Great video! I have a question, Why after choosing the pair (13* 4C2) cant you multiply by 48C3 ? What makes the answers so different?
@alishehper1
2 жыл бұрын
because the other 3 cards have to. be of different values; else it will not be "a pair" anymore.
@ericatwood856
5 жыл бұрын
Great Video! I am doing my statistics class and I watched the lecture video on this and it didn't make any sense at all. Your explanation was so clear and understandalbe. Thank you!
@shobhasundar8232
3 жыл бұрын
You have explained very well. Thank you A math teacher from Bangalore🙏
@utpalkumar2770
5 жыл бұрын
After one pair, there can three card of same type in that case ur answer is wrong
@ddstar
4 жыл бұрын
Sir, I know these videos are old, but can you do a video of expected combinitions? You have a hand 9H, TH, JH, QC, AC. What is combinitions of drawing a straight?
@RCLIP106
4 жыл бұрын
If we consider two pair as a subset of one pair, then this solution is correct. Otherwise this solution is wrong. Or am I missing something?
@MathematicsTutor
4 жыл бұрын
Here is correct solution: kzitem.info/news/bejne/1KBs0paEfoJ6aWk Thanks
@XxNinjaLimeXX
4 жыл бұрын
I can't find an answer :(. Can anyone tell me how the probability equation would change if we allowed for getting more aces? Like not excluding other better hands just the most basic what are the chances of getting 2 or greater of the same rank in a hand of 5
@jiyang4602
Жыл бұрын
Where did 4^3 came from?
@MathematicsTutor
Жыл бұрын
3 of any 12 with 4 suits. Hope that helps. Thanks
@jiyang4602
Жыл бұрын
@@MathematicsTutor thank you. I understand it now
@lifestyles2482
5 жыл бұрын
Do u mean the flop containing a pair or in Texas Holdem where u are dealt 2 whole cards that make a pair on the flop (5cards)?
@MathematicsTutor
5 жыл бұрын
Here we have 5 cards, 2 of them of same value like 2 Queens as explained. Thanks
@lifestyles2482
5 жыл бұрын
Anil Kumar this does not answer my question. I’m guessing it has less to do with cards in hand.
@MathematicsTutor
5 жыл бұрын
@@lifestyles2482 I do not know the rules of Texas Holdem. In the video, we are talking about dealing 5 cards and we need to know the probability of getting exactly two of the same kind. So, if we have 2 Kings, then the other three can be any other card. Thanks
@lifestyles2482
5 жыл бұрын
Anil Kumar ok. It would be even more interesting if u did a video on probability regarding making a pair on the flop and also probability of improving the hand on the turn and the river.
@MathematicsTutor
5 жыл бұрын
@@lifestyles2482 I will surely do. Thanks
@solpath_
Ай бұрын
I thought the question was any pair drawn from hand to a flop. My answer was 19.2% Shouldn't the question be: "What is the probability of getting one specific pair in a hand of five?"
@HankC9174
2 жыл бұрын
but the ' other ' 3 cant be a pair also yes ?
@Rob_Lerch
6 жыл бұрын
Silly question here... What does the "C" mean... Like when you type it into the calculator, what is the math behind the "C"?
@MathematicsTutor
6 жыл бұрын
Thanks. C is for combinations. Hope this video can help kzitem.info/news/bejne/rqCtyGmqb6aIbG0 Thanks
@jangbahadur8635
4 жыл бұрын
A card bara hota h y k card
@mbk928
5 жыл бұрын
Thanks Clear and easy
@gorkemcinar8600
Жыл бұрын
Kumar means gamble in turkish😂
@MathematicsTutor
Жыл бұрын
Good to know. Thanks
@MathematicsTutor
Жыл бұрын
Playlist on Cards: kzitem.info/news/bejne/wKl3uH5qrXurnHo
@MathematicsTutor
Жыл бұрын
Here are some more: kzitem.info/news/bejne/2Zhsromup51jhI4
@mehjamilabonny4160
3 жыл бұрын
wrong answer.
@alvarolealjr
5 жыл бұрын
Not to be racist or anything but *inserts racist comment about accent. jk excellent explanation
@shivronsugrim3389
5 жыл бұрын
Great video! I have a question, Why after choosing the pair (13* 4C2) cant you multiply by 48C3 ? What makes the answers so different?
@MathematicsTutor
5 жыл бұрын
Can't answer your question. However, the approach which I have taken makes more sense. Thanks
@Amantri98
4 жыл бұрын
I think his version prevents another pair from arising.
@joshuagarcia6767
2 жыл бұрын
I think it is because if you multiplied it by 48C3 then there is a possibility that you get another card that is the same as the pair and if you do, then it would no longer be a single pair :)
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