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None of the numbers can be duplicated and order doesn't matter. I tried to do 7/99 and 2 to get it but it doesn't seem right

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    $\begingroup$ This is a matter of psychology (and in fact has been studied by psychologists), because it's an empirical fact that people do not select numbers randomly or even uniformly. Are you interested in the real probability or is this just a way of framing a purely mathematical question? $\endgroup$
    – whuber
    Jan 16, 2020 at 18:04
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    $\begingroup$ real probability answer $\endgroup$
    – MARIE
    Jan 16, 2020 at 19:20
  • $\begingroup$ Could you therefore specify the conditions under which the numbers are selected? For instance, data are available on how people select groups of numbers for lotteries (and some of that won't translate accurately to selecting numbers from 1 to 99, because few lotteries use numbers that go that high), but there's probably not much else known about how they select numbers for other purposes. $\endgroup$
    – whuber
    Jan 16, 2020 at 19:41

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Assuming all numbers are independent and have an equal probability of being chosen... We have to ask, what are the total number of ways there are of choosing 7 numbers? Combination formula.

$$\frac{n!}{k!(n-k)!}$$

99 choose 7.

$$\frac{99!}{7!(92)!}$$

Simplifies to

$$\frac{99*98*97*96*95*94*93}{7!}$$

which is 14887031544 combinations.

The probability of choosing any one of these combinations is $$\frac{1}{14887031544}$$

Which should be your solution, as the probability of either person choosing a particular set is the same.

(edited to remove last unnecessary clac)

Hope that helps.

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