# Lottery problem - dependent vs. independent events?

A lottery sells $10,000,000$ tickets with probability of winning at $\large\left(1\over 5000000\right)$

Let $t$ to represent the number of tickets sold. Using the formula $1-\left(4999999\over 5000000\right)\large^t$, the probability of there being a winner is approximately $86.5\%$ (or for no winner, $13.5\%$). Okay, I understand that.

Using the same numbers, what if the lottery guarantees there will be $\large2$ winners.

This is problematic because the probability of $0$ winners calculates as $13.5\%$, which is impossible since we know that there will be $2$ winners out of $\large t$ tickets sold. Ultimately my question and confusion is how would you calculate the probability of winner/no winner in such a scenario?

Thanks

Edit:

Let me clarify what I'm basically asking. I know the events are not independent because the lottery is guaranteeing exactly 2 winners out of 10 million. So, how do you calculate probabilities of $X$ winners at various points of tickets sold. For example, they have sold 9,000,000 out of the possible 10,000,000 tickets. What is the probability of only one winner so far? The winners are not selected after 10,000,000; they are picked randomly.

I assume for 1 winner it would be $9,000,000 \over 10,000,000$

But what about non-obvious figures, i.e. 2 winners?

Would it simply be $\left(9,000,000\over 10,000,000\right)\large^2$

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(1) you have omitted something in calculating 0.1353; and (2) everything depends on precisely how the lottery proposes to satisfy the guarantee of two winners. –  Pieter Geerkens Sep 1 '13 at 3:52
What did I miss? And the lottery issues the tickets a la raffle. Out of the 10 million sold, 2 are winners. This is determined before they are sold. –  Peeping Tom Sep 1 '13 at 3:55

Suppose the lottery sells $10,000,000$ tickets and draws two numbers as winners. The probability of any given ticket winning is then $\frac 1{5,000,000}$ but two tickets will win. The fallacy in your computation is assuming that the events of each ticket winning are independent. They are not. You are assuming they take each ticket and independently decide whether it wins. The probability of a winner is $1$. The probability that a given ticket will win is $\frac 1{5,000,000}$. You need to be careful with the question you are asking and the independence.
If two winners are guaranteed, the chance of exactly one in the first 9M is $\frac {9M\cdot 1M}{10M \choose 2}$ But if you don't know whether there is zero, one, or two in the first 9M it doesn't change the odds on the last 1M tickets. What new information do you have from the fact that 9M have been sold? –  Ross Millikan Sep 1 '13 at 4:29