Approximating 2-Street Geometric Bet Sizing

Author: Sigma (Twitter: @sigm_4)

At a recent study group session (link), I heard the following claim about calculating 2-street geometric bet sizes for the first time:

You can (approximately) calculate the 2-street geometric bet size by adding 1 to the SPR and dividing by 5.

In other words, with the bet size as $r$ and SPR as $S$,

$$ \begin{align*} r\simeq \frac{1}{5}(S+1) \end{align*} $$

holds true.

For example, when $S=3$, we get $r=\frac{4}{5}$, meaning you should bet 80% pot on both the turn and river (the exact 2-street geometric size is 82.2%).
Below, we will briefly verify the validity of this statement.

When betting $r$ times the pot size on both the turn and river, if we set the pot on the turn to $1$, the river pot becomes $1+2r$, and the river bet amount is $r(1+2r)$.

The total bet amount across 2 streets must equal $S$, so

$$ \begin{align*} r+r(1+2r)=S \end{align*} $$

must hold.
Solving this quadratic equation for $r$,

$$ \begin{align*} r=\frac{-1+\sqrt{1+2S}}{2}=:r_{\mathrm{2e}} \end{align*} $$

Now, assuming SPR is near 4, let $S=4(1+\delta)$ where $\delta \ll 1$. The expression above becomes:

$$ \begin{align*} r_{\mathrm{2e}}&=\frac{-1+\sqrt{9+8\delta}}{2}=\frac{-1+3\sqrt{1+\frac{8}{9}\delta}}{2}\simeq \frac{-1+3(1+\frac{1}{2}\cdot\frac{8}{9}\delta)}{2} \\ &=1+\frac{2}{3}\delta \end{align*} $$

This gives us the approximation. Then,

$$ \begin{align*} \frac{1}{5}(S+1)=1+\frac{4}{5}\delta\simeq r_{\mathrm{2e}}+\frac{2}{15}\delta \end{align*} $$

and since

$$ \begin{align*} \frac{\frac{2}{15}\delta}{r_{\mathrm{2e}}} \simeq \frac{2}{15}\delta \ll 1 \end{align*} $$

as long as $\delta \ll 1$, the approximation formula $r=\frac{1}{5}(S+1)$ is valid (with the $\frac{2}{15}$ factor providing roughly one additional digit of precision).

Computing for specific SPR values:

$$ \begin{array}{cccc} \text{(SPR)} & \text{(Approx. 2e value)} & \text{(Exact 2e value)} & \text{(Relative error)} \\  S & \frac{1}{5}(S+1) & r_{\mathrm{2e}} & \frac{\frac{1}{5}(S+1)-r_{\mathrm{2e}}}{r_{\mathrm{2e}}} \\ \hline 0.5 & 0.3 & 0.207 & 0.448 \\ 1 & 0.4 & 0.366 & 0.0928 \\ 1.5 & 0.5 & 0.5 & 0 \\ 2 & 0.6 & 0.618 & -0.0291 \\ 2.5 & 0.7 & 0.724 & -0.0341 \\ 3 & 0.8 & 0.822 & -0.0277 \\ 3.5 & 0.9 & 0.914 & -0.0155 \\ 4 &  & 1 & 0 \\ 4.5 & 1.1 & 1.08 & 0.0174 \\ 5 & 1.2 & 1.15 & 0.0359 \\ \end{array} $$

As the table shows, the equation $\frac{1}{5}(S+1)=r_{\mathrm{2e}}$ holds exactly not only at $S=4$ but also at $S=\frac{3}{2}$. This means the approximation formula works well across a wide range of SPRs from $\frac{3}{2}$ to around $4$.

(The fact that the formula is exact for only these two SPR values, $S=\frac{3}{2}$ and $S=4$, can be readily verified by solving $\frac{1}{5}(S+1)=r_{\mathrm{2e}}$ for $S$.)

As many of you may have already noticed, the approximation $\frac{1}{5}(S+1)$ is simply a linear interpolation of $r_{\mathrm{2e}}(S)$, viewed as a function of $S$, between the values at $S=\frac{3}{2}$ and $S=4$.
Therefore, by performing a similar interpolation between two different points, say $S=4$ and $S=12$, we obtain an alternative approximation:

$$ \begin{align*} \frac{1}{8}(S+4) \end{align*} $$

This formula is effective for SPRs in the higher range of roughly $4$ to $12$. However, as stacks get deeper, the rate of change in 2-street geometric size decreases. Given the diminishing returns and the increasing computational effort, it may be more efficient to simply memorize the exact 2-street geometric sizes for a few key SPR values.