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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

1 Answer 1

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.

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Yes, exactly! I am asking how to compute dependent vs. independent. For example, once they've sold 9,000,000 tickets what is the probability of 2 winners (or less obviously, only 1 winner), understanding that there will be guaranteed 2 winners out of 10,000,000, and the winners are not picked after 10 million, they are picked randomly? –  Peeping Tom Sep 1 '13 at 4:02
    
Usually the draw is after all the tickets are sold. In that case, you have no information about the tickets that have been sold. But in California we have scratcher lotteries where the winning tickets are determined at the time the tickets are printed. If you know the prizes have been claimed, the chance of winning is then zero. If they have not yet been claimed, you need to assess how quickly the prize will be claimed. You might think that the fact that they have not been claimed means the remaining tickets have the two winners, but maybe the winner has been sold but not claimed. –  Ross Millikan Sep 1 '13 at 4:18
    
I think I answered my own question in my OP edit. You would treat it as discrete. You would divide number of tickets sold by the number of those available and raise it to the power of the number of winners desired (if this is wrong please correct me). Yes if you have additional information then naturally it will be applied; in the case of no info I think what I got is right. –  Peeping Tom Sep 1 '13 at 4:20
    
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
    
Okay, your formula is what I'm looking for. How would you change it for 2 winners? Or help me understand its logic a bit –  Peeping Tom Sep 1 '13 at 4:39

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