Frankly I've seen questions similar to this answered on this site but they don't paint a clear enough image. Though I must confess that I'm a total novice to probability and combinatorics so I might just be stupid.
I have attempted to calculate, at what point is a Full House 100% probable in 7 card poker by reducing the number of cards in the deck. However the answer I recieved by folowing the formulas I've been providded with, produced results which seem incorect.
To my understanding to calculate Full House we use three diferent formulas added together to account for all posible combinations.
Those are:
Two Three of a kinds and a single [3 3 1]
$\binom{13}2$$\binom{4}3^2$$\binom{11}1$$\binom{4}1$
Three of a kind and Two pairs [3 2 2]
$\binom{13}2$$\binom{4}3$$\binom{12}2$$\binom{4}2^2$
Three of a kind, a pair, and two singles [3 2 1 1]
$\binom{13}2$$\binom{4}3$$\binom{12}2$$\binom{4}2$$\binom{11}2$$\binom{4}1^2$
Of course the next step is to compute, then divide by $\binom{52}7$ and get the answer.
But if we start reducing the number of cards in the deck, the numbers produced start getting weird.
For example let's say we have the cards:
5, 6, 9, 10, K, A of Spades
2, 7, 10, J, K of Hearts
2, 5, 6, 7, 10, J, Q, A of Diamonds
4, 5, 6, 9, J of Clubs
This means the deck has:
4 suits
11 ranks
9 pairs
4 Three of a kinds
...and over all 24 cards to chose from.
The issue is, that in this scenario the probability given by the formula is 1.33472 or 133%.
Of this [3 3 1] is equal to 0.9985%
[3 2 2] is 4.6599%
and [3 2 1 1] is 127.8136%
These numbers or at least the third one obviously does not look right. However as the novice I am to probability and combinatorics I can't see just where I'm going wrong.
Is there a diferent technique I should be using, or is there some sort of diference I'm not accounting for? I'd love to understand.
