Fortune Cookie Probability

Assuming that six different numbers are on the fortune, and that they aren't printed in any particular order, I agree, the odds of a "1" being one of the numbers would be 6/99.

Similarly to how you figured that you should multiply the chances of "1" being the lucky number in a particular slot by six because there are six different spots, you can find the probability that "1" and "2" are each in a specific spot, then multiply by the number of different ways you could arrange the two numbers in the list of six. ( 1, 2, ?, ?, ? ) would be one, ( 1, ?, 2, ?, ? ) would be another, ( 2, 1, ?, ?, ? ) and so on...

I believe you end up with [ (1/99) * (1/98) ] * (6*5) = 5/(33*49) = 5/1617. (The first part in brackets being the probability, and the second part being the number of different arrangements) For 13 to also show up, add another number in the same manner, [(1/99)*(1/98)*(1/97)] * (6*5*4) = ...

It's probably worth mentioning that you can only do this because a number can appear in exactly one of any of the six spots. If they were printed in say, increasing order, or could be repeated (or some combination) we would have to think about it a little harder.

This actually makes the last odds you thought of the easiest to calculate. Since we don't care about the other two numbers, and you've picked a spot for each number, the probability's just (1/99)*(1/98)*(1/97). But, if the reason that the three numbers are in those places is because the lucky numbers are always printed from smallest to biggest, it would not be so simple. For example, (1, 2, 5, 9, 13) would work, but (1, 2, 5, 13, 25) wouldn't. Actually, it doesn't seem like it would be too difficult to figure out, 1 being the smallest possible number and 2 being the very next simplifies it a lot, try working it out if you'd like. (my brain feels like syrup in numbers it's been sitting collecting dust for so long lol)