Expected number of throws

$n$ fair coins are thrown all at once. Those that show Head are removed from the game, and the rest are thrown again. This process is repeated until there are no more coins left. What is the expected number of throws? Find a closed form formula, meaning that it may still be expressed as a sum but may not be expressed as a recurrence relation.

Solution Let $f(n)$ be the desired result. There is a $\binom{n}{k} / 2^n$ chance that $k$ of them will show heads, and the process is repeated with $n-k$ coins now. The rest of the game is expected to take $f(n-k)$ throws. If $k=0$ then we still have the same situation as before. If $k=n$, we have 0 throws left. In all of these situations, we do have to spend 1 turn doing the throw. So the recurrence relation is:

$$f(n) = 1 + \sum_{k=1}^{n-1} \frac{\binom{n}{k}}{2^n} f(k) + \frac{1}{2^n}f(n)$$ $$f(n) = \frac{2^n + \sum_{k=1}^{n-1} \binom{n}{k} f(k)}{2^n-1}$$

Given a large $n$, computing this would take $O(n^2)$ time as we have to calculate and store the previous $f(k)$ values, and each take $O(k)$ time. We can manipulate this expression further to come with a summation formula for $f(n)$ that would allow it to be computed in linear time.

First we claim and prove the following lemma. For $1 \leq m < n$

$$\sum_{k=m}^{n} \binom{n}{k}\binom{k}{m} = \binom{n}{m}2^{n-m}$$

Proof: given $n$ white balls, we wish to pick $m$ of them and paint them black, and the rest may stay white or be painted gray. The RHS side is obvious, there are $\binom{n}{m}$ ways to choose black balls, and the rest may be painted white or gray.

To show the LHS, first we pick a number $k \in [m,n]$. We choose $k$ balls and paint them gray. Out of these $k$, we further choose $m$ to be painted black.

Alternatively we can prove the lemma by dividing both sides by $\binom{n}{m}$: $$\sum_{k=m}^{n} \frac{n!}{k!(n-k)!} \frac{k!}{m!(k-m)!} \frac{m!(n-m)!}{n!} = 2^{n-m}$$ $$\sum_{k=m}^{n} \frac{(n-m)!}{(n-k)!(k-m)!} = 2^{n-m}$$ $$\sum_{i=0}^{n-m} \frac{(n-m)!}{(n-m-i)!i!} = 2^{n-m}$$

which is a well-known identity

Given the lemma, now we claim the following statement: let $a_n = \frac{2^n}{2^n-1}$ then: $$f(n) = \binom{n}{1}a_1 - \binom{n}{2}a_2 + \binom{n}{3}a_3 + \dots + (-1)^{n+1} \binom{n}{n}a_n$$ We will show that this formula matches our recurrence relation above by strong induction. Suppose they hold for $1,2,\dots,n-1$, then we first evaluate the following expression: $$\sum_{k=1}^{n-1} \binom{n}{k}f(k) = \binom{n}{1}f(1) + \binom{n}{2}f(2) + \dots + \binom{n}{n-1}f(n)$$ $$= \sum_{k=1}^{n-1} \binom{n}{k} \sum_{m=1}^{k} (-1)^{m+1} \binom{k}{m}a_m$$ we rearrange and collect each $a_m$ terms: $$= \sum_{m=1}^{n-1} (-1)^{m+1}a_m \sum_{k=m}^{n-1} \binom{n}{k} \binom{k}{m}$$ Apply the lemma: $$= \sum_{m=1}^{n-1} (-1)^{m+1}a_m \left(\binom{n}{m}2^{n-m} - \binom{n}{m} \right) = \sum_{m=1}^{n-1} (-1)^{m+1} \binom{n}{m} a_m \left(2^{n-m} - 1 \right)$$ We can plug this expression to our recurrence formula and compare against our claim (induction hypothesis). After multiplying both sides by $2^n-1$, we are left to show that: $$(2^n-1) a_n + \sum_{m=1}^{n-1} (-1)^{m+1} \binom{n}{m} a_m \left(2^{n-m} - 1 \right) = (2^n-1) \left( \sum_{m=1}^{n-1} (-1)^{m+1} \binom{n}{m}a_m + (-1)^{n+1}a_n \right)$$ Collect $a_n$ terms to the LHS, and the rest of the terms to RHS: $$(2^n-1)(1 + (-1)^n)a_n = \sum_{m=1}^{n-1}a_m(-1)^{m+1} c_m$$ where $$c_m = (2^n-1)\binom{n}{m} - \binom{n}{m}(2^{n-m}-1) = \binom{n}{m}(2^n - 2^{n-m}) = \binom{n}{m}2^{n-m}(2^m-1) = \binom{n}{m}2^{n-m} \frac{2^m}{a_m}$$ so what we need to show becomes: $$(2^n-1)(1 + (-1)^n)a_n = \sum_{m=1}^{n-1}a_m(-1)^{m+1} \binom{n}{m}2^{n-m} \frac{2^m}{a_m}$$ $$2^n(1 + (-1)^n) = \sum_{m=1}^{n-1}(-1)^{m+1} \binom{n}{m}2^{n}$$ $$1 + (-1)^n = \sum_{m=1}^{n-1}(-1)^{m+1} \binom{n}{m}$$ $$1 + \sum_{m=1}^{n-1}(-1)^{m} \binom{n}{m} + (-1)^n = 0$$ $$(1 -1)^n = 0$$