Card Hand probability question

[Don1]

Working on a problem and unsure how to set it up. Feels similar to other I believe is solved correctly. Any help appreciated.

Solved problem: 52-card deck, 13-card hand, odds of getting 4 of a kind.

For this I used a Hypergeometric Distribution with the following parameters:
1) population size of 52
2) number of successes in population of 4
3) sample size of 13
4) number of successes in sample of 4

This results in an answer of approximately 0.23%

Unsolved problem: 52-card deck, 13-card hand, odds of getting a 4-card straight flush (4 successive cards in rank of the same suit).

I do not believe I can use a Hypergeometric Distribution for this problem, and a little stumped where to start since the first selected card always can be used (aka is always a success).

Thanks again for any help.

 

[Jameson]

Hi Don,

Welcome to MHB! :)

I think you can use this Wikipedia article that calculates the probability of a 5 card straight flush from 7 cards and make some adjustments for 4 cards out of 13. What do you think after reading that?

 

[soroban]

Hello, Don!

52-card deck, 13-card hand. .Probability of getting 4-of-a- kind.

There are: .$${52\choose13}$$ possible hands.

There are 13 values for the 4-of-a-kind.
We must choose 9 more cards from the other 48 cards: .$${48\choose9}$$ ways.
There are: $$13\cdot{48\choose9}$$ ways to get 4-of-a-kind.
$$P(\text{4-of-a-kind}) \;=\;\dfrac{13\cdot{48\choose9}}{{52\choose13}}$$

 

[soroban]

Hello again, Don!

52-card deck, 13-card hand.

Probability of getting as 4-card Straight Flush (4 cards of successive values in one suit).

There are: .$${52\choose13}$$ possible hands.

There are 4 choices of suit for the Straight Flush.
There are 11 possible sequences for the Straight Flush:
. . $$\text{A-to-4},\;\text{2-to-5},\;\text{3-to-6},\cdots \text{J-to-A}.$$
Hence, there are: .$$4\cdot 11 = 44$$ Straight Flushes.

Therefore: .$$P(\text{Str. Flush}) \;=\;\frac{44}{{52\choose13}}$$

 

[Jameson]

Hmm, I'm not sure if I agree with that soroban. You have accounted for the straight flushes but not for the 9 other cards.

For example we could have A234 (all diamonds) and the rest of the cards could be anything except for the 5 of diamonds (since it would now become a 5-high straight flush) and still have this count as a A-4 straight flush. There are many ways to do this so in the end the numerator will end up being much larger than 44 once we account for all the possibilities of the remaining 9 cards.

 

[awkward]

Hello, Don!


There are: .$${52\choose13}$$ possible hands.

There are 13 values for the 4-of-a-kind.
We must choose 9 more cards from the other 48 cards: .$${48\choose9}$$ ways.
There are: $$13\cdot{48\choose9}$$ ways to get 4-of-a-kind.
$$P(\text{4-of-a-kind}) \;=\;\dfrac{13\cdot{48\choose9}}{{52\choose13}}$$

Hi Don and Soroban,

These problems are harder than they look! It's very easy to mistakenly over-count. For example, in Soroban's solution above, suppose the four-of-a-kind is four aces. Some of the "other 9 cards" includes four twos among the 9 cards. But one of the other possibilities for the initial "four-of-a-kind" is four twos, and some of the "other 9 cards" then includes four aces. We have counted the combination of four aces and four twos multiple times!

As a way out of this, suppose we use the Principle of Inclusion and Exclusion to count the number of hands which do _not_ include four of a kind. Then we can compute the probability of not drawing four of a kind. After that, it's easy to compute the probability of drawing four of a kind.

To that end, there are $N = \binom{52}{13}$ possible hands, which we assume are all equally likely. Let's say a hand has "Property k" if the hand includes four cards of rank k, for $k = 1,2,3, \dots , 13$. We denote the number of hands which have $i$ of the properties by $S_i$. Then
$$S_1 = \binom{13}{1} \binom{48}{9}$$
($\binom{13}{1}$ ways to pick the "property", i.e. the rank for the four of a kind, then $\binom{48}{9}$ ways to pick the remaining cards)
$$S_2 = \binom{13}{2} \binom{44}{5}$$
$$S_3 = \binom{13}{3} \binom{40}{1}$$
$$S_i = 0 \qquad \text{for } i \ge 4$$
By Inclusion/Exclusion, the number of hands which have none of the properties, hence no four-of-a-kind, is
$$N_0 = N - S_1 + S_2 - S_3$$
Then the probability of _not_ getting four of a kind is $p = N_0 / N$, and the probability of getting four of a kind is $1-p$, which, if you do the arithmetic, works out to be 0.0258116.

I'll leave it to you to think about the 4 card straight flush. In other words, I haven't worked that one out yet!

 

[Jameson]

Great points, awkward. I didn't consider that by dealing a hand more than 7 cards you open up to possibility of having two 4-card combinations in the same hand. Since a straight flush beats 4 of a kind we don't need to consider a straight flush plus 4 of a kind combinations separately. It is the same hand as a straight flush and 4 random cards.

Here are the things we need to exclude (might be more if I missed something):

1) A single card changing the rank of the straight flush. If we have a straight flush from A-4 of diamonds then the 5 of diamonds is not able to be chosen or it will become a 2-5 straight flush.

2) Two or three straight flushes in the same hand. We could have A-4 of diamonds, 3-6 of hearts and 10-K of spades. The highest ranking 4-card combination is the one to consider and this is where counting the possibilities gets very tedious.

We need to start with A-4 in one suit and calculate all the other 13 card hands with just one extra straight flush. Do the same for 2-5, 3-6, etc. and for all four suits. Then start with A-4 in one suit and calculate all the other 13 card hands with two extra straight flushes.

 

[awkward]

Hi Jameson,

The question of which hands should be counted as a four-card flush is, I think, ambiguous. The OP didn't say anything about poker, and a "four-card straight flush" is not a standard poker hand. So is a 2-3-4-5-6 of spades a four card straight flush, or is it a five card straight flush and therefore not a four card straight flush? And, for that matter, is J-Q-K-A of spades a four card flush, i.e., can an ace count as both low (1) and high (13)?

Personally, I favor the interpretation that any four consecutive cards of the same suite is a four-card straight flush, and therefore 2-3-4-5-6 of spades should be counted as a four card straight flush. I also favor the interpretation of an ace as 1, not 13. But clearly other interpretations are possible.

To summarize: Is this a question about poker, or not? And if so, what is the definition of "four card straight flush", since that is not a standard poker hand?

 

[Jameson]

I used to play a lot of poker and still enjoy many of the games (Hold 'Em, Omaha, 7-card stud, Razz, etc.), but this isn't a game I'm familiar with so I'm just guessing about some details and you're right that without explicit rules for the problem we don't know how to approach it exactly.

My comment about the 5 of diamonds was not that it makes it a 5 card straight flush, rather that it changes it from a A-4 straight flush to a 2-5 straight flush. The way this game might work in practice is you pick your best 4 cards from the 13 you have, so a 5 card straight flush wouldn't be possible but then again I just don't know.

I have a feeling that the intention of the problem was to calculate at least one 4 card straight flush, in which case we don't have to worry about excluding special cases but without more input from the OP we can't go further. :)