Connecting noodles probability question

Question

I don't know how to solve this. You have 100 noodles in your soup bowl. Being blindfolded, you are told to take two ends of some noodles (each end of any noodle has the same probability of being chosen) in your bowl and connect them. You continue until there are no free ends. The number of loops formed by the noodles this way is stochastic. Calculate the expected number of circles.

score 16 · Accepted Answer · answered Sep 01 '15 at 17:30

16

Let's do it in a more general case. Let the expected number of loops from $n$ noodles be $E(n)$.

Obviously, $E(1)=1$.

For $n>1$, if you pick up two ends, the possibility for those two ends belongs to the same noodle will be $\frac{1}{2n-1}$. Then you have one loop now and there are $n-1$ noodles to keep on going. Otherwise you get no loop and still have $n-1$ noodles to go(The two you pick before are now connected as one). Hence $E(n)=E(n-1)+\frac{1}{2n-1}$

Inductively you may expect that $E(n) = 1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}$.

answered Sep 01 '15 at 17:30

Asydot

858

So roughly spoken we have $E(75)\approx \pi$. – principal-ideal-domain Sep 01 '15 at 20:47
If you are familiar with harmonic numbers you can express $E(n)$ in such terms as $E(n)=H_{2n}-\frac{1}{2}H_n$. – principal-ideal-domain Sep 02 '15 at 20:45
I have some difficulty understanding why the probability of two ends belong to the same noodle for n noodles is not $\frac{n}{C(n,2)}$ – ZEWEI CHU Dec 03 '23 at 23:57
The sample space has $C(2n, 2)$ outcomes, not $C(n,2)$ (n noodles, 2 ends each). So we get $\frac{n}{\binom{2n}{2}}=\frac{1}{2n-1}$ as expected. – BaroqueFreak Mar 18 '24 at 02:05

score 13 · Answer 2 · edited Feb 17 '20 at 10:19

The accepted answer is correct, but I think should be more carefully detailed.

$E(1) = 1$

Pick up two ends. They either belong to the Same Noodle ($S$) or they don't ($\bar{S}$). The $S$ event can be used to "compute expectations by conditioning" in the same spirit of this question. So:

$$E(n) = E(n | S)p(S) + E(n | \bar{S}) p(\bar{S})$$

Now $p(S) = \frac{1}{2n-1}$ and $p(\bar{S}) = 1-\frac{1}{2n-1}$. Now the trick is that $E(n|S) = E(n-1)+1$ (i.e. if you picked the two ends from the same noodle you've added one loop to the bowl, leaving $n-1$ noodles) and $E(n|\bar{S}) = E(n-1)$ (i.e if you've picked two ends from different noodles you didn't increase the number of loops).

$$E(n) = E(n | S)p(S) + E(n| \bar{S}) p(\bar{S}) = \\ = \left[E(n-1)+1\right]\frac{1}{2n-1} + E(n-1)\left[1-\frac{1}{2n-1}\right] = \\ = E(n-1) + \frac{1}{2n-1}$$

Connecting noodles probability question

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