7-Card-Stud starting hands

[kinghippo423]

Hello all,

I posted on another forum a question about 7-Card-Stud stating hands and we all had differents approches to the question that ended us with differents results. I wanted to know some opinions of our solutions and maybe find the correct answer with absolute certainty.

This is the question:

What are the odds to have a pair of tens or better with the first 3 cards. In other words, to have XTT, XJJ, XQQ, XKK, XAA, 222, 333, 444, 555, 666, 777, 888, 999, TTT, JJJ, QQQ, KKK, AAA.

Solution 1 (mine):

52C3 = 22100 3-cards combo
4C3 = 4 possibilities of 3-of-a-kind by rank.
4C2 = 6 possibilities of pairs by rank.

Chances to have 3-of-a-kind:

13(4C3) / 52C3 = 1/425.

Chances to have tens or better without 3-of-a-kind:

5(4C2) * 50 / 52C3 = 15/221.

So, chances to have tens or better are:

1/425 + 15/221 = 338/5525 or 7.02%.

Solution 2 (one forum member):

20/52 * 3/51 +
20/52 * 48/51 * 16/48 * 6/48 +
20/52 * 48/51 * 32/48 * 3/48 +
34/52 * 3/51 * 2/50 +
34/52 * 48/51 * 20/51 * 3/50

In order:

- tens or better that hits with the first 2 cards.

- tens or better that hits on the 3nd card and
2nd card >= T + 2nd card <> 1st card.

- tens or better that hits on the 3nd card and
3nd card = 1st card + 2nd card < T.

- Set of twos through nines.

- tens or better that hits on 2nd and 3nd card
and 1st < T.

Odds : 6.88%

Conclusion:

My solution 7.02%
his 6.88%.

Who's right?

[bigpooch]

[ QUOTE ]
Chances to have tens or better without 3-of-a-kind:

5(4C2) * 50 / 52C3 = 15/221.



[/ QUOTE ]

This should be 5(C(4,2)) * 48 /C(52,3) since you only have
52-4 = 48 kickers.

[Siegmund]

Yes. As bigpooch said, you have counted the trips repeatedly, by allowing 50 instead of 48 kickers.

[kinghippo423]

Of Course ! Thanks for the help. I'm guessing the rest is good.

6.751% <> 6.881% but it's close enough !

[Siegmund]

...and the "one forum poster" meant 32/52, not 34/52, for the first term on his last two lines where your first card is a nine or lower. That should bring the totals into agreement on the 6.75% figure.

[kinghippo423]

You're exactly right. His approch, with 32/52 equals 6.7866205%. Getting really close !

Probabilities can be tricky sometimes.

[kinghippo423]

Is there a site out there that can "teach" me or explained me some technics to attack more complicated problems? Even a site that give problems to solve and being able to see the answer to see our mistakes. I tried but didn't find anything good.

I really want to improve my "probabilities maths" skills !

[infinity235]

Ehm... to be serious, I learned this kind of stuff during high school.

But, you know, I am German.

[jogsxyz]

Try this site.