Probability with the poker

One can calculate the Probabilité of having each type of hand of 5 charts to the Poker

Closed poker: basic hand

The number of combinations of each hand is calculated with the assistance of the combinations

• Is NR the number of values (N=13 for one pack of 52 cards, and N=8 for a set of 32 charts).
• Is S the number of allowed continuations.
• For one pack of 52 cards, by counting the series A-2-3-4-5 with 10-V-D-R-A, S=10.
• For a set of 32 charts, continuations going of 7-8-9-10-V with 10-V-D-R-A, S=4.

Table of synthesis

The first table summarizes the probabilities of obtaining each hand, for one pack of 52 cards and a set of 32 charts. Calculations for pack of 52 cards are made with the “extended fifths”, i.e. the combination A-2-3-4-5 (white fifth) is regarded as a fifth. It is noticed that the order of difficulty of the hands is not the same one for the two plays: the color becomes rarer than the square, and the high chart rarer than a pair. It will be noticed that the hands " servies" , with the top of the brelan, are extremely rare: less than one percent of the hands with 52 charts, and less than 3% to 32 charts. ; Probability of having at least… In practice, the greatest majority of the plays is played in the low zone: Nothing, pair, playable, double pulling even or brelan. These are these hands that it is necessary to study to discuss the risks of openings and the levels of revival. The reading of this table is directly: if the heel is of 52 charts, a player has more than one pair of ace in 14.4% of the distributed plays.

This table takes into account playable pullings: pullings with the color, the fifth bilateral flush, and pullings with the fifth (simple pullings with the fifth are not regarded as " jouables"). It does not take into account the white fifths (but it is without notable incidence on the figures).

The table considers that the " pulling jouable" is higher than the pair. It is overall true, because after exchange pulling makes it possible to gain more frequently than a pair, but if one distinguishes the types of pullings, pulling with the color is from this a little weak point of view, and could be retrogressed in the table.

This table is independent of the number of player, but is not exploited directly thus. The typical use of this table is to answer questions like: Is I have a been useful pair of king, we play four to 32 charts, which the probability a priori so that my hand is the best? For this type of question, the stages of calculation are:

• the probability for a player of having more than one pair of king under these conditions is: 36.8%. He will have less with one probability of 63.2%.
• So that the pair of king is strongest, he is necessary that the first adversary has less AND the second has less AND the third has less. The probability is the product of the three: 63.2% X 63.2% X 63.2% = 25.2%.
• One can thus bet with against three that my pair of kings is not the best hand of the four.

Total

There is 4N charts in the package, there is thus $\backslash \; textstyle$ possible hands of 5 charts.

Fifth flush

A fifth flush is determined by the value of its high chart ($\backslash \; textstyle$ possibilities), and by its color (4 possibilities).

On the whole, $\backslash \; textstyle$.

Square

A square is determined by the value of the square ($\backslash \; textstyle$ possible values), and by the free chart ($\backslash \; textstyle$ possibilities).

The only combination above the square is the fifth flush (possibly royal), and a square cannot be also a fifth flush, whatever the free chart, since the 4 charts of the square have different colors

On the whole, $\backslash \; textstyle$.

Full

A full is determined by the value of the brelan ($\backslash \; textstyle$ possible values), the colors of the 3 charts which compose the brelan ($\backslash \; textstyle$ possible combinations of colors), the value of the pair ($\backslash \; textstyle$ possible values) and the colors of the 2 charts which compose it ($\backslash \; textstyle$ possible combinations of colors).

A full can be neither a square (since there is no free chart), nor a fifth flush (since the 3 charts of the brelan have different colors).

On the whole, $\backslash \; textstyle$.

Color

A color contains 5 charts of different value among NR, each chart having to be same color.

A color can be neither a square, nor a full, since the 5 charts have different values inevitably. A color can be a fifth flush, it thus should be excluded.

On the whole, $\backslash \; textstyle\; \{\backslash \; left\; (\{NR\; \backslash \; choose\; 5\}\; -\; S\; \backslash \; right)\; 4\}$.

Fifth

A fifth is determined by the value of its high chart ($\backslash \; textstyle$ possibilities), and by the colors of the charts which compose it. There is $\backslash \; textstyle\; \{4^5\}$ combinations of colors.

A fifth can be neither a full, nor a square, since the 5 charts have different values inevitably. But it can be a color, and in this case, it is a fifth flush. It is thus necessary to exclude these cases, i.e. 4 combinations of colors among the $\backslash \; textstyle\; \{4^5\}$.

On the whole, $\backslash \; textstyle$.

At least a Fifth (Fifth flush)

It is the possibility of having a fifth or to have a fifth flush better if all the charts are same color.

On the whole, $\backslash \; textstyle$.

Brelan

A brelan is determined by the value of the brelan ($\backslash \; textstyle$ possibilities), the colors of the charts of the brelan ($\backslash \; textstyle$ possibilities), and the 2 charts free.

So that the hand is neither a square, nor a full, it is necessary that the values of the two charts are different one from the other and different from the value of the brelan. Their color is free. ($\backslash \; textstyle$ possibilities).

The hand can be a continuation, since the values of the 3 charts which form the brelan should be different, neither a color (nor a fifth flush), since the colors of the 3 charts of the brelan should be identical.

On the whole, $\backslash \; textstyle$.

Two Pairs

Two pairs are determined by the values of the two pairs ($\backslash \; textstyle$ possibilities), the colors of the two charts of each pair ($\backslash \; textstyle$ possibilities for each one).

Two pairs can be neither a continuation, neither a color, nor a fifth flush since the values of the charts should be different. Two pairs cannot be a square either, since the free chart makes a full as well as possible. Not to make a brelan/full, it is necessary that the free chart has a value different from each of the two pairs (N2 possibilities). Its color is free (4 possibilities).

On the whole, $\backslash \; textstyle$.

At least a pair (Full or double even possible)

A pair is determined by its value ($\backslash \; textstyle$ possibilities), the color of its charts ($\backslash \; textstyle$ possibilities), the values of the 3 free charts ($\backslash \; textstyle\; \{(N-1)\; ^3\}$ possibilities) and their colors.

On the whole, $\backslash \; textstyle$.

At least a pair (Brelan, square, full or double pair possible)

The number of hands not containing a pair is obtained by choosing 5 values among the NR possible: $\backslash \; textstyle$ and for each chart, its color $\backslash \; textstyle\; \{4^5\}$, is $\backslash \; textstyle$.

On the whole, $\backslash \; textstyle$.

A Pair

A pair is determined by its value ($\backslash \; textstyle$ possibilities), the color of its charts ($\backslash \; textstyle$ possibilities).

A pair can be neither a continuation, neither a color, nor a fifth flush since the values of the charts should be different. So that the pair forms neither two pairs, neither a brelan, neither a full, nor a square, it is necessary that the values of the 3 free charts are different between them and different from the value of the pair ($\backslash \; textstyle$ possibilities). Their colors are free ($\backslash \; textstyle\; \{4^3\}$ possibilities).

On the whole, $\backslash \; textstyle$.

High chart

In a hand “High Chart”, each chart has a different value. It is thus necessary to however draw 5 values among NR., these combinations, there is S of it which form continuations, that one should not count. Moreover, each one of these 5 charts can have any color, provided that the 5 charts do not have the same color. There is thus $\backslash \; textstyle\; \{4^5-4\}$ combinations of color. On the whole, there is $\backslash \; textstyle\; \{\backslash \; left\; (\{NR\; \backslash \; choose\; 5\}\; -\; S\; \backslash \; right)\; \backslash \; left\; (4^5-4\; \backslash \; right)\}$ combinations

Closed poker: improvement of a hand

Fifth flush, square, full, color and fifth are little (or not) improvable. One is thus interested in particular in the probabilities of improvement with an initial hand of chart type high, even, brelan or double pair. In the hands “high chart”, one notes in particular the cases where it misses only one chart to form a color or a fifth, which one calls pulling .

Open poker: better hand on 7 charts

In the Texas Hold' EM or the Stud with seven charts, it is a question of forming the best hand of five charts among Sept.

; Details of calculation

That is to say NR the number of values (N=13 for one pack of 52 cards, and N=8 for a set of 32 charts).

Total

There is 4N charts in the package, there is thus $\backslash \; textstyle$ possible hands of 7 charts.

Fifth flush

A fifth flush is determined by the value of its high chart ($\backslash \; textstyle$ possibilities), by its color (4 possibilities), and by the 2 free charts ($\backslash \; textstyle$ possibilities). However, if the fifth flush is not royal, one of the two free charts can improve it, and thus one counts several times the same hand. It is enough to just prohibit the chart above the fifth flush, of the same color, for the 2 free charts. There is then ($\backslash \; textstyle$ possibilities for the 2 free charts.

On the whole, $\backslash \; textstyle\; \{4\; \backslash \; left\; (\{4N-5\; \backslash \; choose\; 2\}\; +\; \{S-1\; \backslash \; choose\; 1\}\; \{4N-6\; \backslash \; choose\; 2\}\; \backslash \; right)\}$.

Square

A square is determined by the value of the square ($\backslash \; textstyle$ possible values), and by the 3 free charts ($\backslash \; textstyle$ possibilities).

The 3 free charts cannot form fifth flush, since one would need 2 charts moreover, and that the 4 remaining charts of the hand have even value.

On the whole, $\backslash \; textstyle$

Full

There are 3 ways of building a full:

1. a brelan, a pair, and two free charts different
2. two brelans and a free chart
3. a brelan and two pairs

For each one of these ways, the hand is determined by:

1. the brelan ($\backslash \; textstyle$ possibilities), the pair ($\backslash \; textstyle$ possibilities), and values of the two free charts (their values are different, their non-object colors, therefore $\backslash \; textstyle$ possibilities).
2. both brelans ($\backslash \; textstyle$ possibilities) and the free chart ($\backslash \; textstyle$ possibilities).
3. the brelan ($\backslash \; textstyle$ possibilities) and the pairs ($\backslash \; textstyle$ possibilities)

None of these combinations can be a square since one prohibits with the pairs to have the same value, and with the free charts to have the same value as the brelans or the pairs. None of these combinations can be a fifth flush, since there are 4,3 and 3 different values.

On the whole, $\backslash \; textstyle$.

Color

There are 3 ways of obtaining a color:

1. 5 of the same charts color, 2 free charts of colors different
2. 6 of the same charts color, 1 free chart of different color
3. 7 of the same charts color

For each one of these ways, the hand is determined by:

1. the 5 charts ($\backslash \; textstyle$ possibilities), their free color and two charts ($\backslash \; textstyle$ possibilities).
2. the 6 charts ($\backslash \; textstyle$ possibilities), their color, the free chart ($\backslash \; textstyle$ possibilities).
3. the 7 charts ($\backslash \; textstyle$ possibilities) and their color.

The possible free charts can form neither full nor square, since at best they form a brelan with one of the 5 charts. They can form all the possible fifths flush. They thus should be withdrawn.

On the whole, $\backslash \; textstyle\; \{4\; \backslash \; left\; (\{NR\; \backslash \; choose\; 5\}\; \{3N\; \backslash \; choose\; 2\}\; +\; \{NR\; \backslash \; choose\; 6\}\; \{3N\; \backslash \; choose\; 1\}\; +\; \{NR\; \backslash \; choose\; 7\}\; -\; \backslash \; left\; (\{4N-5\; \backslash \; choose\; 2\}\; +\; \{S-1\; \backslash \; choose\; 1\}\; \{4N-6\; \backslash \; choose\; 2\}\; \backslash \; right)\; \backslash \; right)\}$.

Fifth

There are 4 ways of obtaining a Fifth with 7 charts:

no pair, 7 values different

The hand is then determined by the value of the fifth (S possibilities), the value of the two free charts ($\backslash \; textstyle$ possibilities), and the color of the 7 charts ($\backslash \; textstyle\; \{4^7\}$ possibilities).

However, if the fifth is not with the Ace, one of the two free charts can improve it, and thus one counts several times the same hand. It is enough to just prohibit the charts above the fifth, for the 2 free charts. There is then ($\backslash \; textstyle$ possible values for the 2 free charts.

This hand can be neither a full nor a square, since the values all are different. It can be a color, possibly a fifth flush, so among 7 colors 5 are identical, which arrives in $\backslash \; textstyle\; \{4\; \backslash \; left\; (1+\; \{7\; \backslash \; choose\; 6\}\; 3+\; \{7\; \backslash \; choose\; 5\}\; 3^2\; \backslash \; right)\}$ combinations of colors. It is thus necessary to restrict the number of combinations of colors to ($\backslash \; textstyle\; \{4^7-4\; \backslash \; left\; (1+\; \{7\; \backslash \; choose\; 6\}\; 3+\; \{7\; \backslash \; choose\; 5\}\; 3^2\; \backslash \; right)\}$.

In all, $\backslash \; textstyle\; \{\backslash \; left\; (\{N-5\; \backslash \; choose\; 2\}\; +\; (S-1)\; \{N-6\; \backslash \; choose\; 2\}\; \backslash \; right)\; \backslash \; left\; (4^7-4\; \backslash \; left\; (1+\; \{7\; \backslash \; choose\; 6\}\; 3+\; \{7\; \backslash \; choose\; 5\}\; 3^2\; \backslash \; right)\; \backslash \; right)\}$.

a pair, 6 values different

the hand is then determined by the value of the fifth (S possibilities), the value of the free chart, ($\backslash \; textstyle$ possibilities), the value of the pair ($\backslash \; textstyle$ possibilities), the colors of the charts of the pair ($\backslash \; textstyle$ possibilities) and the color of the charts of the fifth ($\backslash \; textstyle\; \{4^5\}$ possibilities).

However, if the fifth is not with the Ace, the free chart can improve it, and thus one counts several times the same hand. It is enough to just prohibit the charts above the fifth, for the free chart. There is then ($\backslash \; textstyle$ possible values for the free chart.

This hand can be neither a full nor a square, since except the pair, the values all are different. It can be a color, possibly a fifth flush, so among 7 colors 5 are identical. 2 ways that arrives:

• the 5 charts which do not form part of the pair are same color ($\backslash \; textstyle$ possibilities)
• among the 5 charts which do not form part of the pair, 4 have the same color as one of the 2 charts of the pair ($\backslash \; textstyle$ possibilities).

In all, $\backslash \; textstyle\; \{\backslash \; left\; (\{N-5\; \backslash \; choose\; 1\}\; +\; (S-1)\; \{N-6\; \backslash \; choose\; 1\}\; \backslash \; right)\; \{4\; \backslash \; choose\; 2\}\; \backslash \; left\; (4^5\; -\; \backslash \; left\; (\{4\; \backslash \; choose\; 1\}\; +\; \{5\; \backslash \; choose\; 4\}\; \{2\; \backslash \; choose\; 1\}\; \{3\; \backslash \; choose\; 1\}\; \backslash \; right)\; \backslash \; right)\}$.

two different pairs, 5 values

The hand is then determined by the value of the fifth (S possibilities), the value of the pairs ($\backslash \; textstyle$ possibilities), the colors of the charts of the two pairs ($\backslash \; textstyle$ possibilities) and the color of the three remaining charts of the fifth ($\backslash \; textstyle\; \{4^3\}$ possibilities).

The two free charts form each one a pair with a chart of the fifth, they cannot thus improve it.

This hand can be neither a full nor a square, since except the two pairs, the values are different. It can be a color, possibly a fifth flush, so among 7 colors 5 are identical. These 5 charts inevitably use a color of each pair, plus the 3 charts which do not form part of a pair, since two charts of the same pair have different colors:

• if the two pairs have the same colors (in 1 case on the $\backslash \; textstyle$ of the color of the second pair), 2 possibilities
• if the two pairs have a single joint color (in $\backslash \; textstyle$ case of the color of the second pair), 1 possibility

That is to say $\backslash \; textstyle$

In all, $\backslash \; textstyle\; \{S\; \{5\; \backslash \; choose\; 2\}\; \{4\; \backslash \; choose\; 2\}\; ^2\; \backslash \; left\; (4^3-1\; \backslash \; right)\}$.

a brelan, 5 different values

The hand is then determined by the value of the fifth (S possibilities, the value of the two free charts ($\backslash \; textstyle$ possibilities), the colors of the charts of the brelan ($\backslash \; textstyle$ possibilities), and the color of the four remaining charts of the fifth ($\backslash \; textstyle\; \{4^4\}$ possibilities).

The two free charts form a brelan with one of the charts of the fifth, they cannot thus improve it.

This hand can be neither a full nor a square, since except the brelan, the values are different. It can be a color, possibly a fifth flush, so among 7 colors 5 are identical. These 5 colors inevitably use a color of the brelan, and the 4 remaining charts of the fifth: 3 possibilities.

In all, $\backslash \; textstyle\; \{S\; \{5\; \backslash \; choose\; 1\}\; \{4\; \backslash \; choose\; 3\}\; \backslash \; left\; (4^4-3\; \backslash \; right)\}$

On the whole,

$\backslash \; textstyle\; \{\backslash \; left\; (\{N-5\; \backslash \; choose\; 2\}\; +\; (S-1)\; \{N-6\; \backslash \; choose\; 2\}\; \backslash \; right)\; \backslash \; left\; (4^7-4\; \backslash \; left\; (1+\; \{7\; \backslash \; choose\; 6\}\; 3+\; \{7\; \backslash \; choose\; 5\}\; 3^2\; \backslash \; right)\; \backslash \; right)\; +\; \backslash \; left\; (\{N-5\; \backslash \; choose\; 1\}\; +\; (S-1)\; \{N-6\; \backslash \; choose\; 1\}\; \backslash \; right)\; \{4\; \backslash \; choose\; 2\}\; \backslash \; left\; (4^5\; -\; \backslash \; left\; (\{4\; \backslash \; choose\; 1\}\; +\; \{5\; \backslash \; choose\; 4\}\; \{2\; \backslash \; choose\; 1\}\; \{3\; \backslash \; choose\; 1\}\; \backslash \; right)\; \backslash right)\; +S\; \{5\; \backslash \; choose\; 2\}\; \{4\; \backslash \; choose\; 2\}\; ^2\; \backslash \; left\; (4^3-1\; \backslash \; right)\; +S\; \{5\; \backslash \; choose\; 1\}\; \{4\; \backslash \; choose\; 3\}\; \backslash \; left\; (4^4-3\; \backslash \; right)\}$

Brelan

A brelan is determined by the 5 values (the free brelan and 4 charts, $\backslash \; textstyle$ possibilities), the value of the brelan among those ($\backslash \; textstyle$ possibilities), the colors of the charts of the brelan ($\backslash \; textstyle$ possibilities), and the 4 charts free.

So that this hand is neither a square, nor a full, it is necessary that the values of the four free charts are different two to two and different from the value of the brelan ($\backslash \; textstyle$ possibilities). Their color is free ($\backslash \; textstyle\; \{4^4\}$ possibilities).

This hand can be a continuation, if the 5 values are followed (S possibilities).

This hand can be a color, if the 4 free charts have the same color as one of the charts of the brelan (3 possibility).

On the whole, $\backslash \; textstyle\; \{\backslash \; left\; (\{NR\; \backslash \; choose\; 5\}\; -\; S\; \backslash \; right)\; \{5\; \backslash \; choose\; 1\}\; \{4\; \backslash \; choose\; 3\}\; \backslash \; left\; (4^4-3\; \backslash \; right)\}$

One can also note that some brelans are more probable than others. Quite simply because among the hands which are continuations and contain a brelan, there is less ACE (only included in the continuations with the Ace and those to the 5) that 10 (Continuations with the ACE with the King, the Lady, the Servant or the ten)

Two pairs

Two ways of doing two pairs with 7 charts:

1. 3 pairs and 1 free chart
2. 2 free pairs and 3 charts

For each one of these ways, the hand is determined by:

1. the value of the 3 pairs ($\backslash \; textstyle$ possibilities), colors of the charts of each pair ($\backslash \; textstyle$ possibilities), the value of the free chart ($\backslash \; textstyle$ possibilities) and its color.

Such a hand cannot be a brelan, a full, or a square. It cannot be a continuation since there are only 4 different values. It cannot be a color, since the 3 pairs bring to more the 3 of the same charts color, with the free chart it can have only 4 of the same charts there color.

In all, $\backslash \; textstyle$.

1. the 5 values ($\backslash \; textstyle$ possibilities), values of the 2 pairs among these 5 values ($\backslash \; textstyle$ possibilities, colors of the charts of the two pairs ($\backslash \; textstyle$ possibilities), and colors of the 3 free charts.

Such a hand cannot be a brelan, a full, or a square.

It can be a continuation, if the 5 values are followed (S possibilities).

It can be a color, if the two pairs have at least a joint color and that the 3 free charts have this color:

1. * if the two pairs have the same colors (in 1 case on the $\backslash \; textstyle$ of the color of the second pair), 2
2. possibilities * if the two pairs have a single joint color (in $\backslash \; textstyle$ case of the color of the second pair), 1

possibility That is to say $\backslash \; textstyle$

In all, $\backslash \; textstyle\; \{\backslash \; left\; (\{NR\; \backslash \; choose\; 5\}\; -\; S\; \backslash \; right)\; \{5\; \backslash \; choose\; 2\}\; \{4\; \backslash \; choose\; 2\}\; ^2\; \backslash \; left\; (4^3-1\; \backslash \; right)\}$.

On the whole, $\backslash \; textstyle$.

Pair

A pair is defined by the 6 values of the hand ($\backslash \; textstyle$ possibilities), the value among these 6 which constitutes the pair ($\backslash \; textstyle$ possibilities), the color of the 2 charts of the pair ($\backslash \; textstyle$ possibilities), and the color of the 5 free charts.

Such a hand cannot be two pairs, a brelan, a full or a square.

It can be a continuation, so among the 6 values, at least 5 are followed. ($\backslash \; textstyle$

It can be a color, two possible cases:

• the 5 free charts have the same color (4 possibilities)
• Among the 5 free charts, 4 have the same color as one of the charts of the pair, the color of the fifth chart being free ($\backslash \; textstyle$ possibilities)

On the whole, $\backslash \; textstyle\; \{\backslash \; left\; (\{NR\; \backslash \; choose\; 6\}\; -\; \backslash \; left\; (\{N-5\; \backslash \; choose\; 1\}\; +\; (S-1)\; \{N-6\; \backslash \; choose\; 1\}\; \backslash \; right)\; \backslash \; right)\; \{6\; \backslash \; choose\; 1\}\; \{4\; \backslash \; choose\; 2\}\; \backslash \; left\; (4^5-\; \backslash \; left\; (4+\; \{2\; \backslash \; choose\; 1\}\; \{5\; \backslash \; choose\; 4\}\; 3\; \backslash \; right)\; \backslash \; right)\}$

High chart

In a hand “High Chart”, each chart has a different value. It is thus necessary to draw 7 values among NR: $\backslash \; textstyle$

However, among these combinations, there is $\backslash \; textstyle$ of it which form continuations, that one should not count.

Moreover, each one of these 7 charts can have any color, with condition that there are not at least 5 of them which has the same color i.e. to avoid as:

• the 7 charts have all the same color (4 possibilities)
• Among the 7 charts, 6 exactly have the same color ($\backslash \; textstyle\; \{4\; \{7\; \backslash \; choose\; 6\}\; 3\}$ possibilities)
• Among the 7 charts, 5 exactly have the same color ($\backslash \; textstyle\; \{4\; \{7\; \backslash \; choose\; 5\}\; 3^2\}$ possibilities)

There is thus $\backslash \; textstyle\; \{4^7-4\; \backslash \; left\; (1+3\; \{7\; \backslash \; choose\; 6\}\; +3^2\; \{7\; \backslash \; choose\; 5\}\; \backslash \; right)\}$ combinations of color.

On the whole, there is $\backslash \; textstyle\; \{\backslash \; left\; (\{NR\; \backslash \; choose\; 7\}\; -\; \backslash \; left\; (\{N-5\; \backslash \; choose\; 2\}\; +\; (S-1)\; \{N-6\; \backslash \; choose\; 2\}\; \backslash \; right)\; \backslash \; right)\; \backslash \; left\; (4^7-4\; \backslash \; left\; (1+3\; \{7\; \backslash \; choose\; 6\}\; +3^2\; \{7\; \backslash \; choose\; 5\}\; \backslash \; right)\; \backslash \; right)\}$ combinations.

Category: Poker

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