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carrotsss

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all results can be divided by 52C5 to find the probability

a) one pair means we have exactly two cards of the same number, and no other ones. hence, it’s 13 (picking the pair number) * 4C2 (the suit of the cards) * 12C3 (picking the numbers for the remaining cards, which must each be unique to avoid additional pairs) * 4^3 (picking the suits for each remaining card) I’m not sure if they include full houses as pairs too if they do you’d just add the full house possibilities

b) similar methodology. 13C2 (picking the card of each pair) * (4C2)^2 (picking the suits of the cards in each pair) * (52 - 8) (picking the final card, remembering it cannot be the prior numbers)

c) this means three of the same number. 13 (options for picking the number) * 4C3 (picking the suits for the three cards) * 13C2 (picking the remaining cards, can’t be the same number) * 4^2 (suit for these 2 numbers)

d) very similar to the previous one 13 (number of four-of-a-kind) * 1 (must be one of each suit) * 48 (remaining card)

e) 13 (number for the pair) * 4C2 (suits of the pair) * 12 (number for the three-of-a-kind) * 4C3 (suits of the three-of-a-kind)

f) ace high or low refers to the fact that it can go from A,2,3,4,5 to 10,J,Q,K,A. hence, 10 (starting number possibilities) * 4^5 (suits of each card)

g) 4 (picking suit) * 13C5 (possibilities for numbers within the suit)

h) 4 (there is only one way to get this for each suit)

haven’t checked over any of these so there might be errors but hope this helps
 
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scaryshark09

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carrotsss said:
all results can be divided by 52C5 to find the probability

a) one pair means we have exactly two cards of the same number, and no other ones. hence, it’s 13 (picking the pair number) * 4C2 (the suit of the cards) * 12C3 (picking the numbers for the remaining cards, which must each be unique to avoid additional pairs) * 4^3 (picking the suits for each remaining card) I’m not sure if they include full houses as pairs too if they do you’d just add the full house possibilities

b) similar methodology. 13C2 (picking the card of each pair) * (4C2)^2 (picking the suits of the cards in each pair) * (52 - 8) (picking the final card, remembering it cannot be the prior numbers)

c) this means three of the same number. 13 (options for picking the number) * 4C3 (picking the suits for the three cards) * (52-4)C2 (picking the remaining cards, can’t be the same number) I included house possibilities in this one ig

d) very similar to the previous one 13 (number of four-of-a-kind) * 1 (must be one of each suit) * 48 (remaining card)

e) 13 (number for the pair) * 4C2 (suits of the pair) * 12 (number for the three-of-a-kind) * 4C3 (suits of the three-of-a-kind)

f) ace high or low refers to the fact that it can go from A,2,3,4,5 to 10,J,Q,K,A. hence, 10 (starting number possibilities) * 4^5 (suits of each card)

g) 4 (picking suit) * 13C5 (possibilities for numbers within the suit)

h) 4 (there is only one way to get this for each suit)

haven’t checked over any of these so there might be errors but hope this helps
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for a), they have an answer 16x yours. im guessing they added the full house possibilities, but can you please explain this
 
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Luukas.2

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A full house (like AAAKK) is neither a pair nor a three-of-a-kind - the terms describing hands do not overlap
 
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scaryshark09

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wait yours is correct, did you edit to make it 4^3 instead of just 4, or did i miss that
 
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