How many 5 Card Poker Games will Have 2 Pairs Find Probability

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[@ddstar]

this has been ver helpful for me. thank you

[@cynthiacarrillo9243]

Thank you for you help! This did explain a lot and I like that I have a better understanding of what I am doing.

[@SugarRush150]

thank you!! this video helped me a lot!

[@hamokhachadourian8854]

and if we have more than 52 cards, thus we have these cards (Q diamond + K spade + Q spade) extra over standard 52 cards, so we have 55 cards?

[@hajranaeem4201]

why could not it be like 13.4C2.124C2.44C1/52C5?

[@thoorayamused2528]

THANK YOU SO MUCH!

[@MathematicsTutor]

Thanks for appreciation

[@syed2194]

thank you ji

[@thomasnguyen1150]

thank you

[@ekramtofik-if9kp]

👍👍

[@리뷰쓰려는데귀찮게하]

Why couldn't it be 13C3 * 4C2 * 4C2 * 4C1? (Choose 3 different ranks and choose 2 of each 2 ranks and choose another 1 from the remaining rank) Anyway thanks a lot for the video

[@davidjames1684]

That answer would be too low since even though you are selecting the 3 different ranks, you are not selecting the one that will have the single card from it. For example, suppose the 13C3 picks Q, K, and A. Now which one will you assign the single card to? Your formula doesn't pick that one so your answer is too low by a factor of 3. Put a * 3C1 in your formula and it will then be correct.

[@karlroeth96]

can somebody explain the idea how there are 44 cards left?

[@conradmorris8881]

you cannot choose a card from the same suit as the two pairs you have selected. therefore, there are 11 suits to choose from * 4 cards in each suit = 44 cards left to choose from.

[@analystwaterbombali334]

@@conradmorris8881 what if your two pairs are Jack Spades Jack Heart Jack Diamond Jack Club. It seems there are 12 values to choose from each suit?? shouldnt it be 12*4 = 48?

[@Eastercube]

@@analystwaterbombali334 that would be not (2 pair), but (4 of a kind). for 2 pair we need two different denominations

[@omarreyesortiz5641]

If the question was one pair and three different cards , would the answer be 13c1x4c2x48c1x44c1x40c1?

[@MathematicsTutor]

Related Videos: ruclips.net/p/PLJ-ma5dJyAqrcWkF51fq4E_3nud2m0SF1

[@omarreyesortiz5641]

Thanks so much sir . I learned a lot from those 7 videos , understood all of them but unfortunately the second video no matter how many times I watched it couldn’t get it . When you have a chance please explain

[@MathematicsTutor]

Thanks for your time, interest and feedback. All the Best!

[@acejerouinfigurasin3542]

I don't understand hahaha

[@candy7333]

Just wasted my time, I never understand

[@MathematicsTutor]

it is complicated.