|
|
| the
number of possible hands |
 |
There are 52 cards in a pack,
and 5 cards in a hand, so the answer is any 5 from 52 - or
52C5 in
shorthand. The total comes to 2,598,960. |
|
|
| |
| straight
flush [ includes royal flush ] |
All same suit
All consecutive |
As in Brag, Aces can be counted either high or low when creating straights,
which gives 10 straight flushes for every suit ( A-2-3-4-5 to 10-J-Q-K-A
). So the total is 40. |
|
With 13 denominations, there are 13 possible fours. These, however,
can be combined with any of the remaining 48 cards to complete the
hand, so there are 13 × 48 = 624 ways to get four of a kind.
|
Four of same rank |
3 of one rank
Two of another |
There are 13 possible threes, each of which can come in 4C3
= 4 combinations of suits. There remain 12 denominations to make
up the pair, each of which can come in 4C2 =
6 permutations of suits. The formula is thus:
| Threes |
|
Pairs |
|
No.
of Hands |
| 13
× 4 |
× |
12
× 6 |
= |
3,744 |
|
|
In each of the suits, we can get 13C5 =
1287 flushes. So total flushes = (1287 × 4) = 5,148.
Subtracting the 40 straight flushes leaves a total of 5,108. |
All same suit |
All consecutive |
As Aces can be counted either high or low when creating straights,
there are 10 possible ranks of straights (A-2-3-4-5 to 10-J-Q-K-A).
But each of the five cards can be any one of 4 suits, so there are
in fact 10 × 4 × 4 × 4 × 4 × 4 = 10240 straights. Subtract from
that the 40 straight flushes, and the final total is 10,200. |
3 of one rank |
With 13 denominations and
4C3 = 4 permutations of suits, there are
13 × 4 = 52 threes. But for every possible three, there are
many different combinations of the other two cards needed to make
up the full hand of five cards. Only 12 denominations can be taken
into account ( the 13th would produce a four-of-a-kind ) and the denominations
must be different ( or a Full House would result ). Each of these
cards can come in any of 4 suits, so the full calculation is: |
| Threes |
|
Other
two
denominations |
|
Allow
4 suits:
cards 4 & 5 |
|
Hands |
| 13
× 4 |
× |
66 |
× |
4
× 4 |
= |
54,912 |
|
| There are 78 permutations
of 2 denominations - calculated from 13C2 -
so that is the number of different "pairs of pairs" you
can get. But each pair can come in 6 different combinations of suits,
so the true number for the two pairs in this hand is 78 × 6 ×
6. These in turn can be combined with any one of the 44 remaining
cards of different denominations : |
Two pairs |
Perm
2
denominations |
|
6
suit combs.
for each pair |
|
5th
card |
|
Hands |
| 13C2
= 78 |
× |
6
× 6 |
× |
44 |
= |
123,552 |
|
One Pair |
There are 78
ways to get the pair. The other 3 cards in the hand must all be of
different denominations (both from the pair and from each other) but
can be of any suit. So the three other cards can be made up of 12C3
= 220 permutations. But since any of these three cards can come
from any of the 4 suits, the "other three" can be
made up of (220 × 4 × 4 × 4) combinations of cards
: |
| One
Pair |
|
3
different
denominations |
|
Last
3 cards:
Allow 4 suits |
|
Hands |
| 78 |
× |
220 |
× |
4
× 4 × 4 |
= |
1,098,240 |
|
Highest Card |
The three elements of the
Ace high permutation are:
- The number of Aces ( 4
)
- The perms of denominations
( ranks ) for the other four cards
- The perms of suits for
the other four cards
|
Multiply
a, b, and c - and ignore any permutations which would produce
straights or flushes - and that is the answer. This is it step
by step, for the moment considering only the hands that have
the Ace of Clubs high.
Permutations of Ranks
The cards must all have different denominations, and must be
drawn from the twelve ranks 2 to 13 ( = King ). So there
are 12C4 = 495 perms. Deduct from
that the two combinations which along with the Ace would produce
straights ( i.e. 2-3-4-5 and 10-J-Q-K ), and we are left with
493 perms of ranks.
Permutations of Suits
There are four cards to perm, and 4 suits, so the full perm
is 4 × 4 × 4 × 4 = 256 combinations of suits.
One of these, however, will produce a flush (Club-Club-Club-Club
in this example, where the high card is the Ace of Clubs);
that one can be discounted, leaving the true number of perms
of suits as 255.
Number of Aces High
From the above calculations, it follows that the number of hands
with an Ace of Clubs High is 493 × 255 = 125,715. Multiply
that by 4 to include in the total the other three aces, and
the number of Aces high is calculated as 502,860.
The General Formula
A formula for Highs can be derived from the above computations.
Note that in the case of Ace Highs, the number of straights
is 2, but in every other case the number is 1; also that the
"rank" element in the formula refers to the normal
concept of cards' ranks: e.g. Twos have a rank of 2, Tens
a rank of 10, Jacks a rank of 11, and Aces a rank of 14
| High
Card |
Other
Four Cards |
| No.
of Suits |
|
Perms
of Ranks |
|
Perms
of Suits |
| 4 |
× |
(
rank - 2 )C4 - straights |
|
255 |
|
|
| Hand |
Total |
% |
Blank |
Breakdown
of Highs |
% |
| Straight
Flush |
40 |
0.0015 |
|
Ace |
502,860 |
19.35 |
| Four of
a Kind |
624 |
0.024 |
|
King |
335,580 |
12.91 |
| Full House |
3,744 |
0.14 |
|
Queen |
213,180 |
8.20 |
| Flush |
5,108 |
0.20 |
|
Jack |
127,500 |
4.91 |
| Straight |
10,200 |
0.39 |
|
10 |
70,380 |
2.71 |
| Three
of a Kind |
54,912 |
2.11 |
|
9 |
34,680 |
1.33 |
| Two Pairs |
123,552 |
4.75 |
|
8 |
14,280 |
0.55 |
| One Pair |
1,098,240 |
42.26 |
|
7 |
4,080 |
0.16 |
| Highs |
1,302,540 |
50.12 |
|
TOTAL |
1,302,540 |
50.12 |
| TOTAL |
2,598,960 |
100.00 |
|
|
|
|
NOTES
- The Median hand
lies right between A-K-Q-J-7 and A-K-Q-J-6
- Straight Flushes
will appear on average every 64,974 hands: rather less often
than you see on the Westerns.
|
|
