We determine the probabilities of the hands for mambo poker.
In mambo poker one uses the best 3-card hand which can be formed
from 4 cards. We are going to count the numbers of ways of achieving the
possible hands. Since the hands are based on 3 cards, when we use the word
straight or flush throughout this file, we automatically mean 3-card straights
and 3-card flushes, respectively. Any deviations from this will be named
explicitly.
Since the game is played high-low, we examine both high and low
hands. We consider high hands first. The total the number of possible 4-card
hands is given by \begin{ thaibet168 }{{52}\choose{4}}=
270,725.\end{displaymath}
In order to avoid double counting certain hands, we shall mention
a variety of 4-card possibilities and decide later how the hands should be
valued. For example, there are 4-card hands containing both a straight and a
flush. We shall distinguish them initially in order to make certain, for
example, that straights really beat flushes.
The initial step is to determine all the types of hands to be
counted. They are 4-of-a-kind, 4-card straight flush, 4-card flush containing a
straight flush, 4-card flush not containing a straight flush, 4-card straight
containing a straight flush, 4-card straight not containing a straight flush,
straight flush with a pair, straight flush without a pair, 3-of-a-kind,
straight & flush, straight without a pair, straight with a pair, flush
without a pair, flush with a pair, 2 pairs, 1 pair and high card.
There are 13 possible ranks for the quartet and precisely 1
4-of-a-kind of each rank. Thus, there are 13 4-of-a-kind hands.
There are 8 straight flushes beginning with an A or a Q. Any of 9
cards can be added to each to produce a flush which is not a 4-card straight
flush. There are 40 straight flushes remaining and any of 8 cards can be added
to each. This gives us $(8\cdot 9) + (40\cdot 8) = 392$ 4-card flushes
containing staright flushes.
We employ a technique in this case which will be used in several
other cases leading us to explain it now in detail. There are ${{13}\choose{4}}=
715$ sets $\{x,y,z,w\}$ of 4 distinct ranks one can choose from 13 ranks. Of
these, 11 have the form $\{x,x+1,x+2,x+3\}$. Another 98 have the form
$\{x,x+1,x+2,y\}$, where y is neither x-1 nor x+3, for if x = A or x = Q, there
are 9 choices for y and if x lies between 2 and J, inclusive, there are 8
choices for y. So removing these 109 sets of ranks leaves 606 sets
$\{x,y,z,w\}$ of ranks which contain no straight or 4-card straight
possibilities. Thus, for each of these 606 sets of ranks, there are 4 choices
of suits giving us 2,424 4-card flushes containing neither a straight flush nor
a 4-card straight flush.
There are 8 straight flushes beginning with A or Q. We can add
any of 3 cards to each of them to obtain a 4-card straight containing a
straight flush. To the remaining 40 straight flushes we may add any of 6 cards.
This produces $(8\cdot 3) + (40\cdot 6) = 264$ 4-card straights containing a
straight flush.
4-card straight not containing a straight flush.
There are 11 sets $\{x,x+1,x+2,x+3\}$ of ranks corresponding to
4-card straights. There are 44 = 256 choices of the 4 cards, but some choices
correspond to hands already counted: All in the same suit is a 4-card straight
flush, and either the first 3 or the last 3 in the same suit gives a straight flush.
There are 4 choices for the former and 24 choices for the latter. Removing
these 28 hands already counted gives $11\cdot 228 = 2,508$ such hands.
Straight flush with a pair.
There are 48 straight flushes and any of 9 cards producing a pair
as well. This gives $9\cdot 48 = 432$straight flushes which also contain a
pair. Notice the hand also contains a straight.
Web: https://thaibet168.art.blog/2021/10/13/mambo-poker/
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