Will someone hit this softball out of the park?

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What are the odds of something that statistically should happen 1 in 207 times happening 3 times in 66 occurrences? Let me know.
 
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Short answer:
p=\binom{66}{3}(\frac{1}{207})^{3}(\frac{206}{207})^{63}\approx\frac{1}{263}
The probability for a single, specific arrangement is (\frac{1}{207})^{3}(\frac{206}{207})^{63}
the binomial coefficient tells us how many such arrangements exist (in your case, 45760 different arrangements of 3 "hits" and 63 "misses")
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