- Probability using combinations
- Example: Lottery probability
- Example: Different ways to pick officers
- Probability with permutations & combinations example: taste testing
- Probability with combinations example: choosing groups
- Probability with combinations example: choosing cards
- Probability with permutations and combinations
- Mega millions jackpot probability
We can use combinations (when order does not matter) to find the probability of drawing two aces and two kings in a draw of four cards. Created by Sal Khan.
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- This is the first one in the precalc-probability playlist that tripped me conceptually. I don't understand why/how we expressed it purely in terms of combinations without even a hint of using the multiplication rule of probability (since each draw is a dependent event). But I came up with solution which I find is much easier to grasp.
Firstly, you need to realize that the probability of drawing 4 cards which has 2 aces and 2 kings of a single arrangement is the same for any other arrangement. For example, P(ace, ace, king, king) = P(king, ace, ace, king) = P(ace, king, king, ace). Let's figure out the probability of getting 2 aces and 2 kings of a single arrangement using a simple arrangement.
P(ace, ace, king, king) = (4 / 52) ⋅ (3 / 51) ⋅ (4 / 50) ⋅ (3 / 49)
Secondly, you need to realize that the probability of drawing 4 cards with 2 aces and 2 kings regardless of arrangement is equal to the probability of drawing 4 cards which has 2 aces and 2 kings of a single arrangement times all the ways to arrange 4 cards with 2 aces and 2 kings. So, let's figure out how many ways to arrange 4 cards composed of 2 aces and 2 kings. It's equivalent to figuring out how many ways to choose 2 cards from a hand of 4 kings (king, king, king, king) to turn into aces; it's simply ₄C₂.
Lastly, we multiply those two quantities to get the probability of drawing 4 cards with 2 aces and 2 kings regardless of arrangement.
P(ace, ace, king, king) ⋅ ₄C₂ = 36 / 270725
Sal's expression evaluates to the same number.
I still want to understand Sal's solution though. Any helpful explanation is appreciated.(7 votes)
- I figured it out. To understand it I assume that there are 3 aces and 3 kings to replace the 4 aces and the 4 kings. To choose 2 aces from 3 aces there are 3 combinations(or 6 permutations) and its the same for choosing 2 kings from 3 kings. Now let us find how many combinations can aces and king be together. We assume that the 3 ace combinations are:AB AC BC; the 3 king combinations are: DE DF EF. So the first combination is AB & DF, the second one is AB & DE, third one is AB & EF, forth one is AC & DF, fifth one is AC & DE, sixth one is AC & EF, the seventh one is BC & DF, the eighth one is BC & DE, the ninth one is BC & EF. Now we understand that the pattern is actually just the number of ace combinations multiply by the number of king combinations(in this case 3*3). Therefore, I hope you can understand the reason why it is (4C2)*(4C2) in the formula.(4 votes)
- For anyone who is not an avid cards player, this video is probably confusing.
The key to understanding this question is realizing that there are 4 suits in a standard deck. There are 4 different types of aces and 4 different types of kings.
After conceptualizing it this way, 4C2 makes much more sense as opposed to imagining all of the aces being identical.
For every 2 aces you pick, you have 4C2 scenarios of which kings you pair them with. That is why they are multiplied in the numerator.
And finally, the denominator is 52C4 because that is how many different combinations of cards you can get when drawing 4 cards from a deck of 52.
Hope this helps!(6 votes)
- I am trying to intuitively understand why the numerator should be a product of 4C2 of the aces and 4C2 of the kings instead of the sum.
You'll note that even when Sal wrote the idea down on the board he put '2 aces + 2 kings' not '2 aces * 2 kings'. So at least using a verbal description, this is counter-intuitive since it seems like we are dealing with the addition of two possibilities - the ways of getting 2 aces and the ways of getting 2 kings.
The first question I have to answer is - are these events dependent or independent? It seems like choosing the aces and choosing the kings are independent events, except that removing those first two cards from the deck would change the probability of getting the remaining desired cards slightly, so I guess they are dependent?
If I go back to the Multiplication Rule of Probability, which has not failed us so far, I guess we can say that if picking aces and picking kings are dependent events (are they for sure?) then the probability of both events occurring is
P(A ∩ B) = P(B) * P(A|B)
I can read that as 'the probability of picking 2 aces AND the probability of picking two kings is the probability of picking two aces times the probability of picking 2 kings given that I picked two aces.'
Then I guess I have to remember that the Addition rule of Probability is used in cases where I'm choosing between A OR B? Am I on the right track?(1 vote)
- This video casued me quite a headache, but i think i've got it now... it was on my mind for several days and the other questions/ answers didn't help, so i'll tell my story here in case it helps someone.
Initially, i got the answer wrong when i paused the video to work it out myself but i could't see why. My [incorrect] reasoning was thus:
starting with an analysis of what was going on as Sal advocates rather than going straight for some abstract formula, let's say you start with wanting an Ace: it's 4/52; say you then want your second Ace: 3/51; first King: 4/50; second King: 3/49.
In my mind there is no question that these need to be multiplied with each other, so whatever order you get the cards, the numerator will be 4 x 3 x 4 x 3 (the order doesn't matter as multiplication is commutative); the denominator will be (52 x 51 x 50 x 49) - in that order irrespective of how the cards are drawn (not that it matters as multiplication is still commutative). All fine and correct so far (although it's not the method Sal used). I then reasoned (this is the INCORRECT bit) that with 4 cards there are 4! (i.e., 24) ways to arrange them so 24 ways to draw those 4 cards, giving my incorrect answer of:
(4 x 3 x 4 x 3 x 4!) / (52 x 51 x 50 x 49)
(4 x 3 x 4 x 3 x 4 x 3 x 2 x 1) / (52 x 51 x 50 x 49)
My answer was 4 times larger than the correct answer.
I finally sorted myself out by drawing a tree diagram (not showing all the possibilities, just those i was interested in, which made it much simpler).
It was then clear that there are just 6 ways of picking exactly 2 Kings and 2 Aces (not the 24 that my 4! gave - and that removed my factor of 4 error... the 4! requires us to be considering the specific aces or kings we get). To my way of thinking this is because at this point in the logic there is no distinguishing between the Aces and no distinguishing between the Kings: we don't care about hearts, spades, etc...
The options are, in order of picking the cards:
AAKK, AKAK, AKKA, KKAA, KAKA and KAAK (using A for each Ace and K for each King shows that there is no distinbuishing between the cards of equal rank at this point). Each route has the same probability so they add together to give the combined probability, hence the probability of any individual route is multiplied by 6 (NOT 24).
So, e.g., the probability of AAKK is:
(4/52) x (3/52) x (4/50) x (3/49), and the probability of each permutation of drawing those cards is the same (although in the calculation, the 4s and 3s could come in a different order depending on the order the cards are drawn).
So, by Sal's method, the denominator is the number of ways of choosing any 4 specific cards, i.e., '52 Choose 4', and the numerator is the number of ways available to get the 4 cards we want... and so:
there are (4 Choose 2), i.e., 6, ways to get 2 Aces from 4 Aces (we are now considering the different 'suits' of Ace we might get - or in more general terms, the specific cards we might get) - and for each of those ways there are (4 Choose 2), i.e., 6 (again), ways of getting 2 Kings, so we have to multiply the 2 numbers to get 36.
It's funny: having now understood the answer and the 2 ways i've described of getting there, i've just rewatched the video and realised that Sal explained it perfectly.
I hope that helps someone having similar difficulties.(4 votes)
- I'm a bit confused, can someone explain in a different way why a draw with 2 aces should be 4Choose2? and likewise with kings 4choose2?(1 vote)
- Basically 4Choose2 is telling you how many different combinations there are when picking 2 aces/kings out of the 4 in the deck. You would use combinations instead of permutations here because the order in which you pick them does not matter.(2 votes)
- We're told that a standard deck of 52 playing cards includes four aces, four kings, and 44 other cards. Suppose that Luis randomly draws four cards without replacement. What is the probability that Luis gets two aces and two kings, in any order? So like always, pause this video and see if you can work through this. All right. Now, to figure out this probability, we can think about this, it's going to be the number of, let's call them draws with exactly two aces and two kings, two aces and two kings. And that's going to be over the total number of possible draws of four cards. So number of possible draws of four cards. Now, for many of y'all, this bottom, the denominator here, might be a little bit easier to think about. We know that there's 52 total cards, of which we are choosing four. So we could say 52 choose four, and that will tell us the total number of possible draws of four cards. How many combinations of four cards can we get when we're picking from 52? But the top here might be a little bit more of a stumper. We can think we have exactly two spots for aces, so we're choosing two aces out of how many possible aces? Well, there's four total aces. So if we say four choose two, this is the total number of ways, when you don't care about order, that you can have two out of your four aces picked. And then separately, we can use similar logic to say, all right, there's also four choose two ways of picking two kings out of four possible kings. And now the total number of draws with two aces and two kings, this is going to be the product of these two. And if you're wondering why you can just multiply it, think about it. For every scenario that you have these two aces, you have four choose two scenarios of which kings you're dealing with. So you would take the product of them. And we've already done many examples of computing combinatorics like this, so I will leave you there. If you're so motivated, I encourage you to be, you can actually calculate this value.