The purpose of this formula is to help me to choose the optimal amount to deposit in a room based on the average variance I used to face over all the sessions I've played.

By optimal amount, I mean the equilibrium amount between minimum deposit and maximum deposit.

I play in different rooms and calculate my post-session balance in a excel spreadsheet. It allows me to have these elements:

- Numbered session and date
- Post-session balance for all the rooms
- Total session balance
- % winning session

The idea is to calculate the "average variance" based on the average variance of each rooms I play.

How can I calculate it? Also, do you think it can be accurated (based on a decent sample)?

Thanks

I'm not sure what your question is. The variance of a sample is an "average" in the general English sense of representing all the data in the sample.

If you have different samples and know their individual means and variances, there is a formula to compute the overall variance. If you have k samples with size N_k, mean M_k and variance V_k, the overall variance is:

Sum of (V_k + M_k^2)*N_k/(Sum of N_k) - Sum of (M_k*N_k)^2/(Sum N_k)^2

If you have different samples and know the reason for the differences in variance, for example if some of the samples are at higher limits, then there are other ways to adjust.

Thanks a lot for your input but I have difficulties to understand the formula.
Could you please illustrate your formula by the example I'm going to write below?

Clearly, in my example the result can't be accurate in theory because of the small sample. It is only to get the general idea.

Let's say a player has played 10 online sessions for his whole poker career.
His sessions are played in 3 different rooms: Stars, FTP and Party.
Sometimes, he can play sessions in one room only. Also, we assume that the player doesn't play the same volume in each room. (For example, in one session, a player can play: 20 games in stars, 10 games in FTP and 6 games in party.)

So here if: a player has a bankroll of $2.5k. His main game is MTT and his average BI is $20.
The 20/11/14 he deposited on Stars, FTP and Party respectively: $1000, $400 and $400. Here are the evolution of his 3 bankrolls:

#1: 20/11/14: $1000 / $400 / $400 (Balance)
#2: 24/11/14: $1200 / $350 / $450 (+$200)
#3: 26/11/14: $800 / $350 / $450 (-$400)
#4: 27/11/14: $890 / $280 / $600 (+$170)
#5: 28/11/14: $815 / $240 / $780 (+$65)
#6: 29/11/14: $780 / $290 / $700 (-$65)
#7: 30/11/14: $1800 / $210 / $650 (+$890)
#8: 05/12/14: $1888 / $150 / $400 (-$222)
#9: 06/12/14: $1680 / $100 / $350 (-$308)
#10: 08/12/14: $1690 / $232 / $380 (+$172)

As you can see at #3 the FTP and Party numbers are the same than the previous session. It's just for show you that the player can play sometimes sessions in one single room.

Two questions:
1) Given the variance exposure of the player "over all his poker career", how can we calculate the minimum deposit amount he need to not having to deposit again (or not before a long time).

2) Since we've seen that the player...
- doesn't play the same volume in each room
- doesn't "multi-rooming" necessary every session
- doesn't play MTT 100% of the time (can mix up a bit with cash-game for example)

...does a decent sample size can make this formula accurate?

Many thanks for the help

Variance is not averaged for multiple sessions, it is cumulative. It's just a larger sample than 1 session. And by definition a larger sample is expected to have larger variance.

^ a larger sample is not expected to have larger variance by definition. a larger sample gives you a variance that is more likely to be close to the "true" variance compared to a smaller sample.

If we set up bankroll management rules like for example 30BI for cash-game: we suppose that it's the sufficiant number of BI for a competent player to handle downswing and to generate a profit in the long term. Generally speaking, the bankroll management rules are based on a variance estimation right?
So why can't we calculate the average downswing of a player in the long run?

Sorry, I should have said standard deviation (sqrt of variance) grows with sample size.

What they are suggesting is to treat each site as its own sample and use the variance at each to make your decision or treat all of your data as one large sample and use the variance from the larger sample to make your decision.

What they are saying you can't do is take the variance of each site and then try to 'average' it.

why do you think this? or i don't know what you are talking about

It isn't something I think. As sample size grows X, standard deviation grows by multiple sqrtX. This is by definition.

Just to elaborate above. The standard deviation of a single coinflip is 1/2. The standard deviation for n flips is sqrt(n)*1/2. The sd grows with the square root of the number of flips.

e.g.:
1) Of 100 flips we expect to win 50 with a sd of 5.
2) Of 10000 flips we expect to win 5000 with a sd of 50.

The sd increases in absolute terms, but decreases relative to the total flip count.

I haven't read through the thread yet but I'll lay odds that there's going to be confusion over how people are defining variance.

There are two definitions for variance which are both commonly used in poker. One is more like a statistical definition while the other is a probability definition.

The statistical definition involves an average of past results such as Aaron Brown showed.

The probability definition is simply put the size of the swings you can expect to have but there's no single formula for this because it depends too much on what your analyzing and also it would be very difficult to solve for in many cases such as with tournaments.

The first is useful for determining how lucky you are running as compared to expected results.

The second is to show how much you can be effected by luck and is more related to bankroll management.

Back to the OP's question.

I'm assuming you're talking about variance in cash games which simplifies the problem somewhat, although it's still not as simple as you might think.

Also, while analyzing past results of cash games suggests like you are trying to find the statistical variance involved, it seems to me that you're probably looking for an answer for determining if you are playing within your bankroll which would be the probability variance, however you can probably apply standard deviation that's related to the statistical definition in this case.

Even so, it would probably take a lot of data to solve for the probability definition to get a reliable number.

Perhaps a better way to think about it is how much action you will tend to be giving in each of those rooms. As the action increases the wider your swings will be.

It's my fault, not yours. As TakenItEasy figured out, I was answering a completely different question than you meant; and most of the subsequent responses were really discussing my points, not your question.

There is not an easy answer. You have three small samples. When you have limited data, it's important to use every scrap of it. Unfortunately the data are not directly comparable. At each site, you have a different variance which could be due to stake level, games played, amount of play, style, rake or other factors.

If your distribution of returns had the same shape at each site, you could multiply each result by a constant for each site, compute the overall variance, and then divide back to get the site results. For example, it looks as if you could multiply Stars by 1, FTP by 6 and Party by 3.

However, the distributions do not have the same shape at each site. It could just be an illusion from the small sample, but Stars has a big win with lots of smaller losses (positive skew and leptokurtotic in technical lingo) while Party is the opposite with roughly equal-sized moves, the biggest of which is a loss (negative skew and platykurtic).

There are only two ways to answer your question. The simplest and most reliable is to get bigger samples and treat each site independently. The other way is to model your returns, to figure out why they are different.

Thanks, not sure if I really deserve the credit for this though. I only recall that another thread involving variance produced a pretty big debate which pretty much divided down the middle and mainly just depended on the persons background.

Since I was heavily involved in MTTs when I first joined 2+2, the idea of variance relating to the swings was all I understood at the time because of the insane amount of swings that tournaments produce.

For cash game players who tracked their stats, then variance became about analyzing results vs EV, if I understand it correctly. I only played live cash games and MTTs online so that could still be an inaccurate statement.

The funny thing was that once I realized it was two different ideas being debated, after doing searches for definitions, just about every poker source defined it either one way or the other as if that was the only way to look at it. Only Wikipedia gave both definitions.

Getting back to standard deviation and sample size. Since the standard deviation of a variable is the square root of (statistical) variance, and since variance has more 'accessible' math, let's deal with that first.

There are two basic statistical measures for which variance is often calculated - the sample sum or total and the sample mean. In general the variance of a sample sum of independent random variables (e.g., total profit over n hands) is the sum of the variances so as sample size increases, variance increases and therefore so does the standard deviation.

For the sample mean or average, however, the variance of the mean is the variance of the population divided by the sample size. Therefore variance of the sample mean decreases with sample size and so does the standard deviation, usually called standard error of the mean