Continuous Blackjack (i): introduction and first results.

March 11, 2022 · 6 mins · 1186 words


This is the first of several posts where we will analyze the problem of playing continuous blackjack. In particular, we want to extend the work of Henry Charlesworth and Mu Zhao to other distributions. In this first post, we will reproduce the existing results and propose a way to extend them to other distributions. Then, we will analyze the case of the uniform distribution and show some interesting plots.

Problem Statement

Consider the continuous version of the blackjack game, which has the following rules:

The question is, what’s the best strategy to win this game?


To start let us define \(F(t; a,b)\) to be the probability of landing in the interval \(R=(a, b)\), given that we are actually at \(t\) and we are going to keep playing if \(t < a\). This probability can be split into two parts, (1) the probability that we land on \(R\) in the next turn, and (2) the probability that we don’t land on \(R\) on the next turn but in one of the following ones.

In mathematical terms this is

\[F(t; a,b) = \underbrace{\int_a^b P(x-t)dx}_{(1)} + \underbrace{\int_t^a P(x-t) F(x; a,b) dx}_{(2)}\]

Uniform distribution

In the particular case of \(P=U[0, 1]\)

\[F(t; a, b) = (b - a) + \int_t^a F(x; a, b) dx\]

And differentiating with respect to \(t\) on both sides we get

\[\frac{\partial F(t; a, b)}{\partial t} = -F(t; a, b) \implies F(t; a, b) = Ke^{-t}\]

The constant \(K\) can be obtained using \(F(a; a, b) = b - a\), so \(K=(b-a)e^a\).

\[F(t; a, b) = (b-a)e^{a-t}\]

Notice how the equation doesn’t depend on the particular value of \(a\), \(b\), and \(t\) but the distances between them. Thus, one can define the value \(W = b-a\) as the width of the range, and \(D = t - a\) as the distance from the current location to the lower bound. Then, the equation can be written as

\[F(W, D) = W e^{-D}\]


In this section, we will analyze how to use the results derived in the last section to find the optimal strategy in different scenarios.

1 vs 1

Let’s start with the simplest case: you’re playing against only one other person. The first player gets a point iif the second player busts. If the first players’ score is \(s\), then the second player has a probability \(1 − F(0; s, 1)\) to bust, this means that our probability to win if we stay at \(s\) is \(1 − F(0; s, 1)\). Of course, if we could choose our \(s\) we would choose \(s=1\), but since this is a random process we can’t choose \(s\). The only thing we can choose is at which \(\alpha\) we stop drawing numbers. This \(\alpha\) is defined by the following point: where the probability of winning given that we stick at \(\alpha\) is the same as the probability that we win given that we draw one more number.

Specific case: uniform distribution

For this section let’s assume that \(P = U[0, 1]\) and that all the players start at \(x=0\). In this case, the condition described by the above reasoning is written as

\[1 - F(0; \alpha, 1) = \int_{\alpha}^1 \left( 1 - F(0; x, 1) \right) dx\]

Notice that left side is increasing, while right side is decreasing.

Simulation results

Before moving to theoretical results, let’s try to find the optimal threshold via simulations. To do so, we will try each possible threshold and simulate 50000 games. Then we will check the probability of first player winning as a function of the choosen threshold. The results are plotted in Fig. 1:

Fig. 1 - Probability of first player winning vs stopping threshold.

As you can see, the maximum winning probability is achieved around \(t^* =0.6\), more specifically is \(t^*=(0.565 \pm 0.035)\) 1. Knowing the range where the optimal value lies, let’s repeat the above simulations but only with thresholds between \(0.5\) and \(0.65\). Using this smaller range we get the optimal threshold \(t^* = (0.568 \pm 0.021)\). Finally, the expected winning probability at the optimal threshold is \(p_{win}(t^*) = (0.4278 \pm 0.0011)\). Let’s see now if the analytical results are compatible with the simulations.

Fig. 2 - Probability of first player winning vs stopping threshold.

Analytical results

Using the results from section Uniform Distribution and \(P=U[0, 1]\) , the equation for the optimal \(\alpha\) is

\[1 - (1-\alpha)e^\alpha = \int_\alpha^1 \left( 1 - (1 - y) e^y dy \right) \implies 1 - (1-\alpha)e^\alpha = (1-\alpha)-e^\alpha(\alpha-2)-e\]
Fig. 3 - Plot of the curves that define the optimal threshold.

This is a non-linear equation without a closed form, but it can be solved approximately using numerical methods, and the solution is \(\alpha^* \approx 0.5706\). This means that when the accumulated value of our stack gets bigger than \(0.5706\) we should end our turn. This result is compatible with the range we found in our simulations.


In this first post, we have derived the generic equation for the probability of falling in a range \((a, b)\) given that we’re currently in \(t\). We have also derived the condition that the optimal thresholds have to fulfill. Finally, we have studied the particular case of the uniform distribution - both numerically and analytically - and studied its properties.

In the following posts we will study how all these results change when the distribution \(P\) changes. We will also study the case where we play against \(N\) players instead of only one.

  1. To compute the optimal threshold and its deviation we have run a Monte-Carlo simluation. This is, we have generated 1000 samples of the threshold-vs-probability curve using the expected winning probability and its 99% confidence interval. Then, we have computed the optimal threshold for each of these samples, and from this set of optimal thresholds we have computed the average and standard deviation.