Re: probability homework question
 

OK, now that I'm working with the correct numbers...

There are 54,912 hands with three of a kind exactly (no quads or boat). Along with your trips, you have two cards of two different ranks, of which there are 12 possibilities. The problem can now be simplified down to "there are 12 options, I get 2 random picks, what are the odds I get the two I want".

Let's say you want #1 and #2. Reach into a hat and pull out a number... you have a 1/6 chance of picking a winner on your first pick. Then with your second pick, there are 11 remaining numbers and only 1 winner, so it's 1/11.

54912 x 1/6 x 1/11 = 832.

There are exactly 832 ways to be dealt trips with one higher card and one lower card in 5 card stud.

832/2598960 = 0.032% = you have a 1 in 3123.75 chance to be dealt that hand.

If AAA2K doesn't count, then it's 768 hands, 0.0029%, or 1 in 3384.

If both AAA2K and 222A3 don't count, then it's 704 hands, 0.0027%, or 1 in 3691.7.

cliffnotes: BruceZ is smart, and i r dumb

Re: probability homework question
 

I'm going to correct this and re-post.

Thanks, Pococurante.

Re: probability homework question
 

Buzz: all of your math is right, but I think he meant trips + exactly one rank higher + exactly one rank lower

Re: probability homework question
 

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

Re: probability homework question
 

This was supposed to be easy, not an exercise in looking up published numbers. Let me clarify:

[ QUOTE ]
Count the number of ways to make such a hand, without regard to the order of the cards, and divide by the total number of possible hands C(52,5) = 2,598,960. Consider the number of possible denominations for the set <font color="red">12</font>, the number of ways to make a set for each denomination <font color="red">4</font>, and the number of ways to get the other 2 cards <font color="red">4*4</font>.

[/ QUOTE ]

The numbers in red above assume AAA2K is not allowed, but 222A3 and KKKQA are allowed. Otherwise change the number 12 accordingly.

12*4*4*4 / 2,598,960 =~ 0.03% or 1 in 3384

Done.

Re: probability homework question
 

Hi Bruce - Thanks for your response. I had already resolved to stay out of your hair from now on.[img]/images/graemlins/crazy.gif[/img][ QUOTE ]
and the number of ways to get the other 2 cards <font color="red">4*4</font>.

[/ QUOTE ]If you have chosen the specific lower and higher rank, that's true. Otherwise there are, for threes, for example, 2*4=<font color="blue">8</font> lower cards, excluding aces, and 9*4=<font color="blue">36</font> higher cards, also excluding aces.

And then we have to add in the aces. Where Z can be anything but a three or an ace in A333Z, there are <font color="blue">44</font> cards that can be the Z, not just <font color="red">4</font>.

Buzz

Re: probability homework question
 

[ QUOTE ]
If you have chosen the specific lower and higher rank, that's true. Otherwise

[/ QUOTE ]


Original problem:

[ QUOTE ]
What is the probability of getting 5 cards dealt (normal deck) and three of the cards are the same denomination (trips), and the other two cards are one above and one below...example: 8 8 8 9 7

[/ QUOTE ]

(emphasis added)

Re: probability homework question
 

Hi Bruce - I didn't solve it for the specific example, but for the general case when hero has trips with one card higher and one card lower. For trip eights, it wouldn't only be 88897, but
88896
88895
88894
88893
88892
8889A
and then
888T7
888T6
888T5
888T4
888T3
888T2
888TA
etc.
and then there would be trip sevens, and they're a bit different from trip eights, and then trip sixes which are also a bit different,
etc.

I did it for all of them. For all cases.

It always depends on how you look at things, I guess.

I'll get out of your hair now.

Thanks.

Buzz

Re: probability homework question
 

[ QUOTE ]
Hi Bruce - I didn't solve it for the specific example, but for the general case when hero has trips with one card higher and one card lower.

[/ QUOTE ]

I understand. I didn't just solve it just for the example either, I solved it for the question which I believe was asked, which is for any 3-of-a-kind where the other 2 cards are one RANK above and one RANK below. While the words RANK were not present in the problem statement, I am pretty sure that is what was intended as I simply don't believe that your interpretation, which is "the other two cards consist of one higher and one lower card", is correctly expressed in English by the words "the other two cards are one above and one below". Gene Fish, please tell us which interpretation you intended.

Re: probability homework question
 

[ QUOTE ]
one RANK above and one RANK below

[/ QUOTE ]Hi Bruce - I see another way to read the problem that I had not even considered before.
222A3, 33324, 44435, etc.
(but not 44436 or 44425 or 44426 etc.)
just exactly one rank above the trips and exactly one rank below the trips.

Interesting.

Thanks. I'm going back to Omaha now.

Buzz