How often statistically, should any player at a 9 seat table be getting 4 of a kind? Even if they fold and I don't see it, what's the general odds of 1 in 9 players making 4 of a kind?
The cardinality (whole possible hole + commie cards) is 52 taking 7:
The desired scenarios for you involve freezing 4 cards being equal (13 scenarios) and 3 out of the remaining 48 cards: 13 * 17296 =
The odds are like 0.0017 by dividing desired / cardinality, but this will include situations of 4oak being community cards.
If you want to exclude the in-table 4-of-a-kind (you want it only for you), then you will have to calculate like this:
674274182400 (52 taking 7 in certain order).
13 * 17296 * 5040 (13 pokers, 48 taking 3 free cards, 5040 stands for
7! which contemplates arbitrary sort) -
48 * 47 * 13 * 46 * 120 (scenarios where you don't have one of those cards in your hand, but anyway it is 4oak, while the latter
5! which contemplates arbitrary sort in the commie cards; arbitrary sort is already contemplated in the hole cards when multiplying
48 * 47).
The result is
0.0014, however, another player (only one!) could also have 4oak, if that happens, scenarios are like this:
- XX vs YY with XXYY? :
13 * 6 * 12 * 6 * 44 * 120 (which stands for: 44 as the ?, 120 as 5! for arbitrary flop order, 6 in both cases for suit combination in hands) =
- ?X vs YY with XXXYY :
13 * 4 * 44 * 12 * 6 * 120 (which stands for: 44 as the ?, 120 as 5! for arbitrary flop order, 4 and 6 for suit combination in hands) =
- XX vs ?Y with XXYYY :
19768320 same situation but reciprocal.
The question marks replace bricks. So you want to discard those scenarios when another one has a different 4oak:
674274182400 (52 taking 7 in certain order; this one did not change).
13 * 17296 * 5040 -
Your odds of getting exclusive 4oak and nobody else getting another 4oak is: 0.0013. In the other 0.0001 case the other one will get another 4oak, so by having an exclusive 4oak there's a 1/14 chance the other one has another exclusive 4oak as well.
Please I need a cross-review in this point! I'd like to check if 0.0013 is a good result or I screwed with the calc application. I admit it looks pretty weird to me this 1/14 difference
As a bonus question, if I may, what are the odds of losing with 4 of a kind to a better 4 of a kind?
Now focusing on the conditional analysis on the game table. If your hand/commie already look like this:
- XX with XXYYY: One arbitrary opponent has a chance of having 4oak in 44 / 45 * 22 = 0.04444
- ?X with XXXYY: One arbitrary opponent has a chance of having 4oak in 1 / 45 * 22 = 0.00101
- XX with XXYY?: One arbitrary opponent has a chance of having 4oak in 1 / 45 * 22 = 0.00101
By having those scenario configurations, you have to map the value of X like this:
- 2 is worth 0, 10 is worth 8.
- J Q K A is worth 9 10 11 12.
- Lets call
MAP(X) this mapping that converts the values like I told.
- 12 will be the maximum value here.
Given those three scenarios which could risk of having another 4oak from only one arbitrary opponent, you require an additional condition:
AND the other player beats me with an Y-valued poker, so the values are:
0.04444 * (12 - MAP(X)) / 12 for scenario XX XXYYY.
0.00101 * (12 - MAP(X)) / 12 for scenario XX XXYY?.
0.00101 * (12 - MAP(X)) / 12 for scenario ?X XXXYY.
Disclaimer: These results are probabilistic fractions, which are always in closed interval [0..1]. Since none of these values is zero, you can calculate
H = 1 / value and later say "the chances are 1 in H hands".