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If I use ${13 \choose 1}{4 \choose 3}{50-4 \choose 4}=58.656$ I have to eliminate the frequency in which that two extra cards are a pair, because that would fall into Full House from normal poker (following Wikipedia), so $58.656-3.744=54.912$ which is fine. Now I go one step further.

If I use ${13 \choose 1}{4 \choose 3}{50-3 \choose 4}=61.152$ I have to eliminate the frequency in which that two extra cards are a pair (because of full house), and the frequency in that the remaining card of the suit appears and transform my three of a kind into a four of a kind $61.152-3.744-624=56.784$ which isn't 56.784. I discovered that if I multiply (in the previous formula) the frequency of a full house by 1.5, or I multiply the frequency of a four of a kind by 4 I get exactly 54.912, but I'm not sure why, and that might be what makes my math surpass the 100% (when comparing the frequency/total possible combinations).

If I use ${13 \choose 1}{4 \choose 3}{52-4 \choose 2}=58.656$ I have to eliminate the frequency in which that two extra cards are a pair, because that would fall into Full House from normal poker (following Wikipedia), so $58.656-3.744=54.912$ which is fine. Now I go one step further.

If I use ${13 \choose 1}{4 \choose 3}{52-3 \choose 2}=61.152$ I have to eliminate the frequency in which that two extra cards are a pair (because of full house), and the frequency in that the remaining card of the suit appears and transform my three of a kind into a four of a kind $61.152-3.744-624=56.784$ which isn't 54.912. I discovered that if I multiply (in the previous formula) the frequency of a full house by 1.5, or I multiply the frequency of a four of a kind by 4 I get exactly 54.912, but I'm not sure why, and that might be what makes my math surpass the 100% (when comparing the frequency/total possible combinations).

My problem is that as you can see, when you draw multiple hands, the combinations of for example a pair becomes higher than the total possible combination, a thing that is absurd and can be easily demonstrated with an example hand: 1, 2, 3, 4, 5, 6, 7, there are 7 different cards, none is a pair. Of course, a flush of 2 cards is allowed to get 100% if you draw 5 cards (it is impossible to get 5 different suits because there only exist 4 different), however, it should not reach 100% in a 4-card hand (an easy example would be drawing a spade, a club, a diamond and a heart, none is repeated).

My problem is that as you can see, when you draw multiple hands, the combinations of for example a pair becomes higher than the total possible combination, a thing that is absurd and can be easily demonstrated with an example hand: 1, 2, 3, 4, 5, 6, 7, there are 7 different cards, none is a pair. Of course, a flush of 2 cards is allowed to get 100% if you draw 5 cards (it is impossible to get 5 different suits because there only exist 4 different); however, it should not reach 100% in a 4-card hand (an easy example would be drawing a spade, a club, a diamond and a heart, none is repeated).