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 | 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 | 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 | ||
$$\frac{20}{20}+\frac{20}{19}+\cdots+\frac{20}{2} \approx 20\times(\ln(20) + \frac 12 - 1) \approx 50$$ | $$\frac{20}{20}+\frac{20}{19}+\cdots+\frac{20}{2} \approx 20\times(\ln(20) + \frac 12 - 1) \approx 50.$$ | ||
(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.