# Probability of dealer busting blackjack variant

Let us say that we have a mini blackjack game with only the ace, 2, 3, 4 and 5.

We will need some assumptions for this blackjack variant:

1. The value of the ace is 1 or 6 just like in normal blackjack with 1 or 11.
2. The dealer stands at 9, 10 or 11.
3. Blackjack is a 5 and a ace totalling up to 11.
4. We use infinitely many decks of cards.
5. Dealer busts when the total value of the cards is 12 or higher.

Now my question is how to determine the dealer busting probabilities with 4 cards. Determining for one extra card wasn't hard as 2 to 6 cannot go bust and 9 to 11 stands. The probability of busting with 7 is $$\frac{1}{5}$$ and the probability of busting with a 8 is $$\frac{2}{5}$$. However, with two extra cards it gets significantly harder.

I have written out all the possibilities how, for example, a 3 can go bust with 4 cards. As the only way a 3 can go bust with 4 cards is 3 + 4 + 5, 3 + 5 + 4 or 3 + 5 + 5. Can I just say that if the second card is 4 the dealer only busts with a 5 so we have $$\frac{1}{25}$$ ($$\frac{1}{5}$$ * $$\frac{1}{5}$$) and if the second card is a 5 the dealer busts with a 4 or a 5 as the fourth card, so we have $$\frac{2}{25}$$ ($$\frac{1}{5}$$ * $$\frac{2}{5}$$). Then combining the two to get a probability of 3/25. I do not think this is correct though, so can someone tell me if this is right and if it is not, how I should calculate it. I hope someone can help me and if something is not clear I will further elaborate.

Please note that the third card cannot get the total value to 9, 10 or 11 as the dealer always stands and does not pick a $$4^{th}$$ card.

Edit 1: I just had a thought. Since we have infinte amount of decks, this is an example of sampling with replacement. Does that make my calculations easier?

A general approach is to use recursion. In the terminology of Markov chains, you have two absorbing states (“bust” and “stand”) and want to find the absorption probabilities. Let $$p(t,k)$$ be the probability of eventually busting, given current total $$t$$ with $$k$$ cards. The boundary conditions are $$p(t,k)=0$$ if $$t\in\{9,10,11\}$$ and $$p(t,k)=1$$ if $$t\ge 12$$. Now condition on the value of the next card to obtain $$p(t,k) = \frac{1}{5}\sum_{c=1}^5 p(t+c,k+1).$$ You want to compute $$p(t,4)$$ for all $$t$$.

Note that you need to distinguish between “hard” and “soft” totals, where soft means the ace can be counted as $$6$$ without busting. But I recommend first solving the problem without this complication.

Here are the resulting values of $$p(t,k)$$ for reachable states $$(t,k)$$, rounded to three decimal places and ignoring the dual nature of ace: $$\begin{matrix} t\backslash k &0 &1 &2 &3 &4 &5 &6 &7 &8 &9\\ \hline0 &0.196 & & & & & & & & &\\1 & &0.186 & & & & & & & &\\2 & &0.202 &0.202 & & & & & & &\\3 & &0.235 &0.235 &0.235 & & & & & &\\4 & &0.196 &0.196 &0.196 &0.196 & & & & &\\5 & &0.163 &0.163 &0.163 &0.163 &0.163 & & & &\\6 & & &0. 136 &0.136 &0.136 &0.136 &0.136 & & &\\7 & & &0.280 &0.280 &0.280 &0.280 &0.280 &0.280 & &\\8 & & &0.400 &0.400 &0.400 &0.400 &0.400 &0.400 &0.400 &\\9 & & &0.000 &0.000 &0.000 &0.000 &0.000 &0.00 0 &0.000 &0.000\\10 & & &0.000 &0.000 &0.000 &0.000 &0.000 &0.000 &0.000 &0.000\\11 & & & &0.000 & 0.000 &0.000 &0.000 &0.000 &0.000 &0.000\\12 & & & &1.000 &1.000 &1.000 &1.000 &1.000 &1.000 &1.00 0\\13 & & & &1.000 &1.000 &1.000 &1.000 &1.000 &1.000 &1.000\\\end{matrix}$$

For example, $$p(8,2) = \frac{p(9,3)+p(10,3)+p(11,3)+p(12,3)+p(13,3)}{5} = \frac{0+0+0+1+1}{5} = 0.4.$$

Notice that the results do not depend on $$k$$. Each row in the table is constant. So perhaps I have misinterpreted the problem.

Alternative interpretation: before the game starts, you want to know the probability that the dealer will eventually bust with exactly $$4$$ cards. In that case, the boundary conditions are $$p(t,k)=0$$ if $$t\in\{9,10,11\}$$, $$p(t,k)=0$$ if $$t\ge 12$$ and $$k\not=4$$, and $$p(t,k)=1$$ if $$t\ge 12$$ and $$k=4$$. The recursion remains the same, and you want to calculate $$p(0,0)$$. Here are the results, ignoring the dual nature of ace: $$\begin{matrix} t\backslash k &0 &1 &2 &3 &4 &5 &6 &7 &8 &9\\ \hline0 & 51/625 & & & & & & & & & \\1 & & 13/125 & & & & & & & &\\2 & & 14/125 & 1/25 & & & & & & &\\3 & & 11/125 & \color{red}{3/25} & \color{red}{0} & & & & & &\\4 & & 8/125 & 3/25 & \color{red}{0} & 0 & & & & &\\5 & & 1/25 & 3/25 & \color{red}{0} & 0 & 0 & & & &\\6 & & & 3/25 & \color{red}{0} & 0 & 0 & 0 & & &\\7 & & & 2/25 & \color{red}{1/5} & 0 & 0 & 0 & 0 & &\\8 & & & 0 & \color{red}{2/5} & 0 & 0 & 0 & 0 & 0 &\\9 & & & 0 & \color{red}{0} & 0 & 0 & 0 & 0 & 0 & 0\\10 & & & 0 & \color{red}{0} & 0 & 0 & 0 & 0 & 0 & 0\\11 & & & & \color{red}{0} & 0 & 0 & 0 & 0 & 0 & 0\\12 & & & & 0 & 1 & 0 & 0 & 0 & 0 & 0\\13 & & & & 0 & 1 & 0 & 0 & 0 & 0 & 0\\\end{matrix}$$

The results shown in red are the values you mentioned.

• Hi @RobPratt I am not that familiar with Markov states unforutnately. Is there a way for me to understand and execute this by myself? If no, could you provide more tips, so I can solve it? Thank you very much! Commented Oct 22, 2022 at 14:24
• You do not need to know Markov chains. I just mentioned them for reference. I would be glad to give more details, but please first clarify which interpretation in my answer matches what you want. Commented Oct 22, 2022 at 15:08
• Let's do the first one, since you mentioned that one firstly and I understan dit more clearly I think. Commented Oct 22, 2022 at 15:39
• I have taken a look at the results and I think it indeed went wrong somewhere. Let's say that we have a starting hand of a 3 and a 4, with a value of 7. If we draw only one more card, the only way to bust is with a 5 to get to a value of 12, so $p(7,3)$ should equal $\frac{1}{5}$. In the first interpretation you end up with a probability of $0.280$. Either something went wrong or I misunderstood what $p(7,3)$ means. The second interpretation seems to get more close, however $p(7,4)$ should be $\frac{2}{25}$ as we can draw an ace to get to 8 and then a 4 or 5 to get to a value of 12 or 13. Commented Oct 23, 2022 at 11:33
• I hope you understand what I mean... Commented Oct 23, 2022 at 11:34

Here's a computation of the probability you're looking for in the case when the aces are taken to be bivalent. I assume, as is usual in blackjack, that the dealer must stand if any valid assignment of values to the aces in his hand will give the hand a total in the standing range—that is, from $$9$$ to $$11$$ inclusive in your mini version of the game. As mentioned in RobPratt's answer, the bivalency of the aces makes the calculation a little more complicated, but the basic recursive procedure is essentially the same as the one described in that answer.

Apart from the two so-called "absorbing" states of having stood or busted, there are a number of live states of the game, which can be conveniently described by a pair, $$\ (a,n)\$$, of non-negative integers, where $$\ a\$$ is the number of aces currently in the hand, and $$\ n\$$ is the total value of all the non-aces in it. The value of such a hand could be any of the numbers $$\ n+a+5i\$$ for $$\ i=0,1,\dots,a\$$, so it will be in a live state if and only if none of those numbers lies in the range from $$9$$ to $$11$$ inclusive, and at least one of them lies in the range $$0$$ to $$8$$ inclusive. It turns out that there are $$20$$ of these live states: \begin{align} &(0,0),(0,2),(0,3),(0,4),(0,5),(0,6),(0,7),(0,8),\\ &(1,0),(1,2),(1,6),(1,7),\\ &(2,0),(2,2),(2,5),(2,6),\\ &(3,0),(3,4),(3,5)\ \ \text{and}\\ &(4,4)\ . \end{align} If the dealer's hand is in live state $$\ (a,n)\$$ after $$\ d-1\$$ cards have been dealt, then after the $$\ d^\text{th}\$$ card has been dealt, it will be in one of the states $$\ (a+1,n), \,(a,n+2),\,(a,n+3),\,(a,n+4),$$ or $$\,(a,n+5)\$$, each with probability $$\ \frac{1}{5}\$$. Not all those states will necessarily be live, however. The ones that aren't must be amalgamated together as representing either a stood or a busted state. The probability of entering either of these latter states will be $$\ \frac{s}{5}\$$, where $$\ s\$$ is the number of non-live states that have been amalgamated into it.

The probabilities of entering each of the possible successors of each of the possible states of the game are listed in the following table: $$\begin{array}{c|c|l|} \text{index}&\text{state}&\hspace{3.5em}\text{successors and their probabilities}\\ \hline 1&(0,0)&{\begin{array}{c|c|c|c|c|} \text{successor}&(1,0)&(0,2)&(0,3)&(0,4)&(0,5)\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5} \end{array}}\\ \hline 2&(0,2)&{\begin{array}{c|c|c|c|c|} \text{successor}&(1,2)&(0,4)&(0,5)&(0,6)&(0,7)\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5} \end{array}}\\ \hline 3&(0,3)&{\begin{array}{c|c|c|c|c|} \text{successor}&(0,5)&(0,6)&(0,7)&(0,8)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5} \end{array}}\\ \hline 4&(0,4)&{\begin{array}{c|c|c|c|c|} \text{successor}&(0,6)&(0,7)&(0,8)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{2}{5} \end{array}}\\ \hline 5&(0,5)&{\begin{array}{c|c|c|c|c|} \text{successor}&(0,7)&(0,8)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5} \end{array}}\\ \hline 6&(0,6)&{\begin{array}{c|c|c|c|c|} \text{successor}&(1,6)&(0,8)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5} \end{array}}\\ \hline 7&(0,7)&{\begin{array}{c|c|c|c|c|} \text{successor}&(1,7)&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{1}{5}&\frac{3}{5}&\frac{1}{5} \end{array}}\\ \hline 8&(0,8)&{\begin{array}{c|c|c|c|c|} \text{successor}&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{3}{5}&\frac{2}{5} \end{array}}\\ \hline 9&(1,0)&{\begin{array}{c|c|c|c|c|} \text{successor}&(2,0)&(1,2)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5} \end{array}}\\ \hline 10&(1,2)&{\begin{array}{c|c|c|c|c|} \text{successor}&(2,2)&(1,6)&(1,7)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{2}{5} \end{array}}\\ \hline 11&(1,6)&{\begin{array}{c|c|c|c|c|} \text{successor}&(2,6)&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{1}{5}&\frac{3}{5}&\frac{1}{5} \end{array}}\\ \hline 12&(1,7)&{\begin{array}{c|c|c|c|c|} \text{successor}&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{4}{5}&\frac{1}{5} \end{array}}\\ \hline 13&(2,0)&{\begin{array}{c|c|c|c|c|} \text{successor}&(3,0)&(2,5)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5} \end{array}}\\ \hline 14&(2,2)&{\begin{array}{c|c|c|c|c|} \text{successor}&(2,5)&(2,6)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5} \end{array}}\\ \hline 15&(2,5)&{\begin{array}{c|c|c|c|c|} \text{successor}&(3,5)&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{1}{5}&\frac{3}{5}&\frac{1}{5} \end{array}}\\ \hline 16&(2,6)&{\begin{array}{c|c|c|c|c|} \text{successor}&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{3}{5}&\frac{2}{5} \end{array}}\\ \hline 17&(3,0)&{\begin{array}{c|c|c|c|c|} \text{successor}&(3,4)&(3,5)&\text{stood}\\ \hline \text{probability}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5} \end{array}}\\ \hline 18&(3,4)&{\begin{array}{c|c|c|c|c|} \text{successor}&(4,4)&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{1}{5}&\frac{3}{5}&\frac{1}{5} \end{array}}\\ \hline 19&(3,5)&{\begin{array}{c|c|c|c|c|} \text{successor}&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{3}{5}&\frac{2}{5} \end{array}}\\ \hline 20&(4,4)&{\begin{array}{c|c|c|c|c|} \text{successor}&\text{stood}&\text{busted}\\ \hline \text{probability}&\frac{3}{5}&\frac{2}{5} \end{array}}\\ \hline 21&\text{stood}&{\begin{array}{c|c|c|c|c|} \text{successor}&\text{stood}\\ \hline \text{probability}&1 \end{array}}\\ \hline 22&\text{busted}&{\begin{array}{c|c|c|c|c|} \text{successor}&\text{busted}\\ \hline \text{probability}&1 \end{array}}\\ \hline \end{array}$$

Let $$\ P\$$ be the $$22\times22\$$ matrix whose entry in row $$\ i\$$ and column $$\ j\$$ is the probability that the state with index $$\ j\$$ in the above table immediately follows the one with index $$\ i\$$, and $$\ \pi_k(d)\$$ the probability that the dealer's hand is in the state with index $$\ k\$$ after $$\ d\$$ cards have been dealt. Then we have \begin{align} \pi_k(0)&=\cases{1&if \ k=1\\ 0& otherwise}\\ \pi_j(d)&=\sum_{i=1}^{22}\pi_i(d-1)P_{ij}\ , \end{align} or, in matrix form, \begin{align} \pi(0)&=\big(1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0\big)\\ \pi(d)&=\pi(d-1)P\\ &=\pi(0)P^d \end{align} The probability you apparently want to calculate is the probability, $$\ b_4\$$, that the dealer's hand enters the busted state after exactly $$4$$ cards have been dealt. This is given by \begin{align} b_4&=\sum_{j=1}^{20}\pi_j(3)P_{j\,22}\\ &=\frac{1}{5}\big(\pi_7(3)+\pi_{11}(3)+\pi_{12}(3)+\pi_{15}(3)+\pi_{18}(3)\big)\\ &\hspace{3em}+\frac{2}{5}\big(\pi_8(3)+\pi_{16}(3)+\pi_{19}(3)+\pi_{20}(3)\big) \end{align} Although it would be feasible to carry out this calculation by hand, it would be extremely tedious, and is much easier to do with a computer program or mathematical algebra package.

Using the online Magma calculator I obtained \begin{align} \pi(3)=&\\ &\begin{matrix}\Big(0,&0,&0,&0,&0,&\frac{1}{125},&\frac{3}{125},&\frac{6}{125},&0,&0,&\frac{1}{25},\\ \ \ \frac{6}{125},&0,&\frac{2}{125},&\frac{1}{125},&\frac{1}{125},&\frac{1}{125},&0,&0,&0,&\frac{18}{125},&\frac{2}{125}\Big)\end{matrix} \end{align} and \begin{align} b_4&=\frac{29}{625}\\ &=\frac{1}{5}\left(\frac{3}{125}+\frac{1}{25}+\frac{6}{125}+\frac{1}{125}\right)+\frac{2}{5}\left(\frac{6}{125}+\frac{1}{125}\right)\ . \end{align}