
Probability, lottery...
A lottery ticket(cupon?) with a row of 3 numbers. Those three numbers can vary from 09. A ticket(cupon) might be 5,6,7.
A number is picked randomly among the numbers 09. The number that was picked is then put back into the bowl. The lottery ends when it has three unequal numbers, for instance, it can pull out 5 numbers like this: "1,3,3,1,5", which leaves the lottery result with 1, 3,5 as the winning numbers.
What is the probabilty of winning then you have delievered a cupon consisting of the numbers 2,3,4?
My shot at it:
First number is always 1/9 chance.
Second number is 1/9*1/9 chance.
Third number is 1/9*1/9*1/9 chance.
This probability would be right if the lottery could have 3 numbers, no matter if a number got picked twice, or "trice".
Need a bit help :)

I believe its 1/720. There are 10 possible numbers. You need to match all three. The odds of the first match is 1/10. However after that they will draw a new number if the first number comes up again. So for the second drawing there are basically only 9 possibilities. So the odds of hitting the second number is 1/9. Similarly for the third it is 1/8. So your chances of hitting all three are 1/10 * 1/9 * 1/8 = 1/720.

Haha, yes, actually, I did dream about this answer, that I should just multiply the latter(?) by reducing the 10 to 9, then to 8...
And yes, my bad, 1/10 possibilities on the first, 1/9 on the second one.....
Thanks