Re: triple draw high full house probability
you could make an approximation for 2). if you had just one attempt to draw, then it's (46C1*1C1 + 2C1*3C2 + 10C1*4C2) / 47C2 = 0.1036, this is the chance of getting a full house or better in the first draw. 1 - 0.1036 is the chance for not getting anything good in the first draw, so 1 - (1-0.1036)^3 = 0.2797 would be an approximation which should be good enough for 2).
just for comparison, if you approximated 1) in this way, you would get 1 - (43/47)^3 = 0.234, which is pretty close to 0.239 which is the exact answer.