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 June 26th, 2019, 09:30 AM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 675 Thanks: 88 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? June 27th, 2019, 02:06 PM   #2
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Joined: Sep 2015
From: USA

Posts: 2,649
Thanks: 1476

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.
Attached Images Clipboard01.jpg (59.5 KB, 3 views) June 28th, 2019, 06:33 PM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 675 Thanks: 88 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 June 28th, 2019, 06:38 PM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,649 Thanks: 1476 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. Tags numbers, probability, times Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post enigmatic Probability and Statistics 5 May 22nd, 2019 07:24 PM EvanJ Probability and Statistics 0 November 24th, 2017 10:03 AM EvanJ Probability and Statistics 8 June 3rd, 2016 07:57 PM anicicn Probability and Statistics 5 February 27th, 2015 04:17 AM Mr Davis 97 Probability and Statistics 1 January 26th, 2015 03:58 PM

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