May Comic Contest + Math Question - Page 2
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SilverskY
Korea (South)3086 Posts
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Malongo
Chile3466 Posts
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PobTheCad
Australia893 Posts
never know what 'em ?? 3rd is best , other 2 suck | ||
seppolevne
Canada1681 Posts
and as for the edit question wouldn't it be (99/100)^4 * (1/100)^2 = 1 in 10 410? | ||
veracity
United States3 Posts
On June 02 2009 22:31 inReacH wrote: The reason I decided to make this post is simply because all 3 winners have the exact same initials. ... But the math on this is pretty crazy.. Now given that the first persons name can be anything because we have no expectation about the odds of the first persons name having the initials that it has.. the odds of the following two names being the exact same initials is, according to my math: 1 in 456976 Now this of course makes the (admittedly false) assumption that all initials in the alphabet are equally likely.. I figured that the initials that are used aren't especially more common than other letters so I decided to omit this factor. By using census data on the relative frequencies of first and last names in the US, I came up with a better estimate: the chance of 3 people having the same first and last initials is: 1 in 37,014 Statistics junkies can find some more details on how I arrived at these numbers in the spoiler below. + Show Spoiler + First I used the data at http://www.census.gov/genealogy/names/ to generate some (approximate) tables mapping letters to their frequency as a first, middle, or last initial. (I made the assumption that first and middle initials have the same distribution because there isn't any data for middle names). + Show Spoiler + First/middle initial frequencies (approx.) A 5.86% B 4.19% C 6.52% D 6.71% E 4.16% F 1.58% G 2.96% H 2.04% I 0.66% J 12.08% K 3.70% L 5.07% M 9.07% N 1.65% O 0.41% P 2.83% Q 0.03% R 7.23% S 5.58% T 3.82% U 0.02% V 1.34% W 2.52% X 0.01% Y 0.20% Z 0.11% Last initial frequencies (approx.) A 3.35% B 9.21% C 7.90% D 4.38% E 1.86% F 3.43% G 5.31% H 7.76% I 0.32% J 3.61% K 2.65% L 4.65% M 9.74% N 1.70% O 1.36% P 4.90% Q 0.19% R 5.82% S 9.28% T 3.57% U 0.16% V 1.25% W 6.73% X 0.01% Y 0.57% Z 0.27% From there I was able to come up with some "interesting" statistics. They should be read as "given the random selection of (x) people from the population, the chance that they have the same (y) initial(s)". 2 people: - first: 0.068 (6.8%) -- compare to (1/26 or 3.8%) assuming all initials equally probable - middle: 0.068 (6.8%) - last: 0.062 (6.2%) - first and last: 0.0042 (0.42% / 1 in 234) - first, middle, and last: 0.00029 (0.029% / 1 in 3,423) 3 people: - first: 0.0058 (0.58%) - middle: 0.0058 (0.58%) - last: 0.0046 (0.46%) - first and last: 2.70e-05 (0.00271% / 1 in 37,014) - first, middle, and last: 1.57e-07 (0.000015% / 1 in 6,337,944) All of these calculations just assume that the probabilities of people having certain first, middle, and last names are all independent events and so the results are obtained by multiplying the individual probabilities together. If letter frequencies weren't taken into account, all of the results would just be the fraction (1/26) raised to various powers... the probability of 2 people having the same first initial would be (1/26)^1 and the probability of 3 people sharing all 3 initials would be (1/26)^6. (NOTE: not (1/26)^2 and (1/26)^9 for reasons listed in the OP... essentially the first person's name doesn't matter -- we only care if the next name we see has the same initials as the one before). Cheers. | ||
Vex
Ireland454 Posts
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inReacH
Sweden1612 Posts
On June 03 2009 21:49 veracity wrote: By using census data on the relative frequencies of first and last names in the US, I came up with a better estimate: the chance of 3 people having the same first and last initials is: 1 in 37,014 Statistics junkies can find some more details on how I arrived at these numbers in the spoiler below. + Show Spoiler + First I used the data at http://www.census.gov/genealogy/names/ to generate some (approximate) tables mapping letters to their frequency as a first, middle, or last initial. (I made the assumption that first and middle initials have the same distribution because there isn't any data for middle names). + Show Spoiler + First/middle initial frequencies (approx.) A 5.86% B 4.19% C 6.52% D 6.71% E 4.16% F 1.58% G 2.96% H 2.04% I 0.66% J 12.08% K 3.70% L 5.07% M 9.07% N 1.65% O 0.41% P 2.83% Q 0.03% R 7.23% S 5.58% T 3.82% U 0.02% V 1.34% W 2.52% X 0.01% Y 0.20% Z 0.11% Last initial frequencies (approx.) A 3.35% B 9.21% C 7.90% D 4.38% E 1.86% F 3.43% G 5.31% H 7.76% I 0.32% J 3.61% K 2.65% L 4.65% M 9.74% N 1.70% O 1.36% P 4.90% Q 0.19% R 5.82% S 9.28% T 3.57% U 0.16% V 1.25% W 6.73% X 0.01% Y 0.57% Z 0.27% From there I was able to come up with some "interesting" statistics. They should be read as "given the random selection of (x) people from the population, the chance that they have the same (y) initial(s)". 2 people: - first: 0.068 (6.8%) -- compare to (1/26 or 3.8%) assuming all initials equally probable - middle: 0.068 (6.8%) - last: 0.062 (6.2%) - first and last: 0.0042 (0.42% / 1 in 234) - first, middle, and last: 0.00029 (0.029% / 1 in 3,423) 3 people: - first: 0.0058 (0.58%) - middle: 0.0058 (0.58%) - last: 0.0046 (0.46%) - first and last: 2.70e-05 (0.00271% / 1 in 37,014) - first, middle, and last: 1.57e-07 (0.000015% / 1 in 6,337,944) All of these calculations just assume that the probabilities of people having certain first, middle, and last names are all independent events and so the results are obtained by multiplying the individual probabilities together. If letter frequencies weren't taken into account, all of the results would just be the fraction (1/26) raised to various powers... the probability of 2 people having the same first initial would be (1/26)^1 and the probability of 3 people sharing all 3 initials would be (1/26)^6. (NOTE: not (1/26)^2 and (1/26)^9 for reasons listed in the OP... essentially the first person's name doesn't matter -- we only care if the next name we see has the same initials as the one before). Cheers. Don't your odds compared to my odds make the assertion that J as first combined with C as last is more than 10 times as likely as the average letters as first and last? | ||
mikeymoo
Canada7170 Posts
Now this of course makes the (admittedly false) assumption that all initials in the alphabet are equally likely.. I figured that the initials that are used aren't especially more common than other letters so I decided to omit this factor. Imagine I roll two dice- these dice have these six numbers on them: 1, 2, 2, 3, 3, 3. 1st die: 1/6 (1), 2/6 (2), 3/6 (3) The chance that the second die lines up with the first is not 1/3 (or 12/36), it is 1/36 + 4/36 + 9/36 = 14/36. I just woke up but your assumption makes no sense to me, and the logic behind it even less so. Comics were good. The ghost one was really going somewhere and then... it didn't. | ||
no_comprender
Australia91 Posts
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