Calculating the Probability of a Poker Hand with Exactly Two Suits

[hallwayantics]

Hi was hoping someone out there could help me out with this problem.


http://en.wikipedia.org/wiki/Poker_probability

I have looked over the poker probability wiki page, but I cannot figure out this homework problem:


"What is the probability that the hand dealt contains exactly two suits?"

I can justify several different approaches ... but that just means I'm wrong.

1)

(4 C 2) * ( (26 C 5) - 2*(13 C 5) )

ie choose two suits, then from those 26 cards choose 5. Now subtract the two flush hands.


2)

(4 C 2)(26 C 5) - (4 C 1)(13 C 5)

ie get the number of hands with at most two suits then subtract the number of flush hands. (4 C 1)(13 C 5) is the number of flush hands in poker.

3)

(4 C 2)(26 C 5)



JMM

 

[AC130Nav]

Hi was hoping someone out there could help me out with this problem.


http://en.wikipedia.org/wiki/Poker_probability

I have looked over the poker probability wiki page, but I cannot figure out this homework problem:


"What is the probability that the hand dealt contains exactly two suits?"

I can justify several different approaches ... but that just means I'm wrong.

1)

(4 C 2) * ( (26 C 5) - 2*(13 C 5) )

ie choose two suits, then from those 26 cards choose 5. Now subtract the two flush hands.


2)

(4 C 2)(26 C 5) - (4 C 1)(13 C 5)

ie get the number of hands with at most two suits then subtract the number of flush hands. (4 C 1)(13 C 5) is the number of flush hands in poker.

3)

(4 C 2)(26 C 5)



JMM

I'm not sure what you are doing, but shouldn't it go like this:

Looked at in any order--
The probability that a first card will be in one of the four suits is 1.
The probability that a second card will be in a different suit is (51-12)/51.
The probability that the third card will be in one of the two previous suits is (12+12)/50.
Fourth card 23/49.
Fifth card 22/48.

So the answer is 1 * 39/51 * 24/50 * 23/49 * 22/48.

 

[sylas]

Hi was hoping someone out there could help me out with this problem. http://en.wikipedia.org/wiki/Poker_probability

I have looked over the poker probability wiki page, but I cannot figure out this homework problem:"What is the probability that the hand dealt contains exactly two suits?"

I can justify several different approaches ... but that just means I'm wrong.

1)

(4 C 2) * ( (26 C 5) - 2*(13 C 5) )

ie choose two suits, then from those 26 cards choose 5. Now subtract the two flush hands.

This one gives the number of hands correctly.

2)

(4 C 2)(26 C 5) - (4 C 1)(13 C 5)

ie get the number of hands with at most two suits then subtract the number of flush hands. (4 C 1)(13 C 5) is the number of flush hands in poker.

This gives the number of hands incorrectly. The problem is that (4 C 2)(26 C 5) doesn't count the number of hands with at most two suits. Thank of it this way. You have 6 ways of picking two suits. You count all the hands made up of just those two suits.

But note, a spade flush gets counted 3 times in this method. It is counted for SD, SC and SH.

So the number hands with exactly two suits is given by the first method, which I calculate as 379236

Method three doesn't try to take proper account of the flush hands, so we'll just pretend it isn't there.

AC130Nav, your method isn't even close. For example, you don't consider cases where the first two cards are the same suit. The original question had the right idea.

Cheers -- sylas

PS. Welcome to physicsforums to you both!

 

[hallwayantics]

Thank you, Sylas. The first answer is looking more and more correct to me.

I'm still not quite clear on why the second answer is no good...

It is counted for SD, SC and SH.​

I'm not sure what you mean by this. A flush in spades would mean only spades, why the Diamonds, Clubs, and Hearts?


@AC130

Yeah, that's why I linked to the wiki page, b/c there are different ways to approach probability but I like the nCr way of looking at it.
Thanks for the response.

This one gives the number of hands correctly.



This gives the number of hands incorrectly. The problem is that (4 C 2)(26 C 5) doesn't count the number of hands with at most two suits. Thank of it this way. You have 6 ways of picking two suits. You count all the hands made up of just those two suits.

But note, a spade flush gets counted 3 times in this method. It is counted for SD, SC and SH.

So the number hands with exactly two suits is given by the first method, which I calculate as 379236

Method three doesn't try to take proper account of the flush hands, so we'll just pretend it isn't there.

AC130Nav, your method isn't even close. For example, you don't consider cases where the first two cards are the same suit. The original question had the right idea.

Cheers -- sylas

PS. Welcome to physicsforums to you both!

 

[sylas]

Thank you, Sylas. The first answer is looking more and more correct to me.

I'm still not quite clear on why the second answer is no good...

It is counted for SD, SC and SH.​

I'm not sure what you mean by this. A flush in spades would mean only spades, why the Diamonds, Clubs, and Hearts?

The second method considers 6 ways of picking two suits. The suits are SHDC, so the six ways are SH, SD, SC, HD, HC, DC.

Now, given a pair of suits, you count up all the hands made up of those two suits only (and possibly even one of them). So let SH be the set of all hands made up of spades and hearts only.

Any spade flush hand will be counted as a hand in SH, and a hand in SD, and a hand in SC.

Now #SH, of the number of hands in SH, is (26 C 5).

but 6 * #SH is not the number of hands in (SH + SD + SC + HD + HC + DC). Those six sets are not disjoint! They overlap. Any flush hand will appear in three of the six sets.

It follows that #(SH + SD + SC + HD + HC + DC) = 6.#SH - 2.#FLUSH = 6(26 C 5) - 8(13 C 5)

Hence, in method two, where you have (4 C 2)(26 C 5) (for the number of hands with two or less suits) you should have (4 C 2)(26 C 5) - 8(13 C 5)

Now from this count, you remove all the flushes, which is removing another 4(13 C 5).

The final answer becomes (4 C 2)(26 C 5) - 12 (13 C 5)

And that gives the same result as method 1 again.

Cheers -- Sylas

 

[sylas]

The two methods which give the right answer for the number of hands are:

(4 C 2) (26 C 5) - 3 (4 C 1) (13 C 5)

and

(4 C 2) ( (26 C 5) - 2 (13 C 5) )

It's easy to show they give the same result.

But how about this method? Any hand having exactly two suits must have more of one suit that the other. It has 3 of this suit, and 2 of the other; or it has 4 of this suit and 1 of the other.

There are 4 ways to pick the suit which appears most often in the hand. There are 3 ways to pick the other suit, which appears less often.

The total number of hands is therefore:

4 * 3 * ( (13 C 3) (13 C 2) + (13 C 4) ( 13 C 1) )

Try it. You should get the same number, although it is by no means obvious just from the expression themselves. Combinatorics is full of cases like this, where the same count can be obtained with two methods that otherwise seem quite unrelated.

Cheers -- sylas

 

[hallwayantics]

Yes, it is neat how combinatoric problems can be approached from different angles. I'm enjoying it much more than the regular set theory about inclusion/exclusion intersection/union, proofs etc. Somehow combinatorics makes me feel like I'm doing math back in grade school again (back when I was good at it!)

Thanks again!

The two methods which give the right answer for the number of hands are:

(4 C 2) (26 C 5) - 3 (4 C 1) (13 C 5)

and

(4 C 2) ( (26 C 5) - 2 (13 C 5) )

It's easy to show they give the same result.

But how about this method? Any hand having exactly two suits must have more of one suit that the other. It has 3 of this suit, and 2 of the other; or it has 4 of this suit and 1 of the other.

There are 4 ways to pick the suit which appears most often in the hand. There are 3 ways to pick the other suit, which appears less often.

The total number of hands is therefore:

4 * 3 * ( (13 C 3) (13 C 2) + (13 C 4) ( 13 C 1) )

Try it. You should get the same number, although it is by no means obvious just from the expression themselves. Combinatorics is full of cases like this, where the same count can be obtained with two methods that otherwise seem quite unrelated.

Cheers -- sylas