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- Homework Statement:
- If we have a 6-sided die (it is a fair die), how many throws would we expect to make before we see a repeated number?

- Relevant Equations:
- Probability

Hi,

I found a similar problem online, but there wasn't any solution. The question actually had a larger number of sides (e.g. 50 or 100, etc.), but I thought the underlying solution method should be the same for any arbitrary ##n##-sided die.

If we have a 6-sided die (it is a fair die), how many throws would we expect to make before we see a repeated number?

I have been thinking about this problem for a while, and am very stuck on how to proceed. I did write a Python script to simulate this game for different ##n##, and the graph aligned with intuition (larger ##n## required more throws until repetition). I remember reading that this problem was similar to the

If I follow that birthday paradox line of reasoning, I suppose I could do:

[tex] P(\text{repetition on 5th throw}) = 1 \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{4}{6} = \frac{(6)! \cdot 4}{(6-4)! \cdot 6^5} [/tex]

Thus this could be generalized to the probability of a repetition on the ##k##-th throw:

[tex] P(\text{repetition on k-th throw for n sided die}) = \frac{(n)! \times (k-1)}{(n-(k-1))! \cdot n^k} [/tex]

Is the above correct? I am not sure I am fully convinced by the abiThis question was an interview question, so how could one find the expected number without a calculator - are there tricks or approximations?

Any help would be greatly appreciated.

I found a similar problem online, but there wasn't any solution. The question actually had a larger number of sides (e.g. 50 or 100, etc.), but I thought the underlying solution method should be the same for any arbitrary ##n##-sided die.

**Question:**If we have a 6-sided die (it is a fair die), how many throws would we expect to make before we see a repeated number?

**Attempt:**I have been thinking about this problem for a while, and am very stuck on how to proceed. I did write a Python script to simulate this game for different ##n##, and the graph aligned with intuition (larger ##n## required more throws until repetition). I remember reading that this problem was similar to the

*birthday paradox*conceptually, however I haven't been able to use that to solve the problem.If I follow that birthday paradox line of reasoning, I suppose I could do:

[tex] P(\text{repetition on 5th throw}) = 1 \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{4}{6} = \frac{(6)! \cdot 4}{(6-4)! \cdot 6^5} [/tex]

Thus this could be generalized to the probability of a repetition on the ##k##-th throw:

[tex] P(\text{repetition on k-th throw for n sided die}) = \frac{(n)! \times (k-1)}{(n-(k-1))! \cdot n^k} [/tex]

Is the above correct? I am not sure I am fully convinced by the abiThis question was an interview question, so how could one find the expected number without a calculator - are there tricks or approximations?

Any help would be greatly appreciated.