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: 133784560
The desired scenarios for you involve freezing 4 cards being equal (13 scenarios) and 3 out of the remaining 48 cards: 13 * 17296 = 224848.
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:
- cardinality:
674274182400 (52 taking 7 in certain order).
- scenarios:
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 120 is 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 here.
In this 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) = 29652480.
- ?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) = 19768320.
- 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:
- cardinality:
674274182400 (52 taking 7 in certain order; this one did not change).
- scenarios:
13 * 17296 * 5040 - 29652480 - 19768320 - 19768320 = 902154240.
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".
