# Continuous Blackjack (ii): deriving basic equations from two other perspectives.

March 24, 2022 · 3 mins · 616 words

In the last post, we derived some basic equations of the continuous blackjack game. In this post, we will show other ways - maybe more intuitive - to derive the same results.

# Maximizing winning probability

The first method to get the optimal threshold is the most direct one: given the probability of winning $p(\alpha)$ find its maximum. This is a problem that any high schooler is used to solving, so we believe that this approach can be more intuitive than the previous one. Now the problem comes to determine $p(\alpha)$. Let’s define it first using words, and after we will translate it to maths.

\[p(\alpha) = p_1(\alpha) \times p_2(\alpha)\]where

- $p_1(\alpha)$ is the probability of 1st player landing $(\alpha, 1)$.
- $p_2(\alpha)$ is the probability of 2nd player going bust while trying to reach $s$, given that 1st player has a score $s$, summing over all the possible $s$.

The value of $p_1(\alpha)$ is simply $F(0; \alpha, 1)$ - as we have derived in our first post. And $p_2(\alpha)$ is

\[p_2(\alpha) = \int_{\alpha}^{1} \frac{1}{1 - \alpha} (1 - F(0; s, 1)) ds\]where $\frac{1}{1 - \alpha}$ is the probability of 1st player landing in $s$. And $(1 - F(0; s, 1))$ is the probability of the 2nd player going bust given that the 1st player is in $s$. For $n+1$ players we just need to compute the probability of all the $n$ players going bust, which is $(1 - F(0; s, 1))^n$

Finally, we get

\[p_n(\alpha) = F(0; \alpha, 1) \int_\alpha^1 \frac{1}{1-\alpha} [1 - F(0; s, 1)]^n ds\]And now $\alpha^*$ is given by

\[\frac{d p_n(\alpha)}{d \alpha} \bigg|_{\alpha=\alpha^*} = 0\]## Solving the equation

Let’s now derive the optimal threshold by differentiating and equaling to zero. First of all, using $F(0; \alpha, 1) = (1 - \alpha)e^\alpha$ we can simplify $p_n$ to

\[p_n = e^\alpha \int_\alpha^1 (1 - F(0; s, 1))^n ds\]Now using Leibniz integral rule we have

\[\frac{d p_n(\alpha)}{d \alpha} = e^\alpha \left(\int_\alpha^1 (1-F(0;\alpha,1))^nds - (1-F(0;\alpha,1))^n\right)\]Hence, the optimal value is given by

\[(1-F(0;\alpha,1))^n = \int_\alpha^1 (1-F(0;s,1))^n ds\]which is the same equation we derived in our last post.

# Maximizing expected payoff

Another way to derive the equation for the best strategy is to maximize the expected payoff. In general, the expected value of a random variable $X$ with a pdf $f(x)$ is $\left< x\right> = \int_{-\infty}^{\infty} xf(x) dx$. In our case $x = (1 - F(0; x, 1))$ is the probability of the other player going bust. On the other hand, density $f$ can be obtained using

\[F(0; x, y) := P(S(x) \leq y) = \int_0^y f(S(x))dx\]where $S(x)$ is the sum obtained for a threshold $x$. Therefore,

\[f(S(x)) =\frac{\partial F(0; x, y)}{\partial y}\]and finally, our expected outcome is

\[\left< p_{win}(\alpha) \right> = \int_\alpha^1 (1 - F(0; s, 1)) \frac{\partial F(0; \alpha, y)}{\partial y} ds\]In the particular case of $P=U[0, 1]$ we know that $F(0; x, y) = (y - x)e^x$. Thence, the expected outcome is

\[\left< p_{win}(\alpha) \right> = e^\alpha\int_\alpha^1 (1 - F(0; s, 1)) ds\]which is the same equation as the one derived in the last section.

# Conclusions

In this post, we have derived two new ways to determine the optimal threshold for the continuous blackjack game. This may seem like a waste of effort since similar equations were derived in the last post. However, we believe that knowing how to derivate the same result through different methods is always a good investment of time. If you don’t agree with this last assertion we recommend you to read this blog post.

If you have some other way to derive the same results feel free to add them in the comments!