Let's now look at the probability that 2 players end up with straight
flushes. For simplicity let us use hearts as the suit but it is clear
there is nothing special about hearts. If there are 3 hearts in the board
and 2 players have straight flushes, the only possibility is that the ranks
of the hearts on board have the form
,
1 player has *x*-2and *x*-1 of hearts, and the other player has *x*+3 and *x*+4 of hearts.
Thus, the possible range for *x* is
.
Let's count the
number of boards which look like this. There are 4 choices for suit, 8
choices for the rank *x* and
choices for the other
2 cards comprising the board. The product of these 3 numbers is 23,712.

For each such board, there are 4 hearts which must go to 2 players in a
fixed way in order for 2 straight flushes to arise. There are
choices for the completion of 1 set of hearts to a hand,
and
for the completion of the other 2 hearts to
a hand. This produces 740,460 ways for 2 players to have hands which
give 2 players straight flushes for a fixed board of the type under
discussion. For the other 8 players, we are choosing 32 cards from 39
and then partitioning them in (5) ways to give us semi-deals. Thus, the
number of semi-deals with 2 players having straight flushes, when there
are 3 suited cards on board, is the product of 23,712,
,
(5) and 740,460. We get

16,010,830,956,990,783,872,058,420,189,870,000,000 | (9) |

semi-deals.

Now let us assume there are 4 hearts on board and 2 players have straight
flushes. We have to determine the number of rank sets for 4 hearts which
allow 2 players to have straight flushes. One possibility is that there
are 4 successive ranks of the form
,
where
.
There are 7 sets of that form. Another possibility is 3
successive ranks together with an isolated rank, that is, a rank set of
the form
,
where
.
The range for
*x* is from 3 through 10. When *x*=3 or *x*=10, there are 3 ranks which
*y* cannot take on, but for any other value of *x*, there are only 2
ranks from which *y* is excluded. Thus, there are 62 rank sets with 3
successive ranks.

Now consider 2 successive ranks occurring but not 3 consecutive ranks.
The only way 2 players can have straight flushes in this case is if the
rank set has the form
,
where *y*<*x*-1 and *z*>*x*+2. The
possible range for *x* is
.
There are 2 possible values
for *y*, namely, *x*-2 or *x*-3, and 2 possible values for *z*. Thus,
there are 28 rank sets.

Finally, 4 hearts on board with no successive ranks do not allow 2
players to have straight flushes so this pattern cannot occur. Thus,
we have 97 legal rank sets with 4 suited cards on board. There are 4
choices for the suit, 39 choices for the fifth card, 740,460 ways for
2 players to have hands which give 2 straight flushes, and the number of
semi-deals in (5) for the remaining 8 players. We take the product of
all these numbers to give us

664,292,310,442,510,596,237,641,250,000 | (10) |

semi-deals for which 2 players have straight flushes when there are 4 suited cards on board.

We now move to the case there are 5 suited cards on board. The primary
work is to determine how many rank sets there are which allow 2 players
to have qualifying straight flushes. If there are 5 successive ranks,
then 2 players can have qualifying straight flushes. In this case, the
range for the smallest card is from 3 through 8. Hence, there are 6 rank
sets being counted here. Now suppose there are 4 successive ranks
*x*,
*x*+1,*x*+2,*x*+3 and a fifth rank *y*. The possible range for *x* is
.
When *x*=3 or *x*=9, there are 3 values excluded for
*y*. In the other cases, there are only 2 values excluded for *y*. So
we have 47 rank sets which work when there are 4 successive ranks.

Now suppose the rank set has the form
.
Clearly,
must be the case as there is not room to fit 2 qualifying
straight flushes otherwise. When *x*=3 or *x*=10, *y*,*z* may be any 2
ranks chosen from a 7-set so there are 21 choices. When *x*=4 or *x*=9,
there are 27 choices because we cannot choose both A and 2 in the former
case, or both K and A in the latter case. For all other values we can
choose any 2 from an 8-set giving us 28 choices. Altogether we have
208 rank sets which work when there are 3 successive ranks.

Now we may assume the longest sequence of successive ranks is 2. First
consider the situation in which there are 2 such sequences, that is, we
have ranks *x*,*x*+1 and *y*,*y*+1. Without loss of generality we may assume
*x* is the smallest of the 4 ranks. If the fifth rank *z* is smaller than
*x*, then one qualifying player must form a straight flush using *z*,*x*,*x*+1but NOT hold *x*+2 in her hand as no other player would be able to have a
straight flush. So the 2 possibilities are *x*-3,*x*,*x*+1 and *x*-2,*x*,*x*+1with the player holding *x*-2,*x*-1 and *x*-3,*x*-1, respectively. This
then forces *y*,*y*+1 to be either *x*+3,*x*+4 or *x*+4,*x*+5, where the second
player's cards are uniquely determined in both cases for her to have a
qualifying straight flush. Since *x*-3 can be as small as ace and *x*+5can be as big as ace, the possible range for *x* is 4 through 9 and we
get 24 rank sets. Next, if *z* lies between *x*+1 and *y*, then we can
work out that the possibilities are
*x*,*x*+1,*x*+3,*x*+5,*x*+6,
*x*,*x*+1,*x*+3,*x*+6,
*x*+7,
*x*,*x*+1,*x*+4,*x*+6,*x*+7, and
*x*,*x*+1,*x*+4,*x*+7,*x*+8. Again there are 24
possibilities. If *z* is larger than *y*+1, we get a reflection of the
first case giving us another 24. So we have 72 rank sets with 2
sequences of successive ranks.

The next case is only 1 sequence of successive ranks *x*,*x*+1. The other
3 isolated ranks cannot all be smaller than *x* or all larger than *x* as
it is easy to see this makes it impossible for 2 players to have straight
flushes. Assume exactly 1 rank is smaller than *x*. It is easy to see
it must be either *x*-3 or *x*-2. The larger rank is either *x*+3 and
something of rank *x*+5 and bigger, or *x*+4 and something bigger than
*x*+5. Actually listing them all leads to finding there are 68 like this.
By symmetry there are another 68 when exactly 1 rank is larger than *x*+1.
This gives us 136 rank sets in this case.

Finally, we move to rank sets with no successive ranks. There can be 2
players with straight flushes only when the rank set has the form
*x*,*x*+2,*x*+4,*x*+6,*x*+8. The range for *x* is ace through 6 so there are
6 rank sets of this type. Adding all of the rank sets gives us 475
rank sets. Since there are 4 possible suits, the total number of rank
sets allowing 2 qualifying straight flushes when there are 5 suited cards
in the board is 1,900. As before, there are 740,460 ways of producing
2 hands which yield the 2 straight flushes for each fixed board. We then
multiply by (5) to take into account the remaining 8 hands. This yields

83,409,687,406,870,878,459,656,250,000 | (11) |

semi-deals for which 2 players have qualifying straight flushes when there are 5 suited cards on board.

Adding (9), (10) and (11) yields

16,010,831,704,692,781,721,439,894,887,167,500,000 | (12) |

semi-deals for which 2 players have qualifying straight flushes. Dividing (12) by (3) yields a probability of .00002762184811 that 2 players have qualifying straight flushes. This is roughly 1 in 36,000.