strategy - This Dice Blackjack game - is there Nash equilibrium?

This dice Blackjack is game for two - bettor and dealer. It is played with fair six-sided dice.

Bettor: Bettor starts the game. Bettor can roll dice as many times he/she desires while the sum of rolls is lower than 21.

Dealer: Dealer plays after bettor. So dealer knows what sum bettor achieve. Dealer can roll dice as many times he/she desires while the sum of rolls is lower than 17.

Other rules: Winner is the one with lower penalization $ = \begin{cases} 21 - \sum,& \text{if } \sum \leq 21\\ 100, & \text{otherwise} \end{cases}$

where $\sum$ is sum of player rolls. If tie, dealer wins. Goal is to achieve as many wins as possible in huge number of games.

Example game:

Bettor starts:

1. roll: 6, sum=6, no reason to stop


2. roll: 5, sum=11, no reason to stop


3. roll: 5, sum=16, risk to overcome 21 in next roll is acceptable for this guy


4. roll: 4, sum=20, time to stop, risk is too high

Dealer:

1. roll: 5, sum=5, no reason to stop


2. roll: 5, sum=10, no reason to stop


3. roll: 6, sum=16, 16 < 20, so the risk of overcome 21 is not important


4. roll: 2, sum=18, still 18 < 20, but 17 was reached, dealer cannot continue and loose the game

Hypothesis (Nash equilibrium): There is simple ideal strategy for bettor - (for example stop rolling at sum of 18). Dealer can also have simple ideal strategy, if she/he expects that the bettor plays the ideal strategy.

Question: If the dealer plays the ideal strategy according to the hypothesis above, can a bettor do better with a strategy different from the optimal one from the hypothesis above?

In other words - can bettor use his/her strategy to fool dealer and get better outcome of the game?

If anything is not clear, please comment, I will do my best to improve the question.

Answer

PARTIAL ANSWER (Until I understand it deeply)

Since we roll in this game with a dice and no limited and numbered cards included, it is the same game rolling a dice until 6. So the game becomes just roll a dice until you get as close as possible to 6. If you exceed you lose.

It is possible to show this with statistics, but I just wrote a program not to think over much since there is no no-computer tag in the question. but I believe it is also possible to do this with statistics.

For example, let say bettor tend to stop when he gets 1 or over. That means bettor will roll only once whatever so:

• if bettor gets 1, he will lose whatsoever since the dealer will roll at least 1 and gets equality and win. Dealer lost chance is $0$


• if bettor gets 2 on his roll, he will stop again, dealer will lose only if he gets 1 then 6. so the chance is $1/6*1/6=1/36$.


• if bettor gets 3 on his roll, dealer will lose only if he get $1+6, 2+6, 2+5,1+1+6$. That means $1/6*1/6+1/6*2/6+1/6*1/6*1/6$


• if bettor gets $4, 1+6, 2+(5,6),3+(4,5,6), 2+1+(4,5,6), 1+1+1+6, 1+2+(4,5,6),$ etc.

and this goes on like this, at the end you will sum these and find dealer's lost or bettor's win chance, that's why I just put this into a code :)

As a result,

if dealer plays until getting 17 or more and assuming bettor knows this, the the best strategy for bettor will stop after getting 18 of course.

The code is here here you can run it if we wish.

and the best strategy for dealer will of course play until he/she got more than bettor and this will change the bettor stopping point, which becomes 19 but the probability to win for bettor becomes much less than before $33.15\%$. so there is no better strategy for bettor whatever sum dealer will stop. always playing until 19 is the best!