Chocolates, bugs, and socks: Difference between revisions

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Every chocolate Lindsay eats, she gets a picture of a bug, randomly selected from a set of 20 bug pictures. She has collected 19 of the bug pictures.
Every chocolate Lindsay eats, she gets a picture of a bug, randomly selected from a set of 20 bug pictures. She just collected 19 of the bug pictures. How many chocolates do you think she's eaten?


How many chocolates has she probably eaten?
When she's collected 7 unique bugs, each chocolate brings with it a $\frac{13}{20}$ chance of getting an 8-th bug. How many chocolates does she have to eat before that happens, and with what variance? Well, for any event with probability $p$ of success, the expected number of trials before the first success is $\langle n\rangle =\sum_{n=1}^\infty n(1-p)^{n-1}p = \frac 1p$, and the expected square of the number of trials before success is $\langle n^2\rangle =\sum_{n=1}^\infty n^2(1-p)^{n-1}p = \frac{2-p}{p^2}$, so the variance is $\langle n^2\rangle - \langle n\rangle^2 = \frac 1p - 1$. So the expected number of chocolates Lindsay ate is
$$\frac{20}{20}+\frac{20}{19}+\cdots+\frac{20}{2} \approx 20\times(\ln(20) + \frac 12 - 1) \approx 50$$
$$\frac{20}{20}+\frac{20}{19}+\cdots+\frac{20}{2} \approx 20\times(\ln(20) + \frac 12 - 1) \approx 50.$$
She should expect to eat another 20 chocolates before getting that last bug.
(This was my back-of-the-envelope guess using that $\ln(20)\approx 3$, and that the Euler-Mascheroni constant is about 0.5. The actual value is very close to 52.) The number of chocolates eaten to get from 5 to 6 bugs is independent of the number of chocolates eaten to get from 6 to 7 bugs, so the variances add too:
$$(\frac{20}{20} - 1) + (\frac{20}{19} - 1) $$
exact same number!
 
Note: she should expect to eat another 20 chocolates before getting that last bug.


What's the 90% confidence interval around that? For a rough answer, we need to know the variance in the number of trials needed before a success for a probability $p$ event. If that's $v(p)$, then the variance for our number is $v(\frac{20}{20})+v(\frac{20}{19})+\cdots+v(\frac{20}{2})$. You can compute $v(p)$ because we know the distribution for number of trials before success is an exponential. Do that. Also, what about getting the exact 90% CI?
What's the 90% confidence interval around that? For a rough answer, we need to know the variance in the number of trials needed before a success for a probability $p$ event. If that's $v(p)$, then the variance for our number is $v(\frac{20}{20})+v(\frac{20}{19})+\cdots+v(\frac{20}{2})$. You can compute $v(p)$ because we know the distribution for number of trials before success is an exponential. Do that. Also, what about getting the exact 90% CI?

Revision as of 21:39, 5 March 2015

Every chocolate Lindsay eats, she gets a picture of a bug, randomly selected from a set of 20 bug pictures. She just collected 19 of the bug pictures. How many chocolates do you think she's eaten?

When she's collected 7 unique bugs, each chocolate brings with it a $\frac{13}{20}$ chance of getting an 8-th bug. How many chocolates does she have to eat before that happens, and with what variance? Well, for any event with probability $p$ of success, the expected number of trials before the first success is $\langle n\rangle =\sum_{n=1}^\infty n(1-p)^{n-1}p = \frac 1p$, and the expected square of the number of trials before success is $\langle n^2\rangle =\sum_{n=1}^\infty n^2(1-p)^{n-1}p = \frac{2-p}{p^2}$, so the variance is $\langle n^2\rangle - \langle n\rangle^2 = \frac 1p - 1$. So the expected number of chocolates Lindsay ate is $$\frac{20}{20}+\frac{20}{19}+\cdots+\frac{20}{2} \approx 20\times(\ln(20) + \frac 12 - 1) \approx 50.$$ (This was my back-of-the-envelope guess using that $\ln(20)\approx 3$, and that the Euler-Mascheroni constant is about 0.5. The actual value is very close to 52.) The number of chocolates eaten to get from 5 to 6 bugs is independent of the number of chocolates eaten to get from 6 to 7 bugs, so the variances add too: $$(\frac{20}{20} - 1) + (\frac{20}{19} - 1) $$ exact same number!

Note: she should expect to eat another 20 chocolates before getting that last bug.

What's the 90% confidence interval around that? For a rough answer, we need to know the variance in the number of trials needed before a success for a probability $p$ event. If that's $v(p)$, then the variance for our number is $v(\frac{20}{20})+v(\frac{20}{19})+\cdots+v(\frac{20}{2})$. You can compute $v(p)$ because we know the distribution for number of trials before success is an exponential. Do that. Also, what about getting the exact 90% CI?


How many socks do I have? I've just pulled 10 socks out of a pile of clean laundry one at a time. I've made 2 pairs and have 6 singletons. Based on this information, how many pairs of socks are there total (assuming each sock has a unique partner)? Maximum likelihood answer: 11 pairs

If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is $$\binom np \binom{n-p}{k-2p} \binom{2n}{k}^{-1}$$ (first pick the $p$ pairs, then $k-2p$ singletons). Of course, I usually have a pretty good prior on how many pairs of socks are in my laundry.


Coin problem from Richard.