Question on variance of game and bet size

1) Bet size is given by the equation Ev/var x Current Bankroll.
So, if I have an advantage of 2% during a particular round and I have $5000,I should bet $80 (0.02 / 1.25 x 5000). Assuming 1.25 is the variance of the game. Is this correct?
2) How is the variance of computed?

a) Based on variance of player advantage expressed as a percentage

For example, for simplicity let us take 4 samples. Player advantages are 2%, 1%, -1%, -2%. Then variance of this will be 3.33.

b) Based on variance of player advantage expressed as a fraction
This would be the variance of 0.02, 0.01, -0.01 and 0.02. This would be 0.000333.

3) If a game has 2% advantage off the top and only one round is dealt before shuffle then variance is 0. How does the equation work then?

1) Bet size is given by the equation Ev/var x Current Bankroll.
So, if I have an advantage of 2% during a particular round and I have $5000,I should bet $80 (0.02 / 1.25 x 5000). Assuming 1.25 is the variance of the game. Is this correct?

Yes, this is the optimal bet size, meaning that if consistently bet this way, you will maximize your bankroll growth. A few notes: what I said implies that you recompute this value on each bet, the bet size might not be an even chip size, and if you don't resize your bets, you have a 13.5% risk of ruin.

) How is the variance of computed?

a) Based on variance of player advantage expressed as a percentage

For example, for simplicity let us take 4 samples. Player advantages are 2%, 1%, -1%, -2%. Then variance of this will be 3.33.

b) Based on variance of player advantage expressed as a fraction
This would be the variance of 0.02, 0.01, -0.01 and 0.02. This would be 0.000333.

3) If a game has 2% advantage off the top and only one round is dealt before shuffle then variance is 0. How does the equation work then?

Variance is a mathematical term used in probability and statistics. It is the square of the standard deviation. What it is is a measure of how much the actual result varies from the expected result (in other words, luck).

Suppose you have 1% advantage on your next bet. Your expected value = (bet)(1%), but the actual result will be (-1)(bet) if you lose, (+1)(bet) if you win, and (0)(bet) if you push. Also, you could double, split and resplit, get a blackjack, or surrender, giving you a result anywhere from (-8)(bet) to (+8)(bet). To calculate the variance: p1(result1-1%)^2+p2(result2-1)^2+p3(result3-1)^2 + ..., where p1, p2, p3,... are the probabilities of getting the result.

Say we have a simple game (not actual bj values) where the p(lose)=48%, p(bj)=10%, p(win,nobj)=36%), and p(push)=6%. Your variance would equal=(48%)(-1-1%)+(10%)(1.5-1%)^2+(36%)(1-1%)^2+(6%)(0-1%)^2= 1.07.

Thanks for the example. Based on what you are saying, the variance changes from round to round. It is not a constant for a given set of game rules.
Also, in your example, you use the expected earnings as percentage. But in Wong's example in professional blackjack, chapter 14, "Finer Points", in section Risk, he calculates it based on ratio (i.e 0.01 instead of 1%).

Thanks for the example. Based on what you are saying, the variance changes from round to round. It is not a constant for a given set of game rules.

It does vary from round to round, but very little. You will see this if you check out QFIT's calculator on http://www.qfit.com/cvcxonlineviewer.htm, For instance, 6 Deck, S17, DAS,LS, 78 penetration, the SD (square root of variance) is the following per TC:

<=-1 1.142
0 1.134
1 1.133
2 1.133
3 1.135
4 1.140
5 1.145
6 1.145
...

Also, in your example, you use the expected earnings as percentage. But in Wong's example in professional blackjack, chapter 14, "Finer Points", in section Risk, he calculates it based on ratio (i.e 0.01 instead of 1%).

I've not read "Finer Points," but the two are equivalent, i.e., a advantage of 1%, means your expected value = (0.01)(amount bet). For most it's easier to understand 1% than 0.01.