Probability of winning blackjack dice game?

Question

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.

Answer 1

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

n   p_n
==  ===========
0   1
1   0.166666667
2   0.194444444
3   0.226851852
4   0.264660494
5   0.308770576
6   0.360232339
7   0.253604395
8   0.268094017
9   0.280368945
10  0.289288461
11  0.293393122
12  0.290830213
13  0.279263192
14  0.283539659
15  0.286113932
16  0.287071430
17  0.286701925
18  0.285586725
19  0.284712810
20  0.238168945
21  0.190912335
22  0.143226680
23  0.095381442
24  0.047597788

Answer 2

Your comments on my other answer suggests that the bank plays knowing what the player has. This would not happen in a casino, partly because there may be more than one player, but if you apply that peeking here then you get the same probabilities for the player:

Player      19      20      21  Bust
        0.2847  0.2382  0.1909  0.2862

Given the player's score, the probabilities for the banker's score are

Player      19      20      21
Bank            
19      0.2847      
20      0.2382  0.2856  
21      0.1909  0.2384  0.2860
Bust    0.2862  0.4760  0.7140

So given the player's score, the probabilities for the outcome are

Player      19      20      21  Bust
PlayWin 0.2862  0.4760  0.7140  
Draw    0.2847  0.2856  0.2860  
BankWin 0.4291  0.2384          1

So multiplying these by the probabilities of the player's score and adding them up gives

Player wins 0.3312
Draw        0.2037
Bank wins   0.4651

With the approximation these would be close to $\frac{146}{441}$, $\frac{90}{441}$, $\frac{205}{441}$.

The player's expected loss is about $0.133966$, rather more than the other answer, because this time the banker is peeking at the player's score.