June 14th, 2017, 07:55 AM  #1 
Newbie Joined: Jun 2017 From: Houston Posts: 4 Thanks: 0  What is the probability of getting the same number twice?
I randomly pick 24 numbers. Each number ranges from 1 to 255. It is possible to pick the same number in the range 1255 multiple times. For instance, Random pick #1: 10 Random pick #2: 7 ...etc... Random pick #24: 245 What is the probability that within the 24 picks any 2 numbers are identical? For instance, pick #1 and pick #20 are both 10 Thank you. 
June 14th, 2017, 08:44 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,817 Thanks: 760 
Obviously "with replacement", right? ONLY 2 numbers are identical? Does >2 (like 3 7's) qualify? 
June 14th, 2017, 11:11 AM  #3 
Newbie Joined: Jun 2017 From: Houston Posts: 4 Thanks: 0 
Yes, I think it is with replacement. When I pick the 1st number, that number had a 1/255 probability of being chosen. When I pick the 2nd number, it also had a 1/255 probability of being chosen...etc...

June 14th, 2017, 11:20 AM  #4 
Newbie Joined: Jun 2017 From: Houston Posts: 4 Thanks: 0 
Question: "ONLY 2 numbers are identical? Does >2 (like 3 7's) qualify?" Yes, I'm asking about only 2 of the 24 numbers being identical (ex.: two 7's) As a separate exercise, it would also be valuable to know the probability of: i) finding 3 identical numbers (ex.: three 7's) ii) finding 2 pairs of identical numbers (ex.: two 7's and two 8's) I'm wondering if i) > ii) ... 
June 14th, 2017, 12:53 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,817 Thanks: 760  
June 14th, 2017, 02:22 PM  #6 
Newbie Joined: Jun 2017 From: Houston Posts: 4 Thanks: 0 
Yes

June 14th, 2017, 05:16 PM  #7 
Senior Member Joined: May 2016 From: USA Posts: 916 Thanks: 366 
Think about a similar but simpler problem. Numbers 1 through 6. Pick three. How many ways can you get All numbers different = $6 * 5 * 4 = 120.$ Two numbers the same (be careful: the different number can come first, second or third) = $6 * 5 * 3 = 90.$ Three numbers the same = $6.$ $120 + 90 + 6 = 216.$ Let's check. 6 ways to get first. 6 ways to get second. 6 ways to get third. So total = $6 * 6 * 6 = 36 * 6 = 216.$ Probability all three different = $\dfrac{120}{216} = \dfrac{5}{9}.$ Probability two the same = $\dfrac{90}{216} = \dfrac{5}{12}.$ Probability three the same = $\dfrac{6}{216} = \dfrac{1}{36}.$ Any ideas now? 
June 16th, 2017, 04:50 AM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,817 Thanks: 760 
Jeff, do you consider (as example) 7,7,7 as nonpairs, or as 3 pairs, thus does not qualify? In other words: does only a pair + 22 different other numbers qualify? 
June 16th, 2017, 08:29 AM  #9 
Senior Member Joined: May 2016 From: USA Posts: 916 Thanks: 366 
Denis Being serious about a question that does not seem to be well thought out, I'd start by asking what is the probability of at least one repeat. Within that population of at least one repeat, there is a myriad of possibilities. You could have anywhere from one to twelve repeats each of length two, or eight repeats each of length three, or nine repeats, six of length three plus three of length two. $\text {P(no repeats) } = \displaystyle \prod_{j=1}^{24}\dfrac{256j}{255} \approx 0.327.$ $\text {P(at least one repeat) } \approx 1  0.327 = 0.673.$ Breaking that 67% into meaningful components is work. You don't assume that I can do THAT do you? EDIT: I suppose the best way to attack the 67% is to analyze in terms of number of unrepeated numbers. At 22 unrepreated numbers, the only possibility would be one repeat of length two. At 21 unrepeated numbers, the only possibility would be one repeat of length three. At 20 unrepeated numbers, there would be two possibilities, one repeat of length four or two repeats of length two. I expect the probabilties would drop rapidly so you could most likely address the bulk of the 67% with not too many options. Last edited by JeffM1; June 16th, 2017 at 08:39 AM. 
June 16th, 2017, 09:50 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,817 Thanks: 760 
Well, whatever... IF, IF, IF the 24 draws produce: 1 pair no other number matching that pair 22 remaining are ALL different then probability = ~.39 

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