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Finding e by throwing darts on a post by Mike Lawler makes reference to the following quote on John Allen Paulos A Numerate Life:

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I looked up a similar exercise on the first link solved by N.Taleb, regarding the probability of leaving half of the checkerboard empty of darts, and it involves compositions, as well as the multinomial distribution with equal probabilities. He solves it with Mathematica - so I guess numerically.

In the case of the number e, I guess there has to be a close formula, like a limit explaining this. A law of sorts, but which one? How is this proven?

Antoni Parellada
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    Poisson distribution? – Angina Seng Nov 18 '17 at 20:56
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    The key is that we have a binomial distribution for the number of pennys on each square. If $n$ is large and $p$ is small, we can approximate this very good with a poisson-distribution with parameter $np$. In the given case, we have $np=1$, so we have approximately a poisson-distribution with parameter $1$, hence $P(X=0)\approx P(X=1)\approx\frac{1}{e}$ – Peter Nov 18 '17 at 21:01
  • The problem is that the distribution of the number of pennies on each square is not independent. For instance, if $X_n$ is the number of pennies on the nth square, then $0 = \mathbb{P}(X_1=1 \mid X_2=64) \neq \mathbb{P}(X_1 = 1)$, so how can we apply either a binomial or Poisson distribution? – B. Mehta Nov 18 '17 at 21:03
  • But the choices of squares is independent and the probability for having no penny is equal for every square. – Peter Nov 18 '17 at 21:06
  • Related: https://math.stackexchange.com/questions/2419736/ Here, for each single square, we have n=64 attempts which will hit with probability 1/64. The probability of no hit is roughly 1/e. (In the linked question and answers, they compute the complementary probability to be 1-1/e.) – Torsten Schoeneberg Nov 18 '17 at 21:47

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Let $N$ be the number of squares and $U$ the number of squares having no penny. We draw $N$ random numbers from $1$ to $N$.

For each square, the probability that this specific square has no penny, is $(\frac{N-1}{N})^N$ , which is approximately $\frac{1}{e}$, if $N$ is large.

Now set $X_j=1$ , if $j$ has no penny and $X_j=0$ otherwise. Then, the sum $X_1+\cdots+X_N$ is the number of squares with no pennys.

Since $E(X_1)=\cdots=E(X_n)\approx\frac{1}{e}$ and the expectations add, we have $E(U)\approx N\cdot \frac{1}{e}=\frac{N}{e}$

Peter
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  • The number of squares with exactly one penny is also approximately $\large \frac{N}{e}$, if $N$ is large. – Peter Nov 18 '17 at 21:19
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    Isn't the question asking for $E\left(\frac{N}{U}\right) = e$, which is different to $\frac{N}{E(U)}$ (by, eg, Jensen)? – B. Mehta Nov 18 '17 at 21:25
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    Strictly speaking, yes. But since we can expect $U\approx \frac{N}{e}$, we will get $\frac{N}{U}\approx e$. Of course, it is not predictible which value $U$ will have, and we have to consider the standard deviation of $U$. If $N$ is very large, we have a rather small standard deviation compared to the expectation, so we will get a reasonable approximation. – Peter Nov 18 '17 at 21:33