Let's say we have two ordered 4-tuple and each has 4 numbers like $t_1: (n_1,n_2,n_3,n_4)$ and $t_2: (m_1,m_2,m_3,m_4)$. range for $n_i$ and $m_i$ is $[0,100]$ (whole numbers so no decimals). I want to know what is the probability that the two of them are same if the numbers are selected from the given range i.e. $n_1=m_1$, $n_2=m_2$, $n_3=m_3$, $n_4=m_4$.

Some notes: We are choosing with replacement and order does matter.

P.S. This is my first question, I may have butchered the norm on how to ask proper questions. The reason I am asking this question is I am writing a program that fails when both sets are same and wanted to know what is the probability of my program breaking.

  • 2
    $\begingroup$ This is too vague. Are you choosing with or without replacement? Does order matter? That is, is $(1,2,3,4)$ the same as $(4,3,2,1)$? $\endgroup$
    – lulu
    Jun 15 at 20:42
  • $\begingroup$ If it was without replacement, then the two sets would never have the same numbers. Am I mistaken, or is this obvious? $\endgroup$
    – Doug
    Jun 15 at 20:51
  • $\begingroup$ @Doug I meant "is each set chosen without replacement", not the entire $8$. In which case the answer (if unordered) would be $1\big / \binom {100}4$. Though I don't believe the OP means "set ", which would normally be unordered, but rather an ordered $4-$tuple. In any case, the OP ought to clarify. $\endgroup$
    – lulu
    Jun 15 at 20:53
  • $\begingroup$ to clarify some stuff, order matters as in (1,2,3,4) is different than(4,3,2,1), choosing with replacement. @Doug now that I think of it 4-tuple would be a better description. $\endgroup$
    – Dabhi
    Jun 16 at 18:20
  • $\begingroup$ And I am quite sure we saw this question and answer pair a few hours ago. $\endgroup$
    – amWhy
    Jun 16 at 18:42


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