Number of 5 high lowball hands in a 7 card hand

Question

I am looking at a probability breakdown of lowball hands (lowest $5$ distinct cards, $12345$ being the best) here: http://www.durangobill.com/LowballPoker/Lowball_Poker_7_cards.html

The website lists the total number of $5$ high hands as $781,824$. Using a standard deck, I thought that the counting methodology would be $(4C_1)^5 \cdot (47C_2)$. This overestimates the amount of hands however. Why is this not the correct way to count?

Answer 1

You're double (and triple, and quadruple) counting because sometimes one or both of the $2$ cards you choose out of $47$ are lower than a 6.

For example, the hand consisting of Ace through 5 of diamonds and the Ace and 2 of hearts is counted 4 times.

The hand consisting of Ace through 5 of spades plus both red Aces is counted 3 times.

The hand consisting of 2 black Aces and the 2 through 6 of hearts is counted twice.

---

To set this up right, let's partition the cards: 20 cards are less than 6, 32 cards are 6 or more.

First let's count all the hands we don't need to double count or half count:

$${4 \choose 1}^5{32\choose2}$$

Next let's tackle the hands with 2 of some value below 6 (the value to be doubled is chosen out of five possible values):

$${5\choose1}{4\choose2}{4\choose1}^4{32\choose1}$$

Now let's tackle the hands with 3 of some value below 6 (note the 32 choose 0 included for completeness):

$${5\choose1}{4\choose3}{4\choose1}^4{32\choose0}$$

Now let's tackle the hands with 2 distinct values below 6 for which there are 2 cards:

$${5\choose2}{4\choose2}^2{4\choose1}^3{32\choose0}$$

Adding these figures together, we get 781824, just as stated on the page you linked.

Answer 2

The count can be organized as follows . . .

If a rank is at most $5$, call it a low rank.

If a hand contains the ranks $1,2,3,4,5$, call it a low hand.

Let

$\;\;{\small{\bullet}}\;\;x_0$ be the number of low hands with no low rank duplicated.

$\;\;{\small{\bullet}}\;\;x_1$ be the number of low hands with exactly one low rank duplicated.

$\;\;{\small{\bullet}}\;\;x_2$ be the number of low hands with two low ranks duplicated.

$\;\;{\small{\bullet}}\;\;x_3$ be the number of low hands with some low rank triplicated.

Then

$x_0 = {\large{{\binom{4}{1}}^5\binom{32}{2}}}=507904$.

$\qquad$Explanation:

$\qquad{\small{\bullet}}\;\;$Choose the $5$ low rank cards: $\;{\binom{4}{1}}^5\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the $2$ remaining cards, not of low rank: $\;{\binom{32}{2}}\;$choices.

$x_1 = {\large{\binom{5}{1}\binom{4}{2}{\binom{4}{1}}^4\binom{32}{1}}}=245760$.

$\qquad$Explanation:

$\qquad{\small{\bullet}}\;\;$Choose the duplicated low rank: $\;\binom{5}{1}\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the $2$ cards for that rank: $\;\binom{4}{2}\;$choices

$\qquad{\small{\bullet}}\;\;$Choose the other $4$ low rank cards: $\;{\binom{4}{1}}^4\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the remaining card, not of low rank: $\;\binom{32}{1}\;$choices.

$x_2 = {\large{\binom{5}{2}{\binom{4}{2}}^2{\binom{4}{1}}^3}}=23040$.

$\qquad$Explanation:

$\qquad{\small{\bullet}}\;\;$Choose the $2$ duplicated low ranks: $\;\binom{5}{2}\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the $2$ cards for each of those ranks: ${\;\binom{4}{2}}^2\;$choices

$\qquad{\small{\bullet}}\;\;$Choose the other $3$ low rank cards: $\;{\binom{4}{1}}^3\;$choices.

$x_3 = {\large{\binom{5}{1}\binom{4}{3}{\binom{4}{1}}^4}}=5120$.

$\qquad$Explanation:

$\qquad{\small{\bullet}}\;\;$Choose the triplicated low rank: $\;\binom{5}{1}\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the $3$ cards for that rank: $\;\binom{4}{3}\;$choices

$\qquad{\small{\bullet}}\;\;$Choose the other $4$ low rank cards: $\;{\binom{4}{1}}^4\;$choices.

Then the total number of low hands is $$x_0 + x_1 + x_2 + x_3 = 507904 + 245760 + 23040 + 5120 = 781824$$ In your count of$\;{\large{{\binom{4}{1}}^5\binom{47}{2}}}=1106944$,

$\qquad{\small{\bullet}}\;\;$Each $x_0$-type hand was counted correctly.

$\qquad{\small{\bullet}}\;\;$Each $x_1$-type hand was counted $2$ times.

$\qquad{\small{\bullet}}\;\;$Each $x_2$-type hand was counted $4$ times.

$\qquad{\small{\bullet}}\;\;$Each $x_3$-type hand was counted $3$ times.

As a check:

$$x_0 + 2x_1 + 4x_2+3x_3 = 507904 + (2)(245760) + (4)(23040) + (3)(5120) = 1106944$$ which was the count you obtained.

Answer 3

The other answers have explained why your solution overcounts, and derived the correct answer by ad hoc methods. Here is a solution using the in-and-out formula. The number of $7$-card hands containing a "wheel" (ace-to-five) is $$\binom50\binom{52}7-\binom51\binom{48}7+\binom52\binom{44}7-\binom53\binom{40}7+\binom54\binom{36}7-\binom55\binom{32}7=\boxed{781824}\ ;$$

$\binom50\binom{52}7$ is the total number of $7$-card hands;

$\binom51\binom{48}7$ is the number of ways you can choose a low rank and a $7$-card hand with no cards of that rank;

$\binom52\binom{44}7$ is the number of ways you can choose two low ranks and a $7$-card hand excluding both of those ranks, and so on.

P.S. Let $E$ be the set of all $7$-card hands. Let $A_i$ be the set of all $7$-card hands containing no card of rank $i;$ thus $A_1$ is the set of all $7$ card hands with no ace, etc. The set of all $7$-card hands containing a wheel is $$E\setminus(A_1\cup A_2\cup A_3\cup A_4\cup A_5).$$ By the in-and-out-formula, the number of hands containing a wheel is equal to $$|E\setminus(A_1\cup A_2\cup A_3\cup A_4\cup A_5)|$$ $$=|E|$$ $$-|A_1|-|A_2|-|A_3|-|A_4|-|A_5|$$ $$+|A_1\cap A_2|+|A_1\cap A_3|+|A_1\cap A_4|+|A_1\cap A_5|+|A_2\cap A_3|+|A_2\cap A_4|+|A_2\cap A_5|+|A_3\cap A_4|+|A_3\cap A_5|+|A_4\cap A_5|$$ $$-|A_1\cap A_2\cap A_3|-|A_1\cap A_2\cap A_4|-|A_1\cap A_2\cap A_5|-|A_1\cap A_3\cap A_4|-|A_1\cap A_3\cap A_5|-|A_1\cap A_4\cap A_5|-|A_2\cap A_3\cap A_4|-|A_2\cap A_3\cap A_5|-|A_2\cap A_4\cap A_5|-|A_3\cap A_4\cap A_5|$$ $$+|A_1\cap A_2\cap A_3\cap A_4|+|A_1\cap A_2\cap A_3\cap A_5|+|A_1\cap A_2\cap A_4\cap A_5|+|A_1\cap A_3\cap A_4\cap A_5|+|A_2\cap A_3\cap A_4\cap A_5|$$ $$-|A_1\cap A_2\cap A_3\cap A_4\cap A_5.$$ Since the deck has $52$ cards, and $4$ cards of each rank, we have: $$|E|=\binom{52}7;$$ $$|A_1|=|A_2|=|A_3|=|A_4|=|A_5|=\binom{48}7;$$ $$|A_1\cap A_2|=|A_1\cap A_3|=\cdots=\binom{44}7;$$ $$|A_1\cap A_2\cap A_3|=\cdots=\binom{40}7;$$ $$|A_1\cap A_2\cap A_3\cap A_4|=\cdots=\binom{36}7;$$ $$|A_1\cap A_2\cap A_3\cap A_4\cap A_5|=\binom{32}7;$$ and so $$|E\setminus(A_1\cup A_2\cup A_3\cup A_4\cup A_5)|=\binom{52}7-5\binom{48}7+10\binom{44}7-10\binom{40}7+5\binom{36}7-\binom{32}7$$ $$=\binom50\binom{52}7-\binom51\binom{48}7+\binom52\binom{44}7-\binom53\binom{40}7+\binom54\binom{36}7-\binom55\binom{32}7=\boxed{781824}$$