Permutations and Combinations - 5 Card Poker Hands

[@shitpostheavy]

Was here at the beginning of the year, now I'm back for my finals. Thanks for this.

[@unagi4254]

SAVED MY LIFE!!! can't thank you enough, I was struggling with this topic until I watched your video once and everything made sense!

[@cienmcqueen2999]

Thank you! You helped me with my homework and my exam that I will be taking in 10 hours. I was stuck for long time on the homework and now I'm nearly done with it thanks to your help. Thank you!

[@FPrimeHD1618]

I'm such a dummy. I was so confused as to why it was 12 choose 2 and didn't stop to think that it would be a 4 of a kind lol. Great video! Thanks for the help!

[@haroonrafique7106]

Thats brilliant, thank you very much for such a clear and concise explanation. None of the text books cover this and deliberately sell the students down the proverbial river! thanks so much, you are a star.

[@Amicoskates01]

really can't thank you enough. This video helped more than all the others I watched!

[@WillRiddle88]

This video just saved my butt for my test this morning. Thank you so much!

[@MrVoayer]

Easily one of the best combinations videos on KZbin. It would have been perfect had it had an explanation for wrong way of thinking like the one for choosing two pairs analogous to choosing one pair: (13C1)(4C2) x (12C1)(4C2) x (11C1)(4C1) ! The similar error is possible in choosing two remaining cards when choosing three of a kind. Explainin wrong way of thinking has strong teaching value and often helps more than giving the right answer ! Anyway, this is valuable video ! Keep on doing great job !

[@goldsphere4108]

pls dont put exclamation marks after a mathematical statement, it confuses me with factorial ^^

[@ankushrajput2125]

How does an exclamation mark confused you with a factorial when it comes after a word looool

[@goldsphere4108]

(13C1)(4C2) x (12C1)(4C2) x (11C1)(4C1) ! It does not?

[@isabelatasende5040]

Could you explain why this is wrong?

[@horaciocastanedachavez5120]

Thank you very much, I have an 8 year old son who likes math a lot so we are doing combinations and permutation and I wanted to teach them with poker so its more fun. My son is doing algebra and a bit of trigonometry and analitical geometry so I hope he has not too much trouble understanding this.

[@brycegriffin9271]

Seems legit

[@vertexforger]

Is your son perhaps Terrance Tao 2.0? lmao

[@zogalz5598]

very useful video you cleared these card problems for me after days of struggling thanks

[@BrianVeitch]

13C1 is just used to represent you choosing one rank (type of card). But it could be any rank. For example, (13C1) means it could be K, 10, Ace, 5, Queen, 2, etc. So (13C1)(4C1) means to choose a rank, and then one card from that rank. (13C1)(4C2) it means a pair from any rank. (13C1)(4C3) means choose a rank and then choose 3 cards from that rank. But (13C1) isn't always used. For example, two pair uses (13C2)(4C2)(4C2). (13C2) means pick any two ranks from the 13 available ranks.

[@ruthcdsilva]

Thanks, this was so well explained!

[@j.deleos1412]

Very nice and clear explanations of porker hand probabilities.

[@jasoncain1747]

Thanks for the quick response Brian, I understand now, great explanation.

[@MrPhillips365]

THANK YOU! YOU LITERALLY SAVED MY LIFE.

[@alexanderjones1129]

This is very interested and well explained. Keep up the good work.

[@gababagonist]

Thanks so much, it's almost impossible to find good info on this topic

[@kafianan6586]

Very well explained... Thanks a ton ♥️

[@shitpostheavy]

Good pace. Thank you.

[@dasnyds00]

Thanks for the videos. They were a huge help!

[@mshahzaib4195]

You my sir are a life saver, I can't thank you enough. I highly appreciate your help. :)

[@MexterO123]

My goodness I wish I saw this video before my probability exam, I sucked at combinations and permutations. :/

[@saffronjam]

Very good video, thanks for explanation

[@TigranTerterian]

im blessed this exists

[@x33rak]

Great video, thank you so much!

[@kamoako]

Great video but I wish you would of wrote 4C1 (4C1)(4C1) instead of 4^3 on the first problem. Because as a newbie its hard to understand what you meant by that.

[@jasoncain1747]

Why doesn't this work for a "two pair" situation (13C1) (4C2)(12C1)(4C2)(11C1)(4C1)?

[@BrianVeitch]

+Jason Cain I agree that counting techniques can be tricky. Your formula is almost correct. You need to divide your value by 2. Doing (13C1) and (12C1) means the order matters. For example you would be counting J,J,4,4,7 as a different result than 4,4,J,J,7. To account for the double counting you need to divide by 2. Another way to avoid the double counting is to use (13C2) as a way to choose the two cards, i.e, J and 4. I hope this helps.

[@susiexsun]

+Brian Veitch thanks for explaining --- I had the exact same question. However, how does this apply to the full house case? Full house uses 13C1 and then 12C1 and then does not divide by two, while KKKQQ is the same as QQKKK. What is missing in my thinking? Thank you for your help!

[@BrianVeitch]

+Susie Sun I've been thinking about this every since Jason Cain asked his question. I'll tell you how I think about this. I recommend physically doing what I write. My written explanation may not be super clear. The last paragraph is the most important part. For the two pair: Imagine you have two hats (each representing a PAIR) in front of you. Place them side by side. Place a Jack in Hat 1 and place an ACE in Hat 2. It should look like this: Hat 1, Hat 2, ?, which looks like JJAA?. Now swap locations: Hat 2, Hat 1, ?, which now looks like AAJJ?. Since the hand does not change, we must divide by 2 to account for this combination being counted twice. This was my reasoning for Jason's question. This would give us (13C1)(12C1) / 2 This isn't wrong but it's not analogous to the full house situation. Here's a different way of thinking about the two pair that would make it analogous to the full house. Instead of swapping Hat 1 and Hat 2's locations, swap the cards prior to placing them in the hats. You place the ACE in Hat 1 and the Jack in Hat 2. This gives us AAJJ?. Since Hat 1 and Hat 2 both represent Pairs it didn't really matter which Hat the Ace or Jack went into. They both end up with 2 ACES and 2 JACKS. That's why most formulas online use (13C2). So you actually swap the cards prior assigning them. Onto the Full House situation: Now imagine we have two hats in front of us. Hat 1 represents the three of a kind and Hat 2 represents the pair. Place a Jack in Hat 1 and Place an ACE in Hat 2. This gives us JJJAA. Here's why the original thinking of the two pair doesn't work for a full house. If you swap Hat 1 and Hat 2's locations, we get AAJJJ. Admittedly this is the same hand. But this isn't what the "math" is doing. I'm not yelling in the next paragraph, just indicating the next line is the most important part of my explanation): IF YOU SWAP THE CARDS PRIOR TO PLACING THEM IN THE HATS, i.e, IF YOU PLACE THE ACE IN HAT 1 and THE JACK IN HAT 2, YOU GET AAAJJ. This is not the same hand. Swapping the JACK AND ACE actually gave you different hands. This is why you don't divide by 2. Swapping the JACK AND ACE prior to placing them in the two pair situation gives you the same hand. That's why we divide by 2.

[@susiexsun]

+Brian Veitch Wow that is an amazing explanation. Thank you! So to generalize, can we say the following? (1) The equation for any one hand is (Number_of_Ranks CHOOSE 1) * (Number_of_Suits CHOOSE Number_of_Spaces) (2) For combinations of multiple hands, we divide by the factorial of the number of containers (i.e. 2!) to de-dup when the two containers are equal. The equivalent of this de-duping is using the combination method since that equation automatically includes '2!' in the denominator. Btw - your video and explanation is by far best I've seen on the internet. Really appreciate what you've done!

[@JustinLiang0]

+Brian Veitch What an excellent analogy.

[@shahzmalik]

You really helped me Thanks buddy !

[@trailertube3209]

thanks a lot man, this helped a lot!!

[@karthikreddy3286]

good teacher

[@samynashabeh]

I really liked the way you make things easier for us. I have actually a question for you & it is that I feel that flush has been accounted in all of these probabilities. Maybe there should be one more condition that we do not obtain 4 cards from the same kind & then the probability of each of the calculated event will be reduced. I don't know, I'm just speculating. Am I right ?

[@BrianVeitch]

There is overlap between the probabilities. For flushes, it's (13 C 5) * 4. In this number, we also have the straight flushes, and the four royal flushes. You'd have to subtract them if you want flushes that aren't straights. But I'm not clear on what your question is.

[@samynashabeh]

Thanks for your quick reply. I meant that while accounting for all those probabilities you calculated in the video, didn't we also accounted for the flushes in our calculation of each probability ? And please excuse me as English is my 3rd languahe.

[@BrianVeitch]

The royal flush is the only one I did that could be a flush. All the other probabilities I discussed required at least a pair (pair, two pair, three of a kind, full house, four of a kind). To get any of those, we need the same card in a different suit.

[@haoheli5138]

Thank u so much! It helps a lot!

[@ivoivich89]

Can someone please help how can one calculate all possible combinations of getting only PAIR in hand QQJKA after throwing away cards J,K and A and keeping only QQ_ _ _. Because in this case only 9 ranks have 4 suits and 3 ranks have 3 suits (because we thrown away J,K and A). I know that there is 11.559 combinations but don't know how to calculate this. Help please :)

[@chunkylover5367]

thank you this helped very much.

[@Kasram]

ive watched this video 10 times, hope i get the w

[@elianpulido3360]

THANK YOU

[@ironmantis25]

I'm applying this to yugioh, I need a consistent deck to win.

[@gauthama4092]

Hi Brian, Thanks for this amazing video. Just one doubt. For full house, why cant we use 13C2*4C3*4C2?. If this is wrong then why?

[@danielm9463]

I had this question too!!! I think this is the answer: imagine we randomly draw 21 cards. We draw 4 aces, 2 jacks, 2 queens, 2 sevens, 3 fives, 3 sixes, 1 two, 1 three, 1 eight, 1 nine, and 1 ten. Specifically, here's what we've drawn: [A♣, A♦, A♥, A♠] [J♣, J♦] [Q♥, Q♠] [7♣, 7♥] [5♦, 5♥, 5♠] [6♣, 6♥, 6♠] [2♣] [3♦] [8♦] [9♥] [10♣] I've grouped them intentionally to show the idea. Now, there are 3 properties we need to categorize the cards into general groups: (a) how many are alike in the set (i.e., the first row is all 4-of-a-kind groups, the second row is all 2-of-a-kind groups, the third row is all 3-of-a-kind groups, the fourth row is all 1-of-a-kind groups), (b) the rank (A, 2, 3, 4, ... Q, K), and (c) the suit (♣♦♥♠). The number of ways to draw 21 cards and get this arrangement (one 4-of-a-kind, three 2-of-a-kinds, two 3-of-a-kinds, and five 1-of-a-kinds) is AxBxCxD where: A = (13C1) x (4C4) B = (12C3) x (4C2)(4C2)(4C2) C = (9C2) x (4C3)(4C3) D = (7C5) x (4C1)(4C1)(4C1)(4C1)(4C1)

[@BrianVeitch]

@@danielm9463 Thanks for replying to him. I have to figure out how to pin these comments with my responses. I know I answered the "why can't it be 13C2" somewhere in the comments. Basically, 13C2 chooses two ranks from the 13. But it doesn't account for the switch. Imagine that I choose two ranks J and A. The full house would be JJJAA. But if I flipped it to have 3 aces, then it would be AAAJJ. These are two different hands. But 13C2 doesn't treat these has different. It will count JJJAA and AAAJJ as the same combination. I have a detailed explanation somewhere we I use an analogy of two hats. It's probably a comment in one of the longer comment threads.

[@danielm9463]

@@BrianVeitch Yes!! I saw that reply you had written, and it's exactly what led me to the insight that there are 3 properties we're using to sort the groups: number-of-a-kind, rank, and suit.

[@gauthama4092]

@@danielm9463 Thanks Daniel for taking time to respond to me on this. This helps

[@danielm9463]

​@@gauthama4092 No problem! I'm slowly grasping it myself, and I think I finally understand it fully. If we have 2 pair, then there are 2 ways to pick the ranks for the 2 pairs. We can use 13P2 or 13C2. If QQJJ counts as being different from JJQQ, then we need to use 13P2 for assigning the ranks to the two pairs because permutations care about the order. But if QQJJ = JJQQ, then we should use 13C2 because combinations don't care about the order. For this problem (counting the distinct poker hands), QQJJ=JJQQ, and so we should use 13C2 rather than 13P2. (13C1)(12C1) is sloppy notation that should be avoided altogether. When we write (13C1)(12C1), what we're *actually* writing is 13*12 or 13P2. (I now avoid XC1 altogether, because it often causes me to incorrectly conceptualize the problem.) So, (13C1)(12C1) double-counts the outcomes. Specifically, 13C1 includes the possibility that our first pick is Q or J. Then, 12C1 (because it encompasses any ranks that are different from the first pick) includes the possibility that the second pick is J or Q, respectively. (That is, if the first pick was Q, then 12C1 includes J. If the first pick was J, then 12C1 includes Q.) So (13C1)(12C1) = (12P2) counts those two outcomes (QJ and JQ) as distinct.

[@grandevista4u]

Brian, how do I calculate the following? My math is not adding up to your answer? And how do I do this on a TI-84?

[@grandevista4u]

never mind..i got it

[@micheled.m4417]

Omg please respond 🙏 HOW???? I don’t know how to calculate it at all :(

[@lborate3543]

I’m over here running around my house looking for the beeping sound! It’s in this video folks, no batteries for the smoke alarm on your side. Check them anyways ;)

[@darkdemonshadow1]

if I have a hand of cards in poker would it be a permutation or combination?

[@BrianVeitch]

A combination

[@BrianVeitch]

These two examples don't fit the general three of a kind scenario. Exactly three kings means the other two cards could be the same (but not Kings). This means you're including some combinations that are a full house. Plus since your 3 of a kind are KINGS, you don't have the (13C1) at the beginning. (13C1) allows for the 3 of a kind to be any card (3 Kings, 3 10s, 3 Aces, etc).

[@johnny404]

really help me out with my HW！

[@traumatic121]

DEAR LORD, THANK YOU.

[@toastiezzz]

yer a life saver

[@sydellasamoah8728]

i trust u man

[@chetrajghaley]

great one

[@annarummathepeople]

thank you so much

[@jamesw.2185]

Awesome!!

[@ShoeAlmighty]

i had trips

[@m22d52]

This works only if you are the only player in a game!

[@BrianVeitch]

M22 :D very true. But as long as no other cards are showing except yours the probability will be what’s in the video.

[@itaishalom9381]

10x

[@OleguitoSwagbucks]

the other 3 cards for a pair could make a flush

[@marcobarco]

Hands down the best explanation of poker probabilities on the Internet. Thank you for the thorough walkthrough!

[@BrianVeitch]

To give some more examples, (13C3) means to grab any three ranks from the 13 available ranks. So (13C3)(4C2)(4C1)(4C3) means grab any three ranks, then choose 2 of the 4 cards for the first rank, then 1 of the 4 cards for the second rank, and 3 of the 4 cards from the third rank. So how you set it up depends on what you're hand looks like. That's why your original examples were set up differently. Just to make sure, rank means a type of card. Ace is a rank, 2 is a rank, 3 is a rank, etc.

[@InvisibleToVisible]

Thank youuuu you're a lifesaver! Almost my whole test was just on this poker card stuff - SO helpful!

[@BrianVeitch]

12C1 x 4C1 x 11C1 x 4C1 = 48 x 44 is more like a permutation because if you switch the last two cards, this will treat it as a new hand. For example, the way you have it written, QQQ5J would be a different hand than QQQJ5. So you have to take 12C1 x 4C1 x 11C1 x 4C1 and divide by 2 to account for the order of the last two cards. (12C1) x (11C1) / 2 = (48 x 44) / 2 is the same as 12C2.

[@sadafbenaf]

Thank you so much for the clearing, I thought I loved math! I feel like a child when it comes to permutation and combinations. So, 13C1 could only be used if we are asked for a pair, 2 pairs, 3 or four of a kind, full house, and royal flush?

[@MexterO123]

Brian why do you multiply the two different cases? I don't understand why.

[@alexanderjones1129]

It is called the multiplication rule. If you have r events and each event has ai possible outcomes, then the total number of possible outcomes is a1 x a2 x a3 ...

[@qyusae6413]

this is saving me in 2023 THANK YOU YO

[@yami9977]

man, you are awesome

[@rixok11]

Thank you! It was a very good video.

[@BrianVeitch]

The second example is a very specific type of 3 of a kind. It won't have the (13C1) because your 3 of a kind are JACKS. Plus your 4th card is fixed as a Queen. This is why there is no (12C2) anymore from my example. So (4C1) for the queen, (4C3) for the 3 Jacks, (44C1) to find the last card from the remaining 44 ranks that is not a Queen or Jack. Hope this helps. Just remember your examples are not general 3 of a kinds. Your formulas for them will change because some cards are fixed.

[@JavierRosales-fq8qd]

Life saver

[@sadafbenaf]

Hi Brian, my book solves (how many different 5-hand cards are possible that consist of exactly 3 kings?) as (4C3)(48C2)=4512. Or, (how many different 5-hand cards are possible that consist of 1 Queen and 3 Jacks?) as (4C1)(4C3)(44C1)=704 How are these different than your way of solving three of a kind @ 9:07

[@valingli]

I tried the two pair example before he said the answer and I got 13C1*4C2*12C1*4C2*11C1*4C1. Why is that wrong? Or, what do I misunderstand if I got that answer?

[@foolio002]

very helpful. took forever to understand by reading text

[@rickyfatty363]

Life saver thank you exactly what i needed

[@brendanmanning9461]

This is the only video that ever made it make sense!!

[@josephrose118]

well It can be delt 4 kings. so

[@gonzalezrj2010]

Great way to explain everything!! Your a life saver!

[@FJ2022]

Thank you

[@Dominic111111111]

Your 3 of a kind example is incorrect. Should be 13C1x4C3x12C1x4C1x11C1x4C1

[@micheled.m4417]

Do they both not work the same? Like what’s the difference? Genuine question.

[@dianacious86]

GREAT VIDEOS, THANK YOU!

[@BrianVeitch]

It's the most common mistake, one I made the first time tried to do this problem. If you calculate it your way, you'll get twice as many combinations. It's because of the (13C1)*(12C1) part. If I remember correctly, this means the order of the two ranks matter. (KK994 is different than 99KK4) We know they are considered the same hand. It's why your number is twice as much as it should be. Instead, use (13C2). This allows you to select 2 ranks, where order does not matter. KK994 = 99KK4