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Raymundo
06-09-2003, 11:10 PM
Many of you may not care about this as it pertains to a best 3 card hand out of 5 cards, but what the hell, maybe one of you will and can verify my results. This started out as an exercise in probability and has turned into a bit of a nightmare.

Possible three card hands are
straight flush
three of a kind
straight
flush
pair
high card

The question is, what should be ranked higher--straight flushes or trips?

Mano
06-11-2003, 01:47 PM
Let's see which there are more of.

For a straight flush, you can have from a 3 high to an A high, so there are 12 straight flushes for each suit, or a total of 48 ways to get a straight flush.

For Trips there are 4 different ways to get trips for each rank, 13 different ranks, so there are 52 ways of getting trips.

Therefore straight flushes are harder to get and should be ranked higher than trips.

Raymundo
06-12-2003, 11:06 PM
Your analysis is quite correct when you are only considering three cards total. But here the situation is a little trickier because it's the best three out of five. I've elaborated below, but I'll warn you that it might not be worth the effort or the time it takes to read it.

For trips, I counted any normal five card poker hand that has trips in it, specifically all quads, full houses, and trips. That added up to
Quads = 624 possibilities
Fulls = 3,744 possiblities
Trips = 54,912 possibilities
TOTAL = 59,280 five card hands that have trips in them.

The sf's were trickier. An sf is completely described by its lowest card. There are 12 sf's from A to Q.
-Sf's beginning with A thru J can get their other two cards from any 48 of the remaining 49 cards (the 49th card would be the next highest card in the sf and so you can't count it twice.) There are four different suits to choose from so
12 sf's x 4 suits x 48C2 blanks = 54,144 sf's
-Sf's beginning with the Q can get their other two cards from all 49 of the remaining cards.
1 sf x 4 suits x 49C2 blanks =4,704
Thus, TOTAL five card hands containing a three card sf is 58,848.

So it appears that there are slightly less sf's than trips. But here's the problem: of the 59,280 hands including trips, 432 of them have both trips AND sf's. Since sf's are being counted higher, it turns out that we can't count them as trips hands, only as sf hands. Deducting 432 then from 59,280 leaves us with 58,848 WHICH IS EXACTLY HOW MANY SF HANDS THERE ARE!

Godfather80
06-15-2003, 05:00 PM
Ray

I'm a math teacher and this is one of the coolest solutions I've ever seen. I'm definitely going to present this problem for extra credit.
Thanks

Copernicus
06-16-2003, 01:21 PM
An interesting result, but I hope you aren't concluding from the last observation that they should be ranked equally are you?

keerok
06-16-2003, 05:10 PM
</font><blockquote><font class="small">In reply to:</font><hr />
-Sf's beginning with A thru J can get their other two cards from any 48 of the remaining 49 cards (the 49th card would be the next highest card in the sf and so you can't count it twice.) There are four different suits to choose from so
12 sf's x 4 suits x 48C2 blanks = 54,144 sf's
-Sf's beginning with the Q can get their other two cards from all 49 of the remaining cards.
1 sf x 4 suits x 49C2 blanks =4,704


[/ QUOTE ]

I think the problem is that you're counting some straight flushes twice. The way you count the A23 SFs also includes the 4 five-card hands A23QK suited, which should be counted as AKQ SFs.

Raymundo
06-16-2003, 08:47 PM
</font><blockquote><font class="small">In reply to:</font><hr />
I think the problem is that you're counting some straight flushes twice. The way you count the A23 SFs also includes the 4 five-card hands A23QK suited, which should be counted as AKQ SFs.


[/ QUOTE ]

I think you may be right! If so, Sf's win by a nose. Let me think about this for a minute...

Raymundo
06-16-2003, 08:51 PM
Not a chance. Until trips prove to be rarer than sf's, even by one combination, they are the losers. It's just amazing to me how the mathematics produces such a coincidence.

Raymundo
06-16-2003, 08:53 PM
Thanks, but I can't take full credit. I based my method on that used by a professor of mathematics name Brian Alspach from Canada. If this sort of thing interests you, check out him webpage at
http://www.math.sfu.ca/~alspach/computations.html

Good luck with your math class. I'm a teacher, too (of science).