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What is the probability of the same event happening to two (or more) different independent people within a specified time frame?

For example, if I wave my hands to my friend, there is a probability of 1/2 that he will wave back after a second. Or, he will wave back after 2 seconds with the same probability. What is the probability of two of my friends waving back at the same time at a specific instant (assuming the same probability and timing of 1-2 seconds)? How about three or more? 

time     p1    p2     probability 
 1s     wave  wave        1/2     -> probability of both p1 and p2 to wave back here! 
 2s     wave  wave        1/2     -> or here

I know the more people there are, the higher the probability for more of my friends to wave back to me at the same time. But how do I prove this mathematically?

Ahmed Al-haddad
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I am not sure that I understood your question right. We can consider the following general model. Your meet $n$ friends and wave hands to them. Then each of them have will wave back, after $1$, $2$, ... or $n$ seconds, with equal probability $p=1/n$. There are $n^n$ possible waving delay sequences $(d_1,\dots, d_n)$, where $d_i$ is a waving delay for each friend, and each of these sequences has probability $n^{-n}$. Then the probability that all friends will wave back to you at the same time is $n^{1-n}$, since there are only $n$ such waving delay sequences, namely, $(1,1,\dots,1), (2,2,\dots,2),\dots, (n,n,\dots,n)$.

Alex Ravsky
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    Thanks Alex, I believe you understood the question. If I use $s$ to denote the seconds, then is it gonna be $n^s$ for the possible delay sequences? I just got a bit lost... Also, how do you envision the solution if I consider delay a continuous variable? For example, within 1 second. – Ahmed Al-haddad May 14 '19 at 09:48
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    @AhmedAl-haddad If we have choices ${1,2,\dots,s}$ for the seconds of wave back then for each friend there are $s$ possible distinct waving delays. Thus, by the rule of product, there are $n^s$ possible waving delay sequences. – Alex Ravsky May 14 '19 at 10:11
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    If delay is a continuous variable, say from $0$ to $1$ seconds, then the set of possible waving delay sequences is a set of points of an $n$-dimensional cube $[0,1]^n$. Waving delay sequences with all friends waving back to you at the same time $t$ correspond to points of the diagonal ${(t,t,\dots,t)\in [0,1]^n: t\in [0,1]}$. If we endow the cube with the uniform probability measure (which is the usual (Lebesgue) measure) then the probability that all friends will wave back to you at the same time will be equal to the measure of the diagonal, which is zero. – Alex Ravsky May 14 '19 at 10:11
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    Thank you Alex. I am not a mathematician, could you please give me references to explore and understand the last comment better ($n$-dimensional cube $[0,1]^n$ part and beyond)? Also, why is the measure of the diagonal zero? – Ahmed Al-haddad May 14 '19 at 10:36
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    @AhmedAl-haddad The idea of $n$-dimensional cube (endowed the uniform probability measure) is geometric probability, “the idea of looking at probability in terms of lengths, area, or volumes. See also the classical meeting probability problem. – Alex Ravsky May 14 '19 at 11:08
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    The diagonal has measure zero (if $n\ge 2$), because the measure is $n$-dimensional volume and the diagonal is one-dimensional segment. I can provide a rigorous (and complicated) proof if this fact, but the intuition about it is similar to the following zero values: the length of a point of a straigth line, the area of a segment of a plane, or the volume of a square in a space. I guess that $n$-dimensional volume of an (at most) $n-1$ dimensional figure is zero. – Alex Ravsky May 14 '19 at 11:08