HoustonHokie

You may not, but I often do

Doggone it - I had just gotten back in the forest and figured it out when I got interrupted to do some real work (the boss-man cometh), and I missed the chance to post the solution first!! Just goes to show that you need to focus on the important (BD), not the urgent (work)

So step 2 is done. Now on to step 3...your arithmetic needs some work. ok, this is kind of foreign to me, but I'll give it a go:

To be really picky about it, maybe Aces full of 6s would be better. Because the Ace appears in a 5-high SF, the 5s are only going to knock out 4 of the SFs in suits where you hold the Ace as well (either 1 or 2 of your Aces). Aces full of 6s knocks out 5 SFs, no matter what the suits. All we can really do is postulate around this, though. The betting question depends on multiple factors: the number of players at the table, the hands they hold, and, most importantly, their BAC . I guess the ideal betting situation would be to knock out as many hands as possible where "rational" people wouldn't bet much (say pair of deuces) while leaving a lot of hands that still lose to your full house, but would encourage betting by the same "rational" people (say three Kings). The question is finding the point at which the threat of losing to a stronger hand is balanced by the probability that you will win a certain pot - and that's a lot more math than I'm willing to do right now .

not that it's going to help me any, but I'm just curious: is there significance to akaslickster's name being misspelled in his ID (akaslister)?

Is your sig a hint?

All makes sense now - I did all my work assuming that I'd put my money on position 1! Should have gone into more detail...

I'd say it has to do with:

I can't imagine answering a bona question without probability and a proof, so here goes: edit:grammar

edit: includes solution!

how about

One of the reasons this is a difficult number to calculate is that there are a significant number of cases where, in a finite number of flips, the desired triplet doesn't occur at all. Consider the OP with 17 coin flips. If you're looking for TTT, any combination of 17 flips where the coin lands heads-up at least 15 times will eliminate the possibility of TTT occurring. There are also numerous other cases where TTT doesn't occur in 17 flips (for example, TTHTTHTTHTTHTTHTT). I haven't carried it out all the way to figure out in how many of the 17-flip cases TTT doesn't occur (or any of the other 7 triplets, for that matter), but what do we say about how many coin flips are required for that triplet to occur in those cases? It's definitely greater than 17, but how much greater? It's relatively easy to go through and figure out how many times TTT first occurs after 3, 4, 5, ..., 17 flips, and then multiply the number of flips by the relative probability of each case and sum to get the overall average number of flips required, assuming the triplet occurs at all. But I still don't know what to do with the cases where TTT doesn't occur at all. One thought would be to allow the number of flips (n) to increase until n approaches infinity, but then I think you get into a situation where for an infinitesimally small probability of occurrence (p = 0.000000....), the triplet occurs after infinite flips, and zero times infinity is undefined. Another thought, and better I think, would be to try to figure it out this way: