I know a little bit about probability but I am not sure how to calculate this:
In a dice game of blackjack, there are two parties. The player and the dealer. The aim of this game is to get as close to $21$ without going over, using six sided dice which has an equal chance of landing on each side. Both parties may use as many dice as they like. If the player goes over 21 then they lose and similarly to casino blackjack, the player's turn is first. For the purpose of this question, assume that the player will always keep (stay) the value of either 19, 20, 21 and would continue if the value is 18 or under. If there is a draw then the game is repeated and there is no winner.
Thanks in advance and I hope this is enough information to draw a reasonable answer.
Best Answer
You have not actually asked a question. I suppose you want to know what the player's expected loss is.
Nor have you said how the bank adjusts to the player's behaviour. For example, what happens if the player stays/sticks at 20 and the bank reaches 19?
Let's assume the bank is also required to follow the same rule as you give for the player, but if they both bust by going over 21 then the bank wins.
Then the probability the player busts is about $0.286205909$ (not that far away from $\frac27$, the limit of probability that the player if throwing indefinitely hits a particular large number). Similarly for the bank. So the probability they both bust is this squared, i.e. about $0.081913822$ (not far away from $\frac{4}{49}$). This is the only unfair part of the game so it is the player's expected loss if she has staked $1$.
To calculate the probability of a bust, try $p_0=1$, $p_n=\frac16 \sum_{i=0}^{i=n-1} p_i$ for $1 \le i \le 6$, $p_n=\frac16 \sum_{i=n-6}^{i=n-1} p_i$ for $6 \le i \le 19$, $p_n=\frac16 \sum_{i=n-6}^{i=18} p_i$ for $19 \le i \le 24$, and then add up $p_{22}+p_{23}+p_{24}$. The values of $p_n$ are about