My question goes like this: how many possible ways can you get 4 of the same value and then 2 of the same of a different value in a 6-card hand out of a full deck?
@@dennisjoseph4528, It is Ok. I have discovered. (Three spaces) (52C1* 3*2)/3! (because) eliminating repetitions * (two spaces) (48C1*3)/2! (eliminating repetitions): Multiply 52* 72 = 3.744 (it makes sense?) :)
These is not correct. Because we have to make that with fullhouse of 5 cards but we have 7 chances to do it. 2 in hand and 5 in table and chose 5, so probability goes up.
What about, using a standard shuffled 52 card deck, after dealing yourself a full house, what are the odds that you deal your opponent a full house? I know that it goes from 52 choose 5 to 47 choose 5, but there are not 3743 full houses still possible because the deck changed...any ideas?
Great question! So we shall assume two players and both get full house. Also remember that they cannot get the same 3 of a kind, however they can get the same pair. Keep that in mind to calculate the answer.
@@MathematicsTutor In full house 2 players can have the same 3 of kind and diferent pairs and they can have same pair and same 3 of kind if is in holdem and each player has 7 cards to chose 5 best cards, they can even have same full house and 2 players can have same pair and diferent 3 of kind for holdem these is not correct and probability is much higher, because probability of poker 4 of kind is lower than full house.
Thank you so much sir! You explain better than my stats teacher.
PLAYLIST: ruclips.net/video/9dMpEK2UF0U/видео.html
Thanks
What a great video! i really enjoyed your straight forward and logical teaching style, and your delivery was great!
Thank you so much!
Thanks for appreciation
Thank you so much ! No other video on youtube helped. But this one perfectly explained it for those who don't know pocket. Thank you !
Thank you sir
Sounds like he said 12 and wrote 22 am I right..at 3:51....
So clear. Thank you, Sir. Good teachers are a blessing.
Thanks for the video. Definitely helped!
Thanks for appreciation.
Very well explained and your notes are so neat. Thank you so much!
Thanks for appreciation.
My question goes like this: how many possible ways can you get 4 of the same value and then 2 of the same of a different value in a 6-card hand out of a full deck?
How do you calculate 4C3 manually
More specifically what would the probability be of getting 4 spades from 5 cards in a random draw?
That was very clear. Thank you sir.
Thank you. It is very helpful.
Great video, what are the odds to make a full house when drawing three cards with pair in hand?
Extremely helpful video!Thank you!
Thanks a lot!
Thank you very much for this explanation. Very helpful.
When you said 12 you wrote 22
Thanks, You are right at 3: 38 I said 12 but wrote 22; Fortunately continued with 12 and so the answer is correct.
Thanks for the feedback.
This is a good example: ruclips.net/video/xb6HUxnk2QY/видео.html
thank you!
Awesome sirji
how would you find the probability on Texas hold'em poker?
fine explanation....
THANK YOU
I did it by (5 choose 3)*(4/52)(3/51)(2/50)(4/49)(3/48)*(13*12) and got the same result, neat
Why can't we find the same using (52C1)*(3C2)*(48C1)*(3C1)
@SALTYYC no need. Thanks 👍
@@dennisjoseph4528 I have tried your method and I could not.. Why? Maybe divided with some number?
@@cynthiareid6416 thanks Cynthia I remember figuring it out then but unfortunately I forgot :(
@@dennisjoseph4528, It is Ok. I have discovered. (Three spaces) (52C1* 3*2)/3! (because) eliminating repetitions * (two spaces) (48C1*3)/2! (eliminating repetitions): Multiply 52* 72 = 3.744 (it makes sense?) :)
Kudos, Thank You so much
Examples with Cards: ruclips.net/p/PLJ-ma5dJyAqqU-jwPEAJls8LaefzviyFQ
Thanks
@@MathematicsTutor can you also do one on de montmort's matching problem
@@alinagiri3152 Thanks, will do that. Here is on Birthday Matching: de montmort's matching problem
These is not correct. Because we have to make that with fullhouse of 5 cards but we have 7 chances to do it. 2 in hand and 5 in table and chose 5, so probability goes up.
No, this is the correct answer.
What about, using a standard shuffled 52 card deck, after dealing yourself a full house, what are the odds that you deal your opponent a full house? I know that it goes from 52 choose 5 to 47 choose 5, but there are not 3743 full houses still possible because the deck changed...any ideas?
Great question! So we shall assume two players and both get full house. Also remember that they cannot get the same 3 of a kind, however they can get the same pair. Keep that in mind to calculate the answer.
@@MathematicsTutor So what would be the answer? thank you
@@MathematicsTutor In full house 2 players can have the same 3 of kind and diferent pairs and they can have same pair and same 3 of kind if is in holdem and each player has 7 cards to chose 5 best cards, they can even have same full house and 2 players can have same pair and diferent 3 of kind for holdem these is not correct and probability is much higher, because probability of poker 4 of kind is lower than full house.
wow ! amazing !
Thanks for appreciation