Probabilities Pick 3 Lottery In a Pick 3 lottery three numbers are drawn from three separate sets of numbers, between 0 and 9.
Matching 2 winning numbers, one of them duplicated, can be done in 3 ways, according to all lotteries websites.
But I found 18 ways to do it.
As an example:
If the winning numbers are 1 2 3, then there are 18 possible ways to win:
112-113-121-122-131-133-211-212-221-223-232-233-311-313-322-323-331-332
Since the probability is equal to the number of favourable outcomes divided by the number of possible outcomes, it will be 18 divided by 1,000. The odds would be 1 in 55.55
Please let me know if I am making a mistake in my calculations.
Thanks and regards,
 A: When you fill in a Pick-3 lottery card, you:


*

*choose three digits between $000$ and $999$, and then

*choose "Straight" or "Box" (or some other option)


"Straight" means that you only win if your digits come up in the exact order.
"Box" means that you win if they come up in any order—but they must be the same digits in the same quantities.
If you choose $123\text{ Box}$ then there are six ways you can win; if you choose $122\text{ Box}$ then there are only three ways you can win, and that is reflected in a doubled payout.
Pick-3 lotteries generally do not offer a payout for the partial matches you describe.
Options like "Straight/Box" and "Combo" are considered equivalent to filling in a multiple bet.
A: Please revisit your answer. Since three numbers are coming from the different lotteries, so the combination of numbers (1,1,2) and (1,2,1) would be dealt from the same combination of number or from the same box. And each box have 9 digits and selecting one from each box resulting in total 1000 numbers. But the only one combination of the boxes would result in the winning match. I think this hint would be more than enough.
A: I think by "matching $2$ winning numbers, one of them duplicated", they mean that you should have exactly the same multiset of numbers as was drawn. So for instance if $1$ was drawn twice and $2$ was drawn twice, you should have two $1$s and one $2$. By "can be done in $3$ ways" I think they mean that given your pick of e.g. two $1$s and one $2$, three of the $1000$ equiprobable outcomes match your pick, since there are three choices for the set from which the $2$ is drawn, and that fixes the draws from all three sets.
