4 ms·

as someone who has kept track of their skittles color distribution over the course of a year, I can attest that there can be wide fluctuations in distribution p

by viggity 14y ago

as someone who has kept track of their skittles color distribution over the course of a year, I can attest that there can be wide fluctuations in distribution percentages. I wish I had kept all the numbers, but IIRC, the "typical" bag would have roughly 55-60 skittles, so the "average" color would have 11 skittles. It was common to see a single color with only 9 skittles. I can't recall ever having a bag have one color with less than 7 of any one color, but I would have on occasional see one color get as many as 18 of one color. Sincerely yours, a gigantic nerd.

bballbackus 14y ago
> as someone who has kept track of their skittles color distribution over the course of a year I mean did you not expect someone ask why? Also, why?
- viggity 14y ago
  1. I like the taste having an even distribution (i'd eat one of each color at the same time). There was a stupid little pattern on how on how I'd eat the extras to get each color to have an equal number. 2. I started a very statistics heavy programming job, so I thought it'd be fun to run some numbers on something novel.
mturmon 14y ago
5 colors, so: P(color) = 0.2 = p Expected number of, say, blue, in N = 55 skittles is: E(N_blue) = Np = 11 Standard deviation: Sdev(N_blue) = sqrt(N p (1-p)) = 2.95 So you would expect to see 11 +/- 3 skittles. (If you take the largest discrepancy of the 5 colors, it's harder to compute.) If you want to look at a large deviation from the expected: P(Nblue < 7) ~= P(Nblue = 6) = choose(55,6) p^6 (1-p)^49 = 3.3% which is pretty significant, especially considering you were looking at large deviations over any of five colors. In the other direction: P(Nblue > 17) ~= P(Nblue = 18) = choose(55,18) p^18 (1-p)^37 = 0.98% which is still pretty significant. With probabilities this small, and events like these that are lightly-correlated, the second-order terms in the relevant Bonferroni inequality will be pretty small, so P(any color = 18) ~= 5 * P(Nblue = 18) ~= 5% (You could do all these calculations more exactly, but they are within a factor of 2.)

by viggity 14y ago

bballbackus 14y ago

> as someone who has kept track of their skittles color distribution over the course of a year I mean did you not expect someone ask why? Also, why?

viggity 14y ago
1. I like the taste having an even distribution (i'd eat one of each color at the same time). There was a stupid little pattern on how on how I'd eat the extras to get each color to have an equal number. 2. I started a very statistics heavy programming job, so I thought it'd be fun to run some numbers on something novel.

mturmon 14y ago

5 colors, so: P(color) = 0.2 = p Expected number of, say, blue, in N = 55 skittles is: E(N_blue) = Np = 11 Standard deviation: Sdev(N_blue) = sqrt(N p (1-p)) = 2.95 So you would expect to see 11 +/- 3 skittles. (If you take the largest discrepancy of the 5 colors, it's harder to compute.) If you want to look at a large deviation from the expected: P(Nblue < 7) ~= P(Nblue = 6) = choose(55,6) p^6 (1-p)^49 = 3.3% which is pretty significant, especially considering you were looking at large deviations over any of five colors. In the other direction: P(Nblue > 17) ~= P(Nblue = 18) = choose(55,18) p^18 (1-p)^37 = 0.98% which is still pretty significant. With probabilities this small, and events like these that are lightly-correlated, the second-order terms in the relevant Bonferroni inequality will be pretty small, so P(any color = 18) ~= 5 * P(Nblue = 18) ~= 5% (You could do all these calculations more exactly, but they are within a factor of 2.)