In order to do that, we can cast the computation above in a slightly different light. Instead of thinking about it as summing all deltas and dividing by total hand count, we can compute first the big blinds earned per hand in each session:

We are looking for the average bb/hand across all sessions. It can be tempting to take the average of the bb/hand values for each session, arriving at 1.1 bb/hand. We know already from above that this is wrong – the true figure should be 0.25 bb/hand.

One does not simply take the average of averages and expect a meaningful result3³ Well, one does that only if the groups that are averaged are of approximately equal size. If we had played roughly the same number of hands in each session, then we could have used the plain average of averages and it would be about the same as the total average.. We have to perform a weighted average. The intuition behind this is that the first session has contributed 51 hands at +2.96, while the third session has contributed just 10 hands at +38.13, so we want the first session to be more influential in our average than the third – because it was in the real data!

Thus, the weights in the weighted mean will be given by what proportion of hands were played in each session:

\[w_i = \frac{h_i}{\sum h_i}\]

So for the first session, the weight would be 51/145=35 %, whereas the third session carries a weight of only 10/145=7 %.

The easiest way to perform the weighted mean is to multiply each bb-per-hand value with its weight, and then sum it all up:

\[\bar{x} = \sum \left( w_i x_i \right)\]

When we run the numbers, this results in a value of – you guessed it – 0.25 bb/hand.4⁴ Proving that this indeed gives the exact same result can be a fun exercise in symbolic manipulation, but excessive for this article.

Now that we are looking at this as a weighted mean, it becomes more obvious how to compute its variance. When we substitute in the expression for the weights5⁵ And apply a notational shorthand where each un-indexed variable is implicitly indexed by its nearest outer sum., we have used the following equation to compute our average bb/hand across all sessions:

\[\bar{x} = \sum \left( \frac{h}{\sum h} x \right)\]

Given this expression,

\[\mathrm{Var}[\bar{x}] = \mathrm{Var}\left[ \sum \left( \frac{h}{\sum h} x \right) \right]\]

At this point we can get to work exploiting the variance laws, which tell us that

• \(\mathrm{Var}(x + y) = \mathrm{Var}(x) + \mathrm{Var}(y)\), and
• \(\mathrm{Var}(kx) = k^2 \mathrm{Var}(x)\)

First, we can move the variance inside the sum, thanks to the first law.

\[\mathrm{Var}[\bar{x}] = \sum \left( \mathrm{Var}\left[ \frac{h}{\sum h} x \right] \right) \]

Then since the denominator \(\sum h\) will be the same for all sessions (it is the total number of hands across all sessions, after all), we can think of that as a constant \(k\) and apply the second law.

\[\mathrm{Var}[\bar{x}] = \sum \left( \frac{1}{\left(\sum h\right)^2} \mathrm{Var}\left[ h x \right] \right) \]

Then we have the fact that multiplication distributes over addition, i.e.

• \(ka + kb = k(a + b)\)

which lets us simplify the variance further to

\[\mathrm{Var}[\bar{x}] = \frac{1}{\left(\sum h\right)^2} \sum \left( \mathrm{Var}\left[ h x \right] \right) \]

And this is where we get stuck. We don’t know the variance of each product \(h_i x_i\). What we can do, however, is assume that sessions and hands are somewhat independent6⁶ If basket players don’t have hot hands, then surely poker players also do not., and treat the between-hands variance as (a) stable, and (b) equal to the between-sessions variance.

If it is stable, we can replace the sum with a multiplication:

\[\mathrm{Var}[\bar{x}] = \frac{ n \mathrm{Var}\left[ h x \right] }{\left(\sum h\right)^2} \]

If the between-hands variance is equal to the between-sessions variance, then the variance of \(h_i x_i\) is the same as the variance we have observed between them for our sessions:

The variance of the last column (feel free to compute it manually or use a spreadsheet) is 16,500. Thus, to get the variance of the number of big blinds per hand, we plug in the numbers we’ve figured out in to the big expression:

\[\mathrm{Var}[\bar{x}] = \frac{ 4 \times 16500 }{145^2} = 3.14 \]

Since this is the per-hand variance, we need to multiply by \(100^2\) to get the variance of the win rate7⁷ Which is expressed as the average per 100 hands, as you may recall.. Then we take the square root of that and we get the standard error of the win rate.

\[\sqrt{3.14 × 100^2} = 178\]

Thus, while the estimation of our win rate was a mighty impressive 25, we see now that the standard error is 180. A very rough 90 % credible interval for our win rate would span

\[25 \pm 1.645 × 180\]

i.e. from −270 to +220.

Put differently, there’s still something like a 44 % chance that the true win rate is less than zero, and we’ve just been lucky in the sessions we’ve had so far.