Texas Hold'em: Probability of Full House

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The discussion centers on calculating the probability of achieving a full house in Texas Hold'em when both hole cards must contribute to the hand. The user notes confusion regarding how to factor in the hole cards and the community cards, specifically mentioning the need to use three of the five community cards. Initial calculations suggest a probability of having a qualifying starting hand of 63/221. The user seeks clarification on whether to multiply the probabilities of drawing from the remaining cards for the community cards. The conversation highlights the complexity of poker probabilities and the importance of understanding the specific conditions of the hand.
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i have looked all over the internet and found a ton of different answers to this question so i will try and ask it here:

in texas holdem
what is the probability of getting a full house after all 5 cards are delt
both of the cards delt to you have to make up the full house
 
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ya i found that but I've always been told not to trust wiki. (i do anyway) But my question was how to figure it out if both your hole cards have to play. so you can't play the 5 cards on the board. you can only play 3 of them. this seems really easy but I am getting confused just as easy
 
sorry, I've realized that I've been lazy and haven't contributed

there are 16 combinations of having A and K
6 ways to have AA
6 ways to have KK

these are the only possible starting hands so there are 28/52 cards to pick from the first card and 27/51 possible for the second card.

so the probability of having a qualifying starting hand is 28/52 x 27/51 = 63/221

for the extra 5 cards on the board there are 26 possibilities with 50 cards remaining.
this is where i get lost
do i just 26/50 x 25/49 x 24/48?

then that number is the probability?
 
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