Hi Gene - Trip aces cannot have any card higher or lower so therefore are excluded.
Assume aces can be high or low, but cannot be counted for both high and low.
There are four ways to have three sixes, and excluding aces,
4*4 ways to have one lower card and 4*7 ways to have one higher card, ergo (4)*(4*4)*(4*7) ways to have trip sixes with one higher and one lower card (but not an ace).
Assuming three sixes plus an ace plus another card, the ace is high if the other card is lower than six but low if the other card is higher than six. So there are (4)*(4)*(4*11) ways for 666AX to exist.
All in all, for trip sixes, there are 4*4*4*4*7 + 4*4*4*11 ways for the hand to exist with one higher rank plus one lower rank.
And that’s the same as it would be for trip nines, but with the highs and lows reversed.
For trip deuces, there are (4)*(4)*(4*11) ways for the hand to exist with one higher rank plus one ace.
Completing the table,<ul type="square">
• twos, kings, 2*4*4*4*(0*11+11)
• threes, queens, 2*4*4*4*(1*10+11)
• fours, jacks, 2*4*4*4*(2*9+11)
• fives, tens, 2*4*4*4*(3*8+11)
• sixes, nines, 2*4*4*4*(4*7+11)
• sevens, eights, 2*4*4*4*(5*6+11) =[/list]Collecting terms and simplifying,
2*4*4*4*(11+21+29+35+39+41)=
128*176=22528
22528 is the number of ways to get dealt trips plus one card from a higher rank plus one card from a lower rank.
Then, since there are 2598960 ways to be dealt five cards from a standard 52 card pack with no joker,
22528/2598960 = 0.00866808
I think the probability is about 0.00866808.
I was in error the first time I posted because I counted two aces for high and also for low, actually figuring eight aces (four high and four low) when I did that. I think it’s right now. Not sure why Bruce Z. and I evidently disagree. Seems like he must be right.
I haven’t looked at Brian Alspach’s site. O.K., now I looked, but Brian doesn't cover this exact problem.
Buzz