An interesting little problem in compound probabilities. Been a very long time since I had to think about anything like this though.
The probability of winning anything at all is the sum of the probabilities of winning any particular draw. Obviously the probability of winning the first draw is 1/3000 if you hold one ticket; I'm sure everybody sees that. The probability of winning the second draw, however, is not 1/2999. In order to win the second draw, you have to lose the first one (I'm assuming tickets aren't replaced), which you will with a probability of 2999/3000. You have to multiply those two fractions together. 2999/3000 is almost 1, but not quite, so the probability of winning on the second draw is a little less than 1/2999. In fact the 2999s cancel and you're left with 1/3000. That'll happen at every draw. The probability is thus as BitWhys stated, 1/3000 for every draw, so in 10 draws it's 10/3000, or 1/300. Makes sense: if there were 3000 draws you're guaranteed to win if winning tickets aren't put back in the draw, so the logic of that argument must add up to a probability of 1 after 3000 steps, and it does. Putting winning tickets back into the pool after each draw makes no practical difference, the probabilities differ only after the fifth decimal place.
Holding multiple tickets seems to complicate it a bit more than is obvious though. Strictly speaking it appears you can't just multiply the results for a single ticket by 2, 3, or 4. I was a little surprised when I found that result, so if anyone can point out a flaw in it I'd be grateful. This one's easier to approach from the other end: what's the probability that you'll lose all 10 draws in a row? If you have four tickets, the probability you'll lose the first draw is 2996/3000. Again assuming no replacement, the probability you'll lose the first and second draws is 2996/3000 times 2995/2999. For three draws, there's a third factor, 2994/2998. Just keep multiplying fractions that way and stop after the 10th one, which will be 2987/2991. There's not as much nice cancellation in that exercise as there is with only one ticket, and you actually do have to do a little multiplication of 4-digit numbers. The answer I get for that is 0.986726. That's the probability of losing all ten draws. The probability of winning anything is thus 1 minus that, which is 0.01327. If it were as simple as just multiplying the result for 1 ticket by four, that number would have been 0.01333, which is also, as somebody said, 1 in 75.
More specifically, the probability of winning with one ticket is a third of a percent, 0.003333 or 1/300, and the probability of winning with four tickets is very slightly less than four times that. I don't see why that should be so, frankly, so I reiterate, if anyone can show something's wrong with my argument... ? Has my creaky old brain blown a gasket somewhere?
