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In some lottery one can buy a ticket by choosing seven distinct numbers each of them from numbers ${\{1, 2, \dots, 45}\}$ (so $1/(45379620)$ is the probability to win the first prize). Every week the number of winners is $0$ or $1$ or $2$. In each of cases, that is when the number of winners are $0$ or $1$ or $2$, is it possible to calculate the number (or interval of numbers on $\mathbb Z^+$) of tickets sold?

Notes:

1- One person can buy many tickets of same or different 7-number s, but (in reality of what happened already) when two people win same the 7-number they are different people.

2- (with a little humor, I should say) I couldn't find 'reverse probability' in the tags!

Thank you.

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  • $\begingroup$ This problem cannot be solv'd without the distribution of tickets sold. $\endgroup$ – zoli Jun 29 '15 at 10:16
  • $\begingroup$ @zoli: what do you mean by "distribution of tickets sold"? choosing numbers by people is supposed to be random. $\endgroup$ – L.G. Jun 29 '15 at 11:04
  • $\begingroup$ @AlphaE: I answer your question in the form of a hint -- if you don't mind. $\endgroup$ – zoli Jun 29 '15 at 11:36
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HINT

The question cannot be answered without information about the distribution of the number of tickets sold.

Let $\nu$ and $w$ denote the number of tickets sold and the number of winners, respectively. We need the number of tickets sold assuming a certain combination of the number of winners, that is, the we need the following conditional probability:

$$P(\nu=N \mid w=0 \text{ or } w=1 \text{ or } w=2 )=\frac{\sum_{i=0}^2P(\nu=N, w=i)}{\sum_{j=0}^2P(w=j)}.$$

For $P(\nu=N, w=i)$ we can only do the following $$P(\nu=N, w=i)=P(w=i\mid\nu=N)P(\nu=N)$$

since we know the probability that there is $i$ winners only if we know the number of tickets sold, that is, if $\nu=N$ is given.

Similarly

$$P(w=i)=\sum_{k=0}^{\infty}P(w=i\mid\nu=N)P(\nu=N).$$

Anyway, in order to solve the problem, we need to know the distribution of the tickets sold; we need to know the probabilities $P(\nu=N)$ for all possible $N$.

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  • $\begingroup$ Thank you. I think I got a little what's going on.. $\endgroup$ – L.G. Jun 29 '15 at 12:18

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