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I know that the probability that k distinct number catch from a set of n number are catching in ascending order is $\frac{1}{k!}$. But i don't know what is the probability if the number can are no distinct. I don't know that there is a formula for this. Thanks for the help.

StCS
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  • I am sorry if the question is write very well, but my english isn't very good. – StCS Jun 01 '23 at 12:04
  • If you have 3, 7, 7, 10 , do you consider it is ascending order ( equals is considered as succes) or not ? – Lourrran Jun 01 '23 at 12:07
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    It depends on the numbers. You should be able to do this if you specify how often each number occurs. – lulu Jun 01 '23 at 12:10
  • And I think that this question has no answer. You select k numbers. Example k=10. If you select k numbers (integers) uniformly in range (1,20), the probability to have twice same number is not nul, and the general formula $\frac{1}{k!}$ cannot be used. But if you select 10 numbers uniformly in range (1,100000000), you can use the general formula $\frac{1}{k!}$ – Lourrran Jun 01 '23 at 12:10
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    To make a well posed problem out of this: Say we are choosing $k$ numbers uniformly and independently from ${1, \cdots, n}$. What is the probability that you choose them in increasing order? – lulu Jun 01 '23 at 12:11
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    That one I would do recursively, conditioned on the first choice. – lulu Jun 01 '23 at 12:13
  • The version posed by lulu can also be approached with stars-and-bars. – JMoravitz Jun 01 '23 at 12:13
  • @Lourran, Yes I consider 3, 7, 7, 10 in ascending order – StCS Jun 01 '23 at 12:35
  • Thanks, now I will serch something about stars-and-bars – StCS Jun 01 '23 at 12:37

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Since you are allowing repeats by specifying that (some/all) numbers are not distinct, obviously they can only be in weakly ascending order. Also note that each selection you thus make can be arranged in weakly ascending order in just one way.

Then suppose we choose a total of $k$ numbers from $n$ distinct numbers available in abundance, then by stars and bars,(note that here $k$ are the balls to be put in $n$ bins, (=$k$ stars, $n-1$ bars)

$\dbinom{k+n-1}{k}$ selections in weakly ascending order against $n^k$ total possible selections,

thus $Pr = \dfrac{\dbinom{k+n-1}{k}}{n^k}$

true blue anil
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  • Thanks for your explanation which I found very clear :) – StCS Jun 01 '23 at 13:34
  • Glad to have been of help ! Cheers ! $;; $:) – true blue anil Jun 01 '23 at 13:37
  • Btw, you could also have used combinations with repetitions instead of stars and bars, as explained in this link, but stars and bars is more popular. https://math.stackexchange.com/questions/4243328/how-do-i-make-sense-of-combinations-with-repetition/4243537#4243537 – true blue anil Jun 01 '23 at 13:41