Chocolates, bugs, and socks - Revision history

Anton: Fix an error in sock likelihood function

2026-01-05T06:05:46Z

Fix an error in sock likelihood function

← Older revision		Revision as of 22:05, 4 January 2026
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	If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is		If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is
	$$\binom np \binom{n-p}{k-2p} \binom{2n}{k}^{-1}$$		$$\binom np \binom{n-p}{k-2p} 2^{k-2p} \binom{2n}{k}^{-1}$$
	(first pick the $p$ pairs, then $k-2p$ singletons). Maximum likelihood answer to the original question: ~~[http://www.wolframalpha.com/input/?i=plot+%28n+choose+p%29++%28%28n-p%29+choose+k-2p%29+%2F+%282*n+choose+k%29+for+k%3D10%2C+p%3D2%2C+n%3D7+to+15~~ 11 pairs] (so $22-10=12$ socks left). Of course, you'd usually have a pretty good prior multiplying this likelihood function.		(first pick the $p$ pairs, then $k-2p$ of the remaining pair from which to pick singletons, and for each singleton select whether you're taking the left or right sock from the pair). Maximum likelihood answer to the original question: 11 pairs (so $22-10=12$ socks left). Of course, you'd usually have a pretty good prior multiplying this likelihood function.

Anton at 16:32, 7 March 2016

2016-03-07T16:32:22Z

← Older revision		Revision as of 08:32, 7 March 2016
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	[[Category:Blog]]		[[Category:Blog]]

	==How many chocolates has Lindsay eaten?==		==How many chocolates has Lindsay eaten?==

Anton at 16:31, 7 March 2016

2016-03-07T16:31:57Z

← Older revision		Revision as of 08:31, 7 March 2016
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			[[Category:Blog]]

	==How many chocolates has Lindsay eaten?==		==How many chocolates has Lindsay eaten?==

Anton: /* How many socks do I have? */

2016-01-03T17:16:28Z

How many socks do I have?

← Older revision		Revision as of 09:16, 3 January 2016
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	If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is		If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is
	$$\binom np \binom{n-p}{k-2p} \binom{2n}{k}^{-1}$$		$$\binom np \binom{n-p}{k-2p} \binom{2n}{k}^{-1}$$
	(first pick the $p$ pairs, then $k-2p$ singletons). Maximum likelihood answer to the original question: [http://www.wolframalpha.com/input/?i=plot+%28n+choose+p%29++%28%28n-p%29+choose+k-2p%29+%2F+%282*n+choose+k%29+for+k%3D10%2C+p%3D2%2C+n%3D7+to+15 11 pairs]. Of course, you'd usually have a pretty good prior multiplying this likelihood function.		(first pick the $p$ pairs, then $k-2p$ singletons). Maximum likelihood answer to the original question: [http://www.wolframalpha.com/input/?i=plot+%28n+choose+p%29++%28%28n-p%29+choose+k-2p%29+%2F+%282*n+choose+k%29+for+k%3D10%2C+p%3D2%2C+n%3D7+to+15 11 pairs] (so $22-10=12$ socks left). Of course, you'd usually have a pretty good prior multiplying this likelihood function.

Anton: Anton moved page A few fun calculations to Chocolates, bugs, and socks without leaving a redirect

2015-03-08T17:48:19Z

Anton moved page A few fun calculations to Chocolates, bugs, and socks without leaving a redirect

← Older revision	Revision as of 09:48, 8 March 2015
(No difference)

Anton: Anton moved page Two fun calculations to A few fun calculations without leaving a redirect

2015-03-08T17:44:39Z

Anton moved page Two fun calculations to A few fun calculations without leaving a redirect

← Older revision	Revision as of 09:44, 8 March 2015
(No difference)

Anton at 17:44, 8 March 2015

2015-03-08T17:44:21Z

← Older revision		Revision as of 09:44, 8 March 2015
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	Note: she should expect to eat another 20 chocolates before getting that last bug.		Note: she should expect to eat another 20 chocolates before getting that last bug.

			==How many unique bugs are there?==
			Suppose Lindsay '''knows''' she's eaten 50 chocolates, among which there were 19 unique bugs, but she doesn't know how many unique bugs there are out there. If she's eaten n chocolates and found k unique bugs, and there are b unique bugs, the likelihood is
			$$
			\binom{b}{k}k^{n-k}b^{-n}
			$$
			(pick the subset of k, which also uses up k of your chocolates, then distribute the remaining n-k chocolates among those k bugs). For n=50 and k=19, the maximally likely answer is that there are [http://www.wolframalpha.com/input/?i=plot+%28b+choose+k%29+*+k%5E%28n-k%29%2F%28b%5En%29+for+n%3D50%2C+k%3D19%2C+b%3D19+to+22 20 bugs].

	==How many socks do I have?==		==How many socks do I have?==
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	If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is		If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is
	$$\binom np \binom{n-p}{k-2p} \binom{2n}{k}^{-1}$$		$$\binom np \binom{n-p}{k-2p} \binom{2n}{k}^{-1}$$
	(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~~. Maximum likelihood answer to the original question: [http://www.wolframalpha.com/input/?i=plot+%28n+choose+p%29++%28%28n-p%29+choose+k-2p%29+%2F+%282*n+choose+k%29+for+k%3D10%2C+p%3D2%2C+n%3D7+to+15 11 pairs]. Of course, you'd usually have a pretty good prior multiplying this likelihood function.		(first pick the $p$ pairs, then $k-2p$ singletons). Maximum likelihood answer to the original question: [http://www.wolframalpha.com/input/?i=plot+%28n+choose+p%29++%28%28n-p%29+choose+k-2p%29+%2F+%282*n+choose+k%29+for+k%3D10%2C+p%3D2%2C+n%3D7+to+15 11 pairs]. Of course, you'd usually have a pretty good prior multiplying this likelihood function.

Anton: /* How many chocolates has Lindsay eaten? */

2015-03-06T06:54:24Z

How many chocolates has Lindsay eaten?

← Older revision		Revision as of 22:54, 5 March 2015
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	==How many chocolates has Lindsay eaten?==		==How many chocolates has Lindsay eaten?==

	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?		Every chocolate Lindsay eats, she gets a picture of a bug, randomly selected from a set of 20 bug pictures. She just collected her 19-th unique bug picture. 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		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

Anton: Anton moved page Fun problems to Two fun calculations without leaving a redirect

2015-03-06T06:44:33Z

Anton moved page Fun problems to Two fun calculations without leaving a redirect

← Older revision	Revision as of 22:44, 5 March 2015
(No difference)

Anton at 06:42, 6 March 2015

2015-03-06T06:42:33Z

← Older revision		Revision as of 22:42, 5 March 2015
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	~~'''~~How many chocolates has Lindsay eaten?~~'''~~ 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 Lindsay eaten?==

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

	~~----~~		==How many socks do I have?==

	~~'''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: [http://www.wolframalpha.com/input/?i=plot+%28n+choose+p%29++%28%28n-p%29+choose+k-2p%29+%2F+%282*n+choose+k%29+for+k%3D10%2C+p%3D2%2C+n%3D7+to+15 11 pairs]~~		I've just pulled 10 socks out of a pile of laundry, making 2 pairs and 6 singletons. How many pairs of socks are there total (assuming each sock has a unique partner)?

	If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is		If there are $n$ pairs of socks, I pull out $k$, and make $p$ pairs, the likelihood is
	$$\binom np \binom{n-p}{k-2p} \binom{2n}{k}^{-1}$$		$$\binom np \binom{n-p}{k-2p} \binom{2n}{k}^{-1}$$
	(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.		(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. Maximum likelihood answer to the original question: [http://www.wolframalpha.com/input/?i=plot+%28n+choose+p%29++%28%28n-p%29+choose+k-2p%29+%2F+%282*n+choose+k%29+for+k%3D10%2C+p%3D2%2C+n%3D7+to+15 11 pairs]. Of course, you'd usually have a pretty good prior multiplying this likelihood function.

	----

	~~Coin problem from Richard~~.