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What's the chance of getting at least a single 1 on a 64 sided die thrown 64 times?

I believe we can calculate that by multiplying the chances of not rolling a 1 in each of the 64 throws, which would be $(\frac{63}{64})^{64}$, then $(1 - 0.36)$ or so gives us approximately a 63% chance of rolling at least a single 1, but logically speaking, shouldn't we anticipate seeing each number roughly once on average if the die is fair?

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  • $\begingroup$ Knowing that the expected number of times you see the $1$ is $1$ tells you little about the actual probability. Your calculation is the way to go. $\endgroup$
    – lulu
    Commented May 23 at 9:49
  • $\begingroup$ Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read its title. $\endgroup$
    – José Carlos Santos
    Commented May 23 at 9:49

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Short answer: your approach of asking what is the probability of not getting a 64 and then using the complement was correct.

So $P(\text{64 at least once})=1-P(\text{no 64s})=1-(\frac{63}{64})^{64}\approx 0.635$

Which is closer to 64% than 63% but your logic remains valid.

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  • $\begingroup$ Will we see many of the numbers exactly once each in 64 throws? How is this calculated? How many numbers will be seen one time? $\endgroup$
    – user1170675
    Commented May 23 at 10:50
  • $\begingroup$ Expected value is a long run average, not a specific statement that something will happen. $\endgroup$
    – Red Five
    Commented May 23 at 11:05
  • $\begingroup$ But there is an expected number of numbers that will be seen once (not sure how to practically compute it though) $\endgroup$
    – doetoe
    Commented May 23 at 18:03

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