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 All the colors in (Posted on 2010-04-26)
Seven sportsmen (named A,B,C,D,E,F and G) buy 7 hats choosing randomly out of 4 available colors (say G,R,W and Y).
What is the probability that all the colors were chosen?

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 approximation (spoiler) | Comment 2 of 21 |
Let
a(n) = probability that there is only 1 color among n hats
b(n) = probability that there are exactly 2 colors among n hats
c(n) = probability that there are exactly 3 colors among n hats
d(n) = probability that there are exactly 4 colors among n hats

Then a(1) = 1
b(1) = c(1) = d(1) = 0

Iteratively,
a(n+1) = a(n)*.25
b(n+1) = a(n)*.75 + b(n)*.50
c(n+1) =                  b(n)*.50 + c(n)*.75
d(n+1) =                                  c(n)*.25 + d(n)

I implemented this in Excel, and got the following table for n = 1 through 20

n      a(n)       b(n)      c(n)       d(n)
1  1.000000
2   .250000 .750000
3   .062500 .562500 .375000
4   .015625 .328125 .562500 .093750
5   .003906 .175781 .585938 .234375
6   .000977 .090820 .527344 .380859
7   .000244 .046143 .440918 .512695
8   .000061 .023254 .353760 .622925
9   .000015 .011673 .276947 .711365
10 .000004 .005848 .213547 .780602
11 .000001 .002927 .163084 .833988
12 .000000 .001464 .123776 .874759
13 .000000 .000732 .093564 .905703
14 .000000 .000366 .070539 .929094
15 .000000 .000183 .053088 .946729
16 .000000 .000092 .039907 .960001
17 .000000 .000046 .029976 .969978
18 .000000 .000023 .022505 .977472
19 .000000 .000011 .016890 .983098
20 .000000 .000006 .012673 .987321

The requested probability, d(7), is approximately .512695.

And yes, I know this can be done exactly using combinatorics.

 Posted by Steve Herman on 2010-04-26 13:43:43

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