In this lottery $7$ balls are chosen from $1-60$. In order to win the main prize, you must select all $7$ right. I calculate the odds of doing this as:


The odds of getting $3$ correct would be


Now suppose for whatever reason you are only allowed to select balls from $1-30$ and cannot pick $31-60$. Does this change your odds of winning? What are your odds of the top prize and what are your odds of getting 3 right?


  • $\begingroup$ Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. $\endgroup$ Commented Feb 12, 2019 at 15:58
  • $\begingroup$ Shouldn't the probability of winning the lottery for $7$ balls with numbers $1-60$ be $$\frac{1}{60·59·58·57·56·55·54}=\frac{1}{1946482876800}$$ $\endgroup$
    – Dr. Mathva
    Commented Feb 12, 2019 at 16:06

1 Answer 1


As far as I understand the question, you select $7$-element subset of $\{1, ... , 30\}$ uniformly, you win if they coincide with $7$-element subset selected from $\{1, ... , 60\}$ uniformly, and those selections are independent. Suppose, $X$ and $Y$ are those subsets respectively. Now if $Y$ is not a subset of $\{1, ... , 30\}$, then the probability of them coinciding is $0$. Also $P(Y = X| Y \subset 30) = \frac{1}{C_{30}^7}$, so here we have $P(X = Y) = P(Y = X, Y \subset 30) = P(Y = X| Y \subset 30)P(Y \subset 30) = \frac{1}{C_{30}^7}\frac{C_{30}^7}{C_{60}^7} = \frac{1}{C_{60}^7}$.

It is quite paradoxical, but that happens to be the same number, as if your choice were completely unrestricted. And it would be that way, no matter to what subset of $\{1, ... , 60\}$ is your choice restricted (provided, of course that this subset contains not less than $7$ elements).

  • 2
    $\begingroup$ It is not paradoxical at all. Assuming the lottery picks the balls at random any restriction on the numbers you can select does not matter. Every selection has an equal probability to win. If I said you had to pick $1,2,3,4,5,6,7$ you would still have the same probability to win. $\endgroup$ Commented Feb 12, 2019 at 16:33

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