Probability of winning blackjack dice game? 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.
 A: 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.
