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Preamble

My question is in the context of a dice game called 31. You have 6 dice and the goal is to have the highest score. If you roll 36, you score 6 points. If you roll 35 you score 5 points and so on up until 30, where you break even. At 29 you lose 1 points and so on.

You roll all the dice at once and after each roll you need to keep at least 1 die on the table. So you roll a maximum of 6 times.

Question
If your first roll gives you 6-5-5-x-x-x (where the Xs ≤ 4), what is the probability of getting at least two 5s if you roll five dice (excluding the 6 that you keep). By "at least two 5s" I mean that 5-5-x-x-x is the bottom of the range and 6-6-6-6-6 is the top of the range.

The question could also be worded "On the 2nd roll, do you keep the 5s and roll 3 dice or do you roll them?"

Thanks.

As to add more clarification to the rules of the games, here's a good exemple provided by @Brams28 int the comment

[...]a roll is rolling all the dice you still have left. You have to keep at least one die after each roll, but a single die could potentially be rolled 6 times. Example: I roll 6,5,4,3,2,1 on the first roll. I decide to keep the 6 and the 5 but reroll the 4 others. Now I get 5,2,2,1. I keep the 5 and reroll the other 3. Now I get 6,3,1. I keep the 6 and reroll the last two. I get 5,4. OK, I keep those two (of course!)

To determine optimal strategy, I'll start from the end.

With the option of keeping/rerolling one die (i.e. the first five are 'locked in') we keep $ge4$ and reroll $le3$ because the expected value $EV_1$ of one $d_6$ is $3.5$

For the option of keeping/rerolling two dice, we need to determine the EV of two $d_6$ with subsequent reroll. To do this, we look at all possible outcomes of two dice and apply the optimal strategy for rerolling one die.
begin{matrix}
6+6&6+5&6+4&6+color{red}3&6+color{red}2&6+color{red}1\
5+6&5+5&5+4&5+color{red}3&5+color{red}2&5+color{red}1\
4+6&4+5&4+4&4+color{red}3&4+color{red}2&4+color{red}1\
color{red}3+6&color{red}3+5&color{red}3+4&3+color{red}3&3+color{red}2&3+color{red}1\
color{red}2+6&color{red}2+5&color{red}2+4&color{red}2+3&2+color{red}2&2+color{red}1\
color{red}1+6&color{red}1+5&color{red}1+4&color{red}1+3&color{red}1+2&1+color{red}1\
end{matrix}
The red numbers are those that we reroll. Replace them with $EV_1=3.5$ and sum each outcome.
begin{matrix}
12&11&10&9.5&9.5&9.5\
11&10&9&8.5&8.5&8.5\
10&9&8&7.5&7.5&7.5\
9.5&8.5&7.5&6.5&6.5&6.5\
9.5&8.5&7.5&6.5&5.5&5.5\
9.5&8.5&7.5&6.5&5.5&4.5
end{matrix}
Add all these together and divide by $36$ and we get $EV_2=frac{296.5}{36}approx8.236$

Now, for optimal strategy we note that keeping a $5$ and rerolling the second die $le3$ gives us an $EV$ of $8.5>EV_2$, so this is good. Also, keeping a $4$ and rerolling the second die $le3$ gives us an $EV$ of $7.5<EV_2$, so this is not good. Finally, keeping $4 4$ gives us $8<EV_2$, so rerolling both dice gives us a slight advantage.

Likewise, we find the optimal strategy for the option of keeping/rerolling three dice by examining all $216$ possible outcomes and apply our optimal strategy for two dice to each. The result is $EV_3approx13.425$. Now keeping a $5$ and rerolling the other two dice $le4$ gives us $5+EV_2=13.236$, which is less than $EV_3$ so our optimal strategy is to reroll all three dice unless we have at least two $5$s, then we reroll the third die $le3$.

Through simulation, I found $EV_4>18.8$, so optimal strategy would be to reroll all four dice unless we already have $5 5 5 5 =20$ or $5 5 5 4=19$

I also found $EV_5>24.4$, so optimal strategy would be to reroll all five dice unless we have five $5$s.

Finally, with optimal strategy throughout, $EV_6>30.1$ for a full round. The creators of this game likely knew this, with a positive score earned only if you exceed this $EV$.

I don't think the answer to first question is much help in answering the second, and I agree with Daniel Mathias's answer to the second—namely that you should roll all 5 dice. If you do that, and play optimally thereafter, your expected score will be $6 + frac{8,569,700}{350,699} - 30 approx 0.44$.

If you keep the 6 and one of the 5s, roll the remaining four dice and play optimally thereafter, your expected score will be $6 + 5 + frac{989,065}{52,488} - 30 approx -0.16$.

If you keep the 6 and both 5s, roll the remaining three dice and play optimally thereafter, your expected score will be $6 + 5 + 5 + frac{13,049}{972} - 30 approx -0.58$.

If you roll $d$ dice, they show the numbers $m_1 ge m_2 ge dots ge m_d$, and you decide to keep $k$ of them and roll the remaining $d-k$ (where $1 le k le d$), then then the ones you should keep are obviously those showing $m_1, m_2, dots , m_k$. If you play optimally thereafter, your expected final sum will be $ES_{d,k} = m_1 + m_2 + dots + m_k + EV_{d-k}$, where $EV_i$ (to purloin Daniel's notation) represents the maximum expected sum you can obtain by proceeding optimally with $i$ dice. Your optimal strategy is therefore to choose the value of $k$ for which $ES_{d,k}$ is a maximum. We therefore get the following recursive equation for the value of $EV_d$:
$$EV_d = sum_{mbox{all $d$-tuples of die faces}}
frac{max_{1le k le d}left(ES_{d,k}right)}{6^d} .$$
Using a Math Studio script to compute the values of $EV_d$ for $d=3 mbox{ to } 6$ (the values $d = 1$ and $2$ are trivial to do by hand), I obtained the following:
begin{eqnarray}
EV_1 &=& frac{7}{2} \
EV_2 &=& frac{593}{72 }approx 8.236\
EV_3 &=& frac{13,049}{972} approx 13.425\
EV_4 &=& frac{989,065}{52,488} approx 18.844\
EV_5 &=& frac{8,569,700}{350,699} approx 24.436\
EV_6 &approx& 30.152 ,
end{eqnarray}
thus confirming Daniel's analysis.

Edit:

It's probably worth giving the following fairly simple explicit description of an optimal strategy.

If you throw the numbers $j_1, j_2, dots , j_d$ when you throw $d$ dice,

• list those numbers in decreasing order as $m_1, m_2, dots , m_d$;




• calculate the quantites
  begin{eqnarray}
  S_0 &=& sum_{i=2}^d m_i ,\
  S_k &=& EV_k + sum_{i=2}^{d-k} m_i mbox{ for $k= 1, 2, dots , d-2$ , and}\
  S_{d-1} &=& EV_{d-1} ;
  end{eqnarray}



• determine the value $k^*$ of $k$ for which $S_k$ achieves its maximum value;


• If $k^* = 0$, keep all the dice and don't throw any more. Otherwise, keep the $d-k^*$ dice $m_1, dots , m_{d-k^*}$ and throw the $k^*$ dice $m_{d+1-k^*}, dots , m_d$ .