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What’s the anticipated selection of rolls except a regular die presentations a bunch you might have already rolled?

As an example, let’s say you roll 1, then 2, after which 1 once more. The die repeated the #1 on the third roll.

Or in case you roll three, 2, 5, 2, then the quantity 2 used to be proven once more, so it took four rolls to look a repeat.

What’s the common selection of rolls except a die repeats a host?

For any other rationalization of the issue and a touch, watch the video.

**How many dice rolls until a repeat?**

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.**Answer To Puzzle : Rolls Before Repeat**

Here is a video explanation.

**Solution to dice rolls until a repeat**

**Text version of solution**

The 1st roll is always a unique number and never a repeat, and the 7th roll is always a repeated number. So the answer should be somewhere in the middle, like 3.5. This is pretty close! The exact answer is 1223/324, which is about 3.77.

We can solve the problem by working backwards. When we have seen all 6 numbers already, then we are guaranteed a repeat number in the 1 more roll.

What if we have seen 5 numbers? There is a 5/6 chance we will see a duplicate–in which case we are done in that 1 roll. And there is a 1/6 chance we will see a new number–in which case we get to the case of having seen 6 numbers, plus we add the roll that got us there. In other words, if E(*x*) is the expected number of rolls after seeing *x* numbers, we have the following formula:

E(5) = (5/6)(1) + (1/6)(1 + E(6))

E(5) = 1 + (1/6)E(6)

E(5) = 1 + (1/6)1 = 7/6

What if we have already seen 4 numbers? There is a 4/6 chance we will see a duplicate–in which case we are done in that 1 roll. And there is a 2/6 chance we will see a new number–in which case we get to the case of having seen 5 numbers, plus we add the roll that got us there.

E(4) = (4/6)(1) + (2/6)(1 + E(5))

E(4) = 1 + (2/6)E(5)

E(4) = 1 + (1/6)(7/6) = 25/18

There is a pattern to the calculations. If we have seen *x* numbers, then there is an *x*/6 chance we see a repeat–and we are done in that 1 roll. Or there is a (1 – *x*)/6 chance we see a new number–in which case we get to the case of having seen *x* + 1 numbers, plus we add the roll that got us there. So we have the formula:

E(*x*) = (*x*/6)(1) + ((1 – *x*)/6)(1 + E(*x* + 1))

E(*x*) = 1 + ((1 – *x*)/6)E(*x* + 1)

We can calculate and find E(3) = 61/36, E(2) = 115/54, E(1) = 899/324, and finally E(0) = 1223/324.

So when we start out–and no numbers have yet been seen–the expected number of rolls until we see a duplicate is E(0) = 1223/324, which is about 3.77.

I give credit to Math StackExchange for suggesting the recursive approach.

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