not be able to calculate it exactly, but you can figure out what option minimizes it in a given hand. Drawing increases variance, for example. If you have a zero EV draw, the fewer outs, the greater the variance will be. Same is true if your draw is +EV. This is an example where minimizing variance is not optimal. If you have a flush draw and opponent bets 1/4 pot, you are getting 5:1 odds — you should call. Minimizing variance would call for a fold, though.

“Since you need to know your EV to calculate variance”

Is this true?

Seems to me some are using win/loss over several sessions and comparing that to the mean, while you’re focused on a hand by hand EV level. This is why I was wondering exactly what is being measured since it seems that people are considering different inputs.

EV has nothing to do with variance as it'll always be somewhat subjective (we need to know villain's exact range in each spot, a solver's EV makes all types of assumptions as well).

In statistics variance is expressed in stdv, and then there are confidence intervals etc, so the guys in this thread are right but they're just being particular for arguments sake and not to answer your question...

If we classify variance as the thing we are actually interested in as poker players.... ie, the likelihood to swing in any certain direction, duration of downswings etc then variance is determined by standard deviation and actually more so, your winrate, which affects your confidence interval.

Standard deviation is determined by the size of the pots you play for (the bigger the pots, the bigger the possible swings) winrate is determined by skill of course.

Players with a 10bb/100 WR and 120 stdv will still experience less variance than players with a 1bb/100 winrate and 75 stdv. The latter can actually lose over a million hands due to variance.

Try this to test https://www.primedope.com/poker-variance-calculator/

No, you do need EV to calculate variance, whether it’s over multiple hands or a single hand. Over multiple hands, variance is a measure of the divergence of your actual results from your true win rate. But what is your true win rate? That’s just the sum of the EVs of each individual hand you play, so EV is still needed to determine variance.

And it’s not true that the player with the 10BB/100 win rate and 120 sd has less variance than the 1BB/100, 75 stdv player. The 1BB/100 player is certainly more likely to suffer downswings, but the 10 bb/100 player will see more and larger divergences from his 10BB/100 win rate — that’s what the higher variance means. Variance is NOT a measure of how likely you are to experience a negative result. It’s simply a measure of the difference between actual outcomes and the expected value of that outcome.

Minimizing variance seems to be something many regard as a positive thing, but that is really not the case. In your example, which stats would you rather have, the 10BB/100 with higher variance or the 1BB/100 with lower variance? The answer should be obvious. How do you get the higher win rate? Obviously by making more +EV plays in hands, regardless of whether they increase variance or not.

If you REALLY want to minimize variance, that’s easy — just fold every hand. The variance of folding is zero; there’s no possibility of different outcomes if you fold PF. Your EV is either 0, -0.5, or -1 bb (depending on your position), and you will realize that outcome every time. Obviously that’s not a profitable way to play poker, though.

I think a source of confusion is the EV concept. It stands for expected value given the current hand situation, not actual profit/loss that eventually occurs. An expected value is essentially the average that will occur after a large sample under constant conditions. If, after a play, you want to say that EV is the actual profit/loss, that would be a divergence IMO.

In any case, statistical variance in a sample of hands is estimated as the sum of the squares of the actual deviations of hand profit/loss from the mean profit/loss, with the latter usually expressed as win-rate.

Example: Player P/L: H1. 4, H2. 1, H3. 3, H4. -3, H5. 0

Win Rate (mean P/L) = (4+1+3-3+0)/5 = 5/5=1.

Variance [(4-1)^2 +(1-1)^2 +(3-1)^2+(-3-1)^2+(0-1)^2]/4

=(9+0+4+16+1)/4 = 30/4 = 7.5 (the division by 4 rather than 5 is to correct for a small sample

I should have put "variance" in the last paragraph in quotes for argument's sake.

My point was that it's the "variance" worth knowing because usually when people ask "what is my expected variance", what they really mean is: "how likely am I to experience downswings and how long can they take" and win rate plays the biggest role in this...

EV is what your true win rate would be if we were able to know everything in every hand (exact villain's range etc). But vs a pool and every hand being a completely different situation, it's about as hard to know as the distance between two specks of dust on stars living in different galaxies....

A live game worth playing at 5/10 or smaller will have mostly multiway pots. (note the "worth playing" qualifier.)

If you're routinely seeing flops with 3 or 4 other people, even if your ranges are stronger than theirs, you mostly still have to make some sort of hand to win the pot.

My guess would be that this is the source of most of the additional variance, more than the factors given in the previous posts, which apply equally to headsup pots.

That’s certainly true. As a model, assume no rake and equal overall skill levels. Then your EV overall will be zero. For a two player game under such conditions, you will win on average with probability 1/2 and (obviously) lose with the same probability. If you bet $x on a hand, your variance is 1/2 x^2 + 1/2 x^2 = x^2.

For three players you win with probability 1/3, lose with probability 2/3, win $2x when you win, and lose $x when you lose. Since EV is zero, the variance is 4x^2/3 + 2x^2/3 = 2x^2, twice as high as the heads up game.

In general, with n players you will win $(n-1)x with probability 1/n and lose $x with probability (n-1)/n. Variance is (n-1)^2 * x^2 * 1/n + x^2 * (n-1)/n. This simplifies to (n-1) * x^2, which increases as n increases.

Defining variance is very important here. I think in order of most likely to have a losing month (playing ~100 hours), it would go:

1. High stakes live player
2. High stakes online player
3. Low stakes online player
4. Low stakes live player