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Assuming that you start out with a root node, and decide with 50% probability whether or not to add two children nodes. If they do, repeat this process for them. How can you find the average length of this random binary tree?

I'm thinking along the lines of

$\displaystyle\lim_{n\to\infty}\sum\limits_{i=1}^n (\frac{n}{2})^2$.

because 1/2 * n represents 50% probability, and I'm squaring it because the tree gets exponentially larger. However, I feel like I've done something terribly wrong (probably because I have). Can anyone give me some help?

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So you would expand your graph 'breadth-first', one generation at a time and the 'length' is the number of the last generation before it dies? Intuitively it seems the $n+1$'th generation would have $2\cdot\tfrac{1}{2}$ times as many nodes as the $n$'th so there would 'on average' be infinitely many generations. But that's just silly intuition. – Myself Mar 3 '11 at 3:54
@Myself: That was my feeling too, that the "average" diverges. Lets see if a proof presents itself. – Eric Naslund Mar 3 '11 at 4:01
+1 @Eric and @Myself. – user17762 Mar 3 '11 at 4:04
@Eric and @Myself are correct that the expected number of nodes is 1 in each generation, but that need not always imply an infinite expected height. To take a different example, suppose that each entire generation either doubled or was wiped out, then the expected number of nodes in each generation would be 1 ad infinitum but the expected number of generations would be 1. – Henry Mar 3 '11 at 6:41

Hint: These problems can be done by what is known in probability as the first step analysis. Assume that the expected length is $L$. Take one step and try to come up with a relation involving $L$ and solve for $L$.

If you find first step analysis difficult, think along the lines of what the probability that the length of the tree is $L$ and then use this to find expectation. For instance, the probability the length is $0$ is $\frac{1}{2}$, the probability the length is $1$ is $\frac{1}{2} \times \frac{1}{4}$ and so on. Try to find a pattern and sum it up.

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isn't 1/2 the probability that the length is 0? – Eric Naslund Mar 3 '11 at 3:56
oops sorry. will edit it accordingly. – user17762 Mar 3 '11 at 3:58
+1 because the second paragraph actually tells how to solve the problem. But I'm not sure what the first paragraph means, don't you need more than just the probability the length will be $L$, but also probabilities that each possible number of nodes will be achieved at level $L$? – Myself Mar 3 '11 at 4:02
Doesn't it get much more complicated. For example the probability the tree is of height 2 is $\frac{1}{4} \times \frac{1}{4} + \frac{1}{16} \times \frac{1}{8}$, and that the tree is of height 3 is something like $\frac{1}{4} \times \frac{5}{32} + \frac{1}{16} \times \frac{7}{64} + \frac{1}{64} \times \frac{1}{32} + \frac{1}{128} \times \frac{1}{256}$. I would not say the pattern was obvious. – Henry Mar 3 '11 at 6:33

Here are some empirical assertions:

If the probability that the tree is of height $n$ is $P_n$ then $P_0 = 1/2$ and $$P_{n+1} = P_n \left( 1 - \sqrt{2 P_n} + \frac{P_n}{2} \right).$$ [This is related to OEIS A076628 where $P_n = 2 b(n+1)^2$]

Since for $n>10$, $$\dfrac{1}{n^2} < P_n < \dfrac{2}{n^2}$$ [asymptotically $P_n$ is $2/n^2$], the expectation of the height of the tree is infinite.

If this is homework, there is probably an easier method.

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I downvoted this accidentally, and now I can't remove my downvote until somebody (not me, apparently) edits it. Sorry – TonyK Mar 3 '11 at 13:43
@TonyK: so I have edited it – Henry Mar 3 '11 at 14:12
can you explain a bit why you got $$P_{n+1} = P_n \left( 1 - \sqrt{2 P_n} + \frac{P_n}{2} \right)?$$ – Qiang Li Mar 4 '11 at 2:32
I am curious as well, where did the recurrence come from? – Eric Naslund Mar 4 '11 at 3:47
It comes from pattern recognition, i.e. working out the early probabilities as 0.5, 0.125, 0.0703125, 0.046417236328125 and then about 0.0333517645485699, 0.0252941656239401, 0.0199249372625329, 0.0161459365039410. The first few are $1/2^1$, $1/2^3$, $9/2^7$, $1521/2^{15}$ where the denominators are $2^{2^{n+1}-1}$ so the next could be $71622369/2^{31}$. The numerators are squares with square roots 1, 1, 3, 39, 8463. Looking at OEIS gives a suggestion which is consistent with the next numbers, and a recurrence $b_{n+1}=b_n-b_n^2$ where $P_n=2b_{n+1}^2$. So I replaced $b$ by $P$. "Empirical" – Henry Mar 4 '11 at 7:38

With probability 1/2, a random binary tree consists only of the root node. Otherwise, it consists of two branches of height 1 with independent random binary subtrees hanging from each of them.

Let $H$ denote the height of the random binary tree and set $h(n):=\mathbb{P}(H>n)$, the probability that the tree's height exceeds $n$. The tree's height exceeds $n+1$ if it has more than the root node, and at least one of the subtrees has height exceeding $n$. It follows that $$h(0)={1\over2},\qquad h(n+1)={1\over 2}-{1\over 2}[1-h(n)]^2,\quad n\geq 0.\tag1$$

Now, $h(n)$ is decreasing and by equation (1) converges to some root of $s={1\over 2}-{1\over 2}(1-s)^2$. In other words, $h(n)\to 0$ so the tree has finite length: $\mathbb{P}(H<\infty)=1$.

To investigate $\mathbb{E}(H)$, let's rewrite equation (1) as $${1\over h(n+1)}- {1\over h(n)}={1\over 2}+{h(n)\over2(2-h(n))}.\tag2$$ Adding these increments and dividing by $N$ gives $${1\over Nh(N)}- {1\over Nh(0)}={1\over 2}+{1\over N}\sum_{n=0}^{N-1} {h(n)\over2(2-h(n))}.\tag3$$ Letting $N\to\infty$ in (3) shows that $Nh(N)\to 2$. In particular, $h(N)>1/N$ for large $N$ and hence
$$\mathbb{E}(H)=\sum_{N=0}^\infty \mathbb{P}(H>N)=\infty.$$

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It is worth noting that my $P_n = h(n) - h(n-1)$ in your terms – Henry Sep 12 '12 at 9:34
@Henry $P_n=h(n-1)-h(n)$ actually, but yeah, the two approaches are pretty similar. – Byron Schmuland Sep 12 '12 at 16:24

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