Using 6 Sided Dice to Get an Even Distribution of Numbers $1$ to $50$ Is this possible?
I have a $6$ sided dice (call it a $d6$).
I want a way of getting an even distribution of the numbers $1$ to $50$ using that dice with no or minimal rerolls.
A way with a lot of rerolls.
A. Roll d6. $1 - 3 $means use table $A$, $4 - 6$ means use table $B$.
B. Table $A$ needs you to roll $2$ dice. First Dice gives you the column and second dice gives you the row. The numbers $1$ to $25$ are in the table, with the other $11$ spaces telling you to reroll. (Table B is similar, but with numbers $26$ to $50$).
The only problem with this is there are $22$ reroll spaces.
What I want is a way to get an even distribution, but with no rerolls or very few.
 A: Rolling three dice gives you a number from $1$ to $216$.  For a start, you can use the numbers $1-200$ and reroll the last $16$ but remember what the number is.  Now one more roll gives you a number $1-96$ by multiplying your remembered number by $6$, subtracting the new roll, and adding $1$.  You can use $1-50$ and reroll $46$ values.  Again remember the remainder, roll again, and you get a number $1-6\cdot 46=1-276$.  You can use $1-250$ and keep $1-26$.  Keep going.  This has to be optimal because you are using all the information you get.  I haven't computed the average number of throws.
A: Probably the best you can do is to convert any $3$ rolls to a number $1-50$ as follows : If the rolls give $a,b,c$ , then the number is $$36(a-1)+6(b-1)+c$$ If $a>2$ , then you can stop since the number must then exceed $50$, so you can ignore all trials where the first roll is more than $2$. If we have $a=2$ , the second roll is not allowed to exceed $3$, again you can save a roll if $3$ is exceeded.
