three-of-a-kind hands. Adding 13 to the
latter number gives us 2,509 hands which are three-of-a-kind hands and
this is the number in the table of the previous article.
Poker Digest Vol. 2, No. 3, January 29 - February 11, 1999
In my previous article I gave a table for the number of high hands in mambo stud and promised to discuss the computation of the numbers. Let me remind you we are counting the number of three-card hands from all possible 270,725 four-card hands; the mambo hands are straight flush, three-of-a-kind, straight, flush, one pair, high card; and the four-card hands are four-of-a-kind, four-card straight flush, four-card flush containing a straight flush, four-card flush not containing a straight flush, four-card straight containing a straight flush, four-card straight not containing a straight flush, straight flush with a pair, straight flush without a pair, three-of-a-kind, straight and flush, straight without a pair, straight with a pair, flush without a pair, flush with a pair, two pair, one pair and high card.
Some readers may wonder why I advocate dividing up the four-card hands as I have done rather than simply counting the three-card hands directly. Let me illustrate why with three-of-a-kind hands. The four-card hands which count as three-of-a-kind when evaluated as three-card hands are four-of-a-kind and three-of-a-kind. So the number is the sum of the number of four-of-a-kind hands and the number of three-of-a-kind hands.
There are 13 choices for the rank of a four-of-a-kind and the four cards are
determined at that point so there are 13 four-of-a-kind hands. To count
three-of-a-kind hands, there are 13 choices for the rank, four choices for
three cards of that rank, and 48 choices for the remaining card. This gives
us
three-of-a-kind hands. Adding 13 to the
latter number gives us 2,509 hands which are three-of-a-kind hands and
this is the number in the table of the previous article.
Now suppose we try to ``directly'' calculate the number of four-card
hands which will be considered three-of-a-kind hands. There are 13
choices for the rank and four choices for the three-of-a-kind of that rank.
This leaves 49 choices for the remaining card. Multiplying gives
.
Of course, this is incorrect as we
have seen. Why? I can only ask the reader to think about it and
try to see that four-of-a-kind hands are counted four times in the last
calculation. So we must subtract 39 to compensate for the overcounting
and this produces 2,509 which is correct. Now if you say to me the
problem occurs because of multiplying by 49 and all we have to do is
consider the possibilities when choosing the last card. But as soon
as you start making provisions for the choice of the last card, you
are essentially dividing the computation into two parts which is
equivalent to partitioning the four-card hands into four-of-a-kind hands
and three-of-a-kind hands. Why not just do it upfront and reduce the
chance of errors?
I shall now run through the remaining four-card hands and provide details for a few of the more interesting cases.
four-card flushes containing straight flushes.
of four distinct ranks one can choose from 13 ranks. Of these, 11
have the form
.
Another 98 have the form
,
where y is neither x-1 nor x+3, for if x = A
or x = Q, there are nine choices for
y and if x lies between deuce and J, inclusive, there are
eight choices for y.
So removing these 109 sets of ranks leaves 606 sets
of ranks which contain no straight or four-card straight possibilities.
Thus, for each of these 606 sets of ranks, there are four choices of
suits giving us 2,424 four-card flushes containing neither a straight
flush nor a four-card straight flush.
of ranks corresponding to four-card straights. There
are 44 = 256 choices of the four cards, but some choices
correspond to hands already counted: All in the same suit is a four-card
straight flush, and either the first three or the last three in the same
suit gives a straight flush. There are four choices for the former and
24 choices for the latter. Removing these 28 hands already counted gives
such hands.
straight flushes with no pairs.
.
We saw above that there are 98 such sets of four ranks. There are four
choices for the suit of y, three choices for the other two cards
of the same suit, and three choices for the other suit. This gives us
such hands.
.
There are 12 such sets. There are three choices for which
rank is paired and six choices for the pair of the chosen rank. The suits
for the remaining two ranks cannot both agree with one of the suits of the
pair or we would have a straight flush. Hence, there are 16 - 2 = 14
choices for the other two cards. This gives us
straights with a pair.
of ranks not admitting four-card straights or straights. Since four-card
flushes have been counted already, we choose three of the ranks
to be suited in four ways, there are four choices for the suit, and
three choices for the remaining suit. We obtain
flushes without a pair.
of ranks not allowing a straight. There are three choices for the
rank of the pair, six choices
for the pair and 14 choices for the remaining 2 cards since we cannot allow
both of the remaining 2 cards to be in the same suit as either suit of the
pair as this would produce a flush. Therefore, there are
hands with a single pair.
of ranks not allowing a straight or four-card straight. We also must
eliminate flushes. Of the 256 choices for the cards, four are all in
the same suit and 48 have three in the same
suit. Eliminating them leaves 204 choices for each set of ranks.
This produces
hands of this type.
From the above information, one easily can obtain the entries of the table in my previous article by coalescing the appropriate four-card hands corresponding to the same type of three-card hand.