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June 26th, 2019, 08:30 AM   #1
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Probability With the Numbers From 1 to 4 Four Times Each

There are 16 balls, with each number from 1 to 4 four times. Balls are drawn without replacement. How many balls need to be drawn to make at least a 50 percent chance that any one number will have all four balls drawn?
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June 27th, 2019, 01:06 PM   #2
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Well...

I've got a method that works but it's pretty brute force.

For each $n$ you come up with a 4 digit integer partition of $n$
such that each digit is at most 3

Then evaluate the probabilities of each of those and sum them.

Subtract the result from 1. This is the probability that at least 1 of the labels

appears 4 times in the $n$ selections.

$p(\{1,2,\dots,16\}) = \left\{0,0,0,\frac{1}{455},\frac{1}{91},\frac{3}{9 1},\frac{1}{13},\frac{329}{2145},\frac{3}{11}, \frac{63}{143},\frac{59}{91},\frac{391}{455},1,1,1 ,1 \right\}$

$n=10 \Rightarrow p(10)=\dfrac{59}{91}$ is the first to be greater than 50%.

These answers do accord with the totally brute force method of having Mathematica gen up all the possible arrangements and counting them, so
I have pretty high confidence in them.

Whether there's a more elegant solution I don't know.
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June 28th, 2019, 05:33 PM   #3
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For the probabilities other than 0 and 1, I gave them the least common denominator of 15,015. I'm wondering if the numerators can be expressed in n P r form or otherwise show a pattern.

1/455 = 33/15,015
1/91 = 165/15,015
3/91 = 495/15,015
1/13 = 1,155/15,015
329/2145 = 2,303/15,015
3/11 = 2,095/15,015
63/143 = 6,615/15,015
59/91 = 9,735/15,015
391/455 = 12,903/15,015
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June 28th, 2019, 05:38 PM   #4
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I looked up the numerator sequence when put over a common denominator on OEIS.

The sequence was embedded in a half dozen sequences but didn't seem to have it's own.
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