# Probability of rolling at least one 6 while rerolling 1's

Scenario: You roll a number of 6-sided dice

Success: Roll at least one 6

Conditions: You can re-roll any 1's you get on the first roll

What are the odds of success for n dice?

Example: 7 dice

Roll:            (1, 1, 1, 4, 4, 4, 5)

Re-roll:      (1, 2, 3)

Final dice: (1, 2, 3, 4, 4, 4, 5)

Result:        Failure

My take on this was: For any number of rolled 1's, "replace" those rolls with rolling n+x dice (where x is the number of 1's rolled) and thus reducing the problem to simple combinatorics, but I didn't get very far.

I suppose there is a simple "trick", so I'm looking for other angles into this problem.

However, if it turns out not to be so simple, please try to be as verbose and layman-friendly as you can.

• Not an answer as such, but I have published open-source Ruby code that can calculate (floating point) probabilities for this kind of dice system. It doesn't give you an explanation of how the value is derived, but you may find it useful if you have a few similar calculations: rubygems.org/gems/games_dice – Neil Slater Aug 18 '13 at 17:59

The best strategy is to reroll any rolled 1, of course. So using that strategy, what is the chance that you do not end up at a 6 with a single dice? The probability for an instant 6 is $1/6$, and for a 1 with a rerolled 6 is $(1/6)^2 = 1/36$. So in total for a single dice, the chance not to get a 6 is $$1 - (1/6 + 1/36) = 29/36.$$

So for $n$ dice, the probability of getting at least one 6 is $$p = 1 - \left(\frac{29}{36}\right)^{\!n}.$$

For small $n$, we get these approximated probabilities: $$\begin{array}{cc} n & p \\ \hline 1 & 19.44\% \\ 2 & 35.10\% \\ 3 & 47.73\% \\ 4 & 57.89\% \\ 5 & 66.08\% \\ 6 & 72.67\% \\ 7 & 77.99\% \\ 8 & 82.27\% \\ 9 & 85.72\% \\ 10 & 88.49\% \end{array}$$

• Well, that's even simpler then I expected. Thanks. – user1853181 Aug 18 '13 at 19:38

The chance of ending up with a $1$ on throwing one die is $\frac1{36}$ since you throw $1$ the first time and re-roll and get $1$ again.

The remaining numbers are (obviously) equally likely - probability therefore $\frac 15 \cdot \frac {35}{36}=\frac 7{36}$ for each of $2,3,4,5,6$.

Probability of not throwing any $6$s with $n$ dice is therefore the probability that they are all something other than a six, which is $\left(\frac {36-7}{36}\right)^n=\left(\frac {29}{36}\right)^n$

The probability of at least one $6$ is then $1-\left(\frac {29}{36}\right)^n$