Chance of a Pair in a Poker Hand

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Need to calculate some more probabilities (Three of a kind, two pair, straight, flush, royal flush). & Then I am on my way to Vegas.

[zadaa]

This helps for holdem, he calculated that in holdem we will have a paired board 42% of the time. That's schwifty

[afridi]

I think there is a phrasing of the problem here, if we wanted to only have one pair in a hand the first 2 pick was perfectly fine, but for the third card there are 12 values and we can pick any four of them, for the 4th card we cannot pick the 12 values again if we do then we have pairs again so for the 4th pick it should be 11 out of 1 and 4 out of 1 and for the 5th card we pick 10 out of 1. this ensures that there are no other pairs of the card.
the problem above ensures there is at least one pair not ONLY ONE PAIR.

[HankC]

does the solution consider having a pair in the ' last 3 ' ? after the first pair, there is a choice of 48 cards for the 3rd card, 44 for the 4th and 40 for the 3rd .

[AkimboFennec]

Nice explanation, thanks. But my question is how can you solve it using permutations. I tried it with permutations and I get a wrong answer

[PanFaworek]

1 thing I don't understand... is it the probability of picking the cards all at once? cause for me it seems like it. Then my question is, how much more complicated is it to calculate the probability of picking the cards one after another?

[theodoresweger]

Extremely well done, I'm still struggling with the extra cards, that's me, and I will get it after a few reviews. I don't believe most people know how to take the extra cards into account, making it one pair and not possible two pair or something else. Thanks, the Royal flush seems the simplest to figure, am I wrong wasn't this considered the dead man's hand, just asking.

[asadnayani]

I have found a wonderful way of looking at counting probabilities. Thank you, Eddie.

[shivronsugrim]

Why after choosing the pair (13* 4C2) cant you multiply by 48C3 ? What makes the answers so different?

[artieboy]

His explanation of the 12c3 is confusing. He says you can't pick any of the remaining Kings (triples, quadruples). But you also can't pick any cards that match themselves. He left that part out, I think. That is why it is 12c3 (any three cards from a set of 12 different ranks) where card 3 is one of four suites 4c1, card 4 is one of four suites 4c1, and card 5 is one of four suites 4c1. Fun stuff can be a headache sometime.

[shanciewagner]

What about subtracting out two pair if you can only get one pair?

[flarey]

i think you need to subtract the number of ways for the full house before multiplying by the pair - the three other cards must not be of the same value

[peigangdangchina]

Why can we not do 4C3 to find the remaining cards? I feel like 12C3 x 4C3 should be the same for the second part but it’s not.

[SFzip]

“A problem well phrased is half solved.”

[fabiogurgel]

Didn´t understand, if i pick one king, three others left, so i should just combine C3, 1 to form a pair not a C4, 2. Sorry for bad english, am not native

[syuliya]

I agree...this is fantastic.... but I disagree that for a Pair we need consideration of the fact that once one card is picked the second card as condition probobility and not funny consderation of the suit.

[robertohurtado-wcyn]

Can you help me calculate how many hands are expected to get 3 cards for a royal, I heard is every 92 hands approx but I dont find a correct way of calcualted, I know there are 40 distinc hand but with the 2 remaining spots should I multiply by (38c2), so I exclude the 5 for the royal and also any 3 of a kind, @eddie woo

[truegamelover]

1: A8
2: JJ
AA8J4
Who won this hand, , please explain

[solapowsj]

A model class in permutations: Chances and uncertainty. Thank you🌹.

[davidjames]

8:26 The more cards you get, the more likely a pair will be in there somewhere is incorrect. If you already have a pair say in the first 2 cards, then getting 3 more cards will NOT increase your chances of getting a pair somewhere in there as you stated. Actually in a 5 card poker hand dealt from a 52 card deck, it is MORE likely that you do NOT get just one pair. 42.2569% chance of getting exactly one pair means 57.7431% or getting something else, so once you get a pair (such as dealing the cards one at a time), then no, there is NOT a higher probability of getting a pair once you already get it and draw more cards. For example, K, K, A, 7, K.