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June 14th, 2017, 08:55 AM   #1
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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 1-255 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

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June 14th, 2017, 09:44 AM   #2
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Obviously "with replacement", right?

ONLY 2 numbers are identical?
Does >2 (like 3 7's) qualify?
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June 14th, 2017, 12:11 PM   #3
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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...
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June 14th, 2017, 12:20 PM   #4
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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) ...
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June 14th, 2017, 01:53 PM   #5
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Quote:
Originally Posted by tguck View Post
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)
So:
2 7's
no other 7's
the remaining 22 numbers are ALL different

Yes?
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June 14th, 2017, 03:22 PM   #6
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Yes
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June 14th, 2017, 06:16 PM   #7
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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?
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June 16th, 2017, 05:50 AM   #8
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Jeff, do you consider (as example) 7,7,7 as non-pairs,
or as 3 pairs, thus does not qualify?
In other words:
does only a pair + 22 different other numbers qualify?
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June 16th, 2017, 09:29 AM   #9
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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.

$\text {P(no repeats) } = \displaystyle \prod_{j=1}^{24}\dfrac{256-j}{255} \approx 0.327.$

$\text {P(at least one repeat) } \approx 1 - 0.327 = 0.673.$

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.
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Last edited by JeffM1; June 16th, 2017 at 09:39 AM.
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June 16th, 2017, 10:50 AM   #10
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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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