dice question : what is best? got into a discussion what is best if you have :
- 5 dices
- 3 throws 
You aim for getting 12345 or 23456 - doesnt matter which one of the 2 combinations.
First throw :  12356.
So we are missing a 4 to get either 12345 or 23456.
What is best for 2nd throw ? 
To keep 235 - and throw 2 dices ( to get a 4 + 1 or 2 )?
Or to keep either 2356 and throw 1 dice ( to get a 4 ) ?
Remember there is 3 throws total so i assume that it might be smartest to keep the 6 and use 2 throw to throw a 4 with the same dice ? it just feels like you have more chances of getting the 4 throwing 2 dices - but ofcourse you would need to do a 41 or 46 - instead of only needing a 4 in 2 throws with 1 dice.
 A: It’s better to keep the $1$ or $6$. If you reroll it, you just need to get it again.
Your chances of getting the straight in a single roll are $\frac16$ if you keep the $1$ or $6$ (since you need one specific number on one die) and $\frac4{36}=\frac19$ if you reroll the $1$ or $6$ (since you need one of the four ordered pairs $(1,4)$, $(6,4)$, $(4,1)$, $(4,6)$).
Since you have two rolls, if you keep the $1$ or $6$, your chances of getting the straight are $1-\left(\frac56\right)^2=\frac{11}{36}\approx31\%$ (since you get it unless you don’t roll a $4$ on either roll).
If you reroll the $1$ or $6$, with probability $\frac19$ you immediately get the straight; with probability $\frac14$ you get neither a $1$ nor a $4$ nor a $6$, and then you have another $\frac19$ chance on the second roll; with probability $\frac7{36}$ you get a $4$ but no $1$ or $6$, and then you have probability $\frac13$ to get the $1$ or $6$ on the second roll; and with probability $\frac{16}{36}=\frac49$ you get a $1$ or $6$ but no $4$, and then you have probability $\frac16$ to get the $4$ on the second roll, so in total your probability to get the straight in two rolls if you reroll the $1$ or $6$ is
$$
\frac19+\frac14\cdot\frac19+\frac7{36}\cdot\frac13+\frac49\cdot\frac16=\frac5{18}\approx28\%\;.
$$
A: Let's evaluate the probability in each case.
I'll start with the second case, keeping $2356$ and hoping to get a $4$.
We could have $4$ on the first throw $\left(\frac16\right)$, or not $4$ on the first throw and $4$ on the second throw $\left(\frac56\times\frac16\right)$. The probability to have a $4$ with two rolls is
$$\frac16+\frac56\times\frac16=\frac{11}{36}\tag{1}$$
First case is a bit longer to evalutate. Since there are only $36$ possibilities with the roll of two dices, the best way to do this is to write all the possibilities and count the favorable outcomes. There is four events to consider.


*

*Having what we need on the first roll, $4$ favorable outcomes $(1,4)$, $(4,1)$, $(4,6)$ and $(6,4)$. Probability: $\frac4{36}$.

*Having $4$ on the first roll, but neither $1$ or $6$, $\left(\frac7{36}\right)$. Then we throw one die to get $1$ or $6$, $\left(\frac2{6}\right)$. Probability: $\frac7{36}\times\frac26=\frac{14}{216}$.

*Having $1$ or $6$, but not $4$, $\left(\frac{16}{36}\right)$. Then we throw one die to get $4$, Probability: $\left(\frac1{6}\right)$. Probability : Probability: $\frac{16}{36}\times\frac16=\frac{16}{216}$.

*Having neither $1$, $4$ or $6$, $\left(\frac9{36}\right)$. Then we need them both on second throw $\left(\frac4{36}\right)$. Probability: $\frac9{36}\times\frac4{36}=\frac{36}{1296}$.


Since thes events are mutually exclusive, the total probability is the sum of the four.
$$\frac4{36}+\frac{14}{216}+\frac{16}{216}+\frac{36}{1296}=\frac{10}{36}\tag2$$
We now compare $(1)$ and $(2)$, clearly
$$\frac{11}{36}>\frac{10}{36}$$
You should keep $2356$ and try to roll a $4$.
