[Discrete Mathematics] Discrete Probability

Рет қаралды 151,219

Күн бұрын

Пікірлер: 41

@akashdigumber1564 4 жыл бұрын

Just in case someone watches this recently and dont understand the Deck part b Here's how i understood it: We do (4C3) because we pick 3 Aces out of 4. There are usually 13 types of 4 cards. But we do (12Types C 1Type) since the Aces Type of cards doesnt have a pair anymore. (12C1) Now since 1 Type of Card (Diamond, Heart...etc) have 4 Cards, we choose 2 cards out of the 1 TYPE we chose earlier...Hence (4Cards C 2Cards) So...Its (4Aces C 1Ace)*(12Types C 1Type)*(4CardsFromType C 1CardFromType) Hope it helped 👌

@nib10 3 жыл бұрын

I think i should learn how to play cards before using them as an example 😵‍💫

@brookeschwartz8470 2 жыл бұрын

Isn't there also a possibility that the player picks up the last Ace? Why isn't that factored in?

@sandeshkunwar3643 2 жыл бұрын

So it is (4 aces C 3 aces bro) not 1 aces

@MrJulius6565 Жыл бұрын

wdym by 1 type of card have 4 cards? do you mean 1 type of card have 4 suits?

@AleemKhan42069 4 жыл бұрын

yo guys my discrete teacher is so boring like he drains so much energy out of people... glad we have youtube and cool people

@potatootter5088 3 жыл бұрын

Great stuff and I understood everything. Good job.

@Havie 6 жыл бұрын

So much better than my professor...

@Noble909 7 жыл бұрын

5:10. I'm confused... My logic would be 4 aces, choose 3 then 4 suits, choose 1 and finally 12 cards, choose 2.

@NathanGrubbs 7 жыл бұрын

I'm still learning this myself, so take this with a grain of salt: If you choose 1 of 4 suits, then two cards from that suit, those two cards can't be a pair. There's only one '2' in any given suit, for instance.

@liamb8649 7 жыл бұрын

Thank you! super helpful video 10/10

@brookeschwartz8470 2 жыл бұрын

For b (at 6:00), isn't there also a possibility that the player picks up the last Ace? Why isn't that factored in?

@CarlC9898 Жыл бұрын

nope, because it doesn't have another Ace it can pair up with so thats why its excluded

@jasonyang6738 6 жыл бұрын

Nice video, thanks for your work.

@robertjimenez5608 3 жыл бұрын

WOOOW SO IT ALL COMES TOGETHERE AT THE END

@KeyserTheRedBeard 3 жыл бұрын

tremendous video TrevTutor. I killed the thumbs up on your video. Maintain up the superb work.

@giorgiog.4025 8 жыл бұрын

As always cool video. But I had a question: i dont really understand why (speaking about the point b of the 52 cards deck problem) the possibility of a pair is 12 over 1 times 4 over 2. if i choose 1 card out of 12 then shouldnt i have to pick only onother 1 out of 3? i mean why if you have already picked a card out of 12 you have to pick 2 cards and not one out of 4? Thanks in advance for your availability :)

@Trevtutor 8 жыл бұрын

12 choose 1 is choosing a #, then 4 choose 2 are all the possible pairs of suits. I can pick 2,3,4,5,6,7,8,9,10,J,Q,K (12 choose 1), and then I can pick either {Hearts, Clubs}, {Hearts, Spades}, }Hearts, Diamonds} ... etc. which is (4 choose 2).

@thegumball1809 5 жыл бұрын

I appreciate you for the videos. Do you have a patreon?

@michaelm.3641 6 жыл бұрын

Excellent video. Thank you!

@munashekamangira9385 7 жыл бұрын

this cards problem is so confusing.... i dont get a single thing

@thomasemond9130 3 жыл бұрын

pull out a calculator and a deck of cards

@billythesunbeltsamurainapi3670 7 жыл бұрын

in the card example why do you multiply the chances of choosing three aces and 2 jacks and not add them ? this is the only spot where i get confused .

@navi4131 7 жыл бұрын

Exactly i get confused. should we add or multiply.

@nadeemjq 6 жыл бұрын

Great video! :)

@monkeyboy7328 8 жыл бұрын

in the cards example, surely when you pick a card the sample space decreases to 51, and after the second card to 50 etc? and i dont understand how do you simplify that to get the actual answer?

@Trevtutor 8 жыл бұрын

You pick 5 cards from 52. Order of the cards doesn't matter. The choose operator expands to show this. 52 Choose 5 = (52!)/(5!47!) = (52x51x50x49x48)/(5x4x3x2x1).

@Cooking_and_Lifestyle. 7 жыл бұрын

thank you mr trev

@toddthans1856 8 жыл бұрын

I was wondering about the poker hand example. It seems like it should be 13 choose 1 denomination * 4 choose 3 suit then since you no longer can pic the ace as a denomination the next event becomes 12 choose 1 and 4 choose 2. then that gets divided by 52 choose 5 for the probability.

@daniela9171 7 жыл бұрын

This was exactly my thought. This video seems totally wrong

@daniela9171 7 жыл бұрын

then all of that / (52 choose 5)