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Example: Lottery probability

What is the probability of winning a 4-number lottery? Created by Sal Khan and Monterey Institute for Technology and Education.

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  • male robot donald style avatar for user achu
    arent there 4! ways we can write the winning numbers . i mean the order doesnt matter so 3,15,46,49 should be the same as 15,3,46,49 but sal says that theres only one way of getting the correct lottery numbers why is that?
    (84 votes)
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    • leaf green style avatar for user William Hunter
      Achu and Naveen,
      Sal could have solved this problem in two ways. He could have taken the number of possible permutations with a favorable outcome and divided that by the total possible number of permutations –or—he could have taken the number of possible combinations with a favorable outcome and divided that by the total number of possible combinations (which is what he did).
      Let us do it both ways, using the permutations first.
      As you mentioned, there a 4! ways of writing the four numbers. Another way to say this is that there are 4! different ways to order the four numbers –or-- there are 4! different permutations of the four numbers that give us the favorable outcome. This can be written as 4*3*2*1.
      The total number of different permutations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!), which can be written as 60*59*58*57.
      Our answer using permutations would be the number of favorable outcomes/the number of possible outcomes which would be (4*3*2*1)/(60*59*58*57). This simplifies to 1/487,635.
      Using combinations, there is only one (1) combination of numbers that gives us that favorable outcome (that one way, achu).
      The number of possible combinations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!*4!). This can be written as (60*59*58*57)/(4*3*2*1) . This number is 487,635.
      Our answer using combinations would be the number of favorable outcomes/the number of possible outcomes which would be 1/487,635. This is the same answer we got using permutations.
      Consider combinations and permutations to be different “units”. You would not say that after moving 3 inches you are halfway to traveling 6 meters, even though 3/6 = ½. You cannot take the number of possible favorable permutation outcomes and divide that by the number of total possible combination outcomes and get the correct probability. Likewise you cannot take the number of possible favorable combination outcomes and divide that by the number of total possible permutation outcomes and get the correct probability.
      (126 votes)
  • aqualine tree style avatar for user Erik
    Is there any reason why I could not solve the problem this way? Because I did and it turned out ok, but I don't always trust my own leaps of logic:

    If I am hoping to draw 4 particular numbers randomly out of 60, then I can say that on my first draw there are 4 numbers that I could hope to get, out of a total of 60, so I begin with
    4/60 as my chances of getting one of those numbers on that first draw.

    On my second draw, there are now 3 numbers left that I want, out of a possible 59 total remaining, so I begin to multiply:
    (4/60) * (3/59)

    The last two draws follow logically: 2 acceptable numbers out of 58, and then 1 acceptable number out of 57. So the probability of drawing any particular set of 4 numbers out of 60, if we cannot draw any number twice, is:

    (4/60) * (3/59) * (2/58) * (1/57)

    I can just calculate this and get the same answer that Sal gets, but what do I make of the fact that this expression is equal to

    4! / (60! / (60-4)!)
    or
    4! / P(60,4)

    I am trying to draw some sort of larger logical conclusion out of this, but am having some trouble not confusing myself. It seems that I could say this represents the number of possible arrangements of any particular set of 4 numbers distributed over, or divided by, the total number of possible ways to draw all sets of 4 numbers out of 60.

    This makes some sense to me, but I tend to come at these things backwards, using intuition and numbers first and then going back over the solution to see whether or not I could claim any larger guiding principles and/or formulas. Can anyone do a better job of explaining why my intuition was right?
    (13 votes)
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    • leaf green style avatar for user ArDeeJ
      This sounds like a tautology but your intuition is right because it is right.
      Let's look at the different ways of arriving to the answer.

      Your way: look at the situation one draw at a time.
      4/60 * 3/59 * 2/58 * 1/57, which is equal to 4*3*2*1 / (60*59*58*57)

      Sal's way:
      4! / (60 P 4)
      which is the same as
      4! / (60!/(60-4)!)
      which is the same as
      4! * (60-4)! / 60!
      which is the same as
      4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1)
      which is the same as
      4*3*2*1 / (60*59*58*57)
      So your and Sal's ways are the same thing just expressed in different ways.

      Here's one more way to look at it:
      There are (60 C 4) ways of choosing four numbers (all different) from 60. Only one combination of them is what we have chosen. So the probability is
      1/(60 C 4)
      which is the same as
      1/( 60!/(4!*(60-4)!))
      which is the same as
      4! * (60-4)! / 60!
      which, as we already have seen, is the same as
      4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1)
      which is the same as
      4*3*2*1 / (60*59*58*57)
      (13 votes)
  • blobby green style avatar for user syedahmed0601
    If one boy and five girls must stand in a line for a school picture and the boy can't stand first or last in line, how many different ways could the children be arranged?
    (1 vote)
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    • blobby green style avatar for user Wei Du
      There are 6 children standing in line, so there're a total of 6! = 720 possible different re-arrangement. but we don't want:

      BGGGGG (there are 5! = 120 different way of rearranging the girls here) or GGGGGB (also 120 different rearrangement of the girls here)

      so the final answer would be 720 - 120 * 2 = 480 different arrangements.
      (10 votes)
  • male robot donald style avatar for user L.Nihil kulasekaran
    If S=1+2+4+8+16+32................ Can this be taken as S=1+2(1+2+4+8+16.......)???
    If it can be taken,
    S=1+2S
    S=-1 Is this a valid answer for such a big question???
    (3 votes)
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    • piceratops ultimate style avatar for user A Highberg
      Your reasoning only works when the sum S is a real number and does not continue on to infinity. For example, if S = 1 + 1/2 + 1/4 + ... + 1/(2^n) + ... and so on forever, then your logic says that S = 1 + 1/2(S), which gives the right answer of S = 2.

      The difference between the two problems is that the sum S = 1 + 2 + 4 + ... continues to get larger forever and does not stop, while S = 1 + 1/2 + 1/4 + ... has a limit that it can never reach, no matter how long you count. By taking your approach, you can short-cut the infinite series and figure out that limit without having to take forever (literally!).
      (5 votes)
  • blobby green style avatar for user LukeSteins
    just wanted to add my 2 cents. I'm having a hard time explaining it all though so would love feedback. For the lottery question, another way to think of it is as below.

    4p4/60p4 = same answer.

    explanation:
    think of this top part of the probability (numerator) as 4p4 since you have 4 numbers to pick from and you want to pick 4 numbers, the number of ways you can pick 4 numbers from 4 numbers is 4*3*2*1. 4p4. This gives you the total number of non-unqiue ways to choose these numbers.

    on the bottom, you know that you have 4 numbers to pick and 60 to pick from. your total number of ways to pick 4 numbers from 60 is 60p4.

    I know that we dont care about order, but you see when you look for probability, the probability of the order not mattering is the same as the probability of the order mattering. because the numerator and denominator increase by factors of 4!. since when you think about it, the numerator is the number of ways total that you can combine 4 numbers divided by the denominator which is the total number of ways to pick 4 numbers of 60. Both top and bottom are including all the different permutations of 4 numbers. so really it's including those groupings of duplicated values. individually looking at each unique grouping's probability is the same as looking at the probability across all groupings.

    Alternate way:
    4c4/60c4 - removes the order. same answer.

    QUESTION: what if we're drawing the numbers, what is the probability of drawing these 4 numbers. Does the math change / probability change?

    Draw 1: 4 c 1
    Draw 2: 3 c 1
    Draw 3: 2 c 1
    Draw 4: 1 c 1

    fundamental counting principle lets us rep drawing 1,2,3,4 as 4c1*3c1*2c1*1c1

    on the bottom you total ways of drawing per draw.
    Draw 1: total ways = 60 c 1
    Draw 2: 59 c 1
    Draw 3: 58 c 1
    Draw 4: 57 c 1

    the probability of drawing is top / bottom.

    QUESTION: should i use P or C? logically speaking which one makes more sense. I know they yield the same answer but which one fits this equation / changes the equation's logic.

    Also, does this any of what i wrote up apply beyond jsut this question? Is this some unique solution methodology.

    is it a rule that says you can't divide a combination by permutation?

    e.g. 4c4/60p4. not allowed

    because i see a lot of people wondering about 1/60*1/59*1/58*1/57 as a possible answer. My response is the top only shows 1 permutation of arranging the 4 numbers and the bottom represents the total number of ways of picking 4 numbers. so it's like 4c4/60p4 which incorrectly represents the solution. I think it's actually ok, its just not ok for this question.

    because the top yes can be 4c4, it represents 1 but really depending on the person answering. they could be incorrect due to 2 different reasons. 1 not realizing that the order doesn't matter so their method of only using 1 / 60 etc only calculates the probability of 1 method of arranging those 60 and then also not realizing that the bottom looks at all possible permutation. OR they are looking at the top correctly as a single combination and not evaluating the bottom correclty and failing to use also combinations below.

    Apologies for so many thoughts, probabilities is really interesting and difficult for me and writing up my thoughts helps me. I hope it helps someone else too.
    (4 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      As long as you’re consistent, you will get the correct answer. It makes more sense to use the permutation method (for both top and bottom) if you think of the numbers as picked one at a time, but it makes more sense to use the combination method (for both top and bottom) if you think of all four numbers as being picked at once.
      (1 vote)
  • marcimus pink style avatar for user Elizabeth Gertz
    I was just wondering what the nCr and nPr buttons on the calculator do. My teacher explaned it, but i forgot what the do and how to use them. Please help!
    (1 vote)
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    • blobby green style avatar for user reardon.skip
      nCr is used for Combinations, while nPr is used in permutations. 'N' represents the total number of items you have to choose from, and 'R' represents the number you choose. So if you had 36C10, that would mean you have 36 items and you can choose 10, regardless of order, since it is a Combination.
      (6 votes)
  • primosaur ultimate style avatar for user Eric Na
    Isn't 59 factorial (!), 60*59*58*57*56*all the way down to 0??
    (1 vote)
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  • duskpin ultimate style avatar for user Fred12
    if in this lottery, picking a number and putting it back is allowed... so that means you can pick a number a multiple of times... what would the probability be then?
    64! / ((64-4!) * 4! ) ?
    (2 votes)
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    • leaf green style avatar for user ArDeeJ
      Well, you'd choose 4 numbers from 60 numbers (1 to 60) and repetition is allowed, the probability of winning would be 1/(60^4/4!) = 4!/60^4 = 1/540000 ≈ 0.000002
      (The 4! is there because the order of the numbers still doesn't matter. If the winning row was 7,34,34,50 and you had 34,7,50,34 you'd still win)
      (3 votes)
  • blobby green style avatar for user captroper
    I think I may have a fundamental misunderstanding of combinations and / or permutations. I tried to solve this problem by doing the following (60! / (56! * 4!)) / (60^4) which is the combinations formula divided by (I thought) the total number of possible outcomes with 60 numbers in 4 slots. Why is that incorrect?
    (2 votes)
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    • aqualine ultimate style avatar for user Timber Lin
      60^4 isn't the total number of possible groups of 4, because the order of the 4 numbers doesn't matter for combinations. 60^4 is the number of permutations, not combinations.
      and 60!/(56!*4!) is the total number of possible combinations of 4 numbers, so it is the sample space, not the # of successes meaning it should be the denominator.
      (2 votes)
  • blobby green style avatar for user ProfessionsNow
    what if you want to know the probability of a number winning excluding some number already played that will not be played again?
    (2 votes)
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    • starky sapling style avatar for user 4x² (soli Deo gloria)
      That's a fun calculation. Say you have 7 different items in a bag. The probability of pulling a certain one out is 1/7. After doing so you now have 6 items. The next time you pull one out the probability will be 1/6. Makes sense?
      This reasoning is used in many situations where multiple items need to be pulled out of the bag in quick succession.
      (1 vote)

Video transcript

To win a particular lottery game, a player chooses 4 numbers from 1 to 60. Each number can only be chosen once. If all 4 numbers match the 4 winning numbers, regardless of order, the player wins. What is the probability that the winning numbers are 3, 15, 46, and 49? So the way to think about this problem, they say that we're going to choose four numbers from 60. So one way to think about it is, how many different outcomes are there if we choose four numbers out of 60? Now this is equivalent to saying, how many combinations are there if we have 60 items? In this case we have 60 numbers, and we are going to choose four. And we don't care about the order. That's why we're dealing with combinations, not permutations. We don't care about the order. So how many different groups of four can we pick out of 60? And we don't care what order we picked them in. And we've seen in previous videos that there is a formula here, but it's important to understand the reasoning behind the formula. I'll write the formula here, but we'll think about what it's actually saying. So this is 60 factorial over 60 minus 4 factorial, divided also by 4 factorial, or the denominator multiplied by 4 factorial. So this is the formula right here. But what this is really saying, this part right here, 60 factorial divided by 60 minus 4 factorial, that's just 60 times 59, times 58, times 57. That's what this expression right here is. And if you think about it, the first number you pick-- there's 1 of 60 numbers, but then that number is kind of out of the game. Then you can pick from 1 of 59, then from 1 of 58, then of 1 of 57. So if you cared about order, this is the number of permutations. You could pick four items out of 60 without replacing them. Now, this is when you cared about order, but you're overcounting because it's counting different permutations that are essentially the same combination, essentially the same set of four numbers. And that's why we're dividing by 4 factorial here. Because 4 factorial is essentially the number of ways that four numbers can be arranged in four places. Right? The first number can be in one of four slots, the second in one of three, then two, then one. That's why you're dividing by 4 factorial. But anyway, let's just evaluate this. This'll tell us how many possible outcomes are there for the lottery game. So this is equal to-- we already said the blue part is equivalent to 60 times 59, times 58, times 57. So that's literally 60 factorial divided by essentially 56 factorial. And then you have your 4 factorial over here, which is 4 times 3, times 2, times 1. And we could simplify it a little bit just before we break out the calculator. 60 divided by 4 is 15. And then let's see, 15 divided by 3 is 5. And let's see, we have a 58 divided by 2 is 29. So our answer is going to be 5 times 59, times 29, times 57. Now this isn't going to be our answer. This is going to be the number of combinations we can get if we choose four numbers out of 60 and we don't care about order. So let's take the calculator out now. So we have 5 times 59, times 29, times 57. It's equal to 487,635. So let me write that down. That is 487,635 combinations. If you're picking four numbers, you're choosing four numbers out of 60, or 60 choose four. Now, the question they say is, what is the probability that the winning numbers are 3, 15, 46, and 49? Well, this is just one particular of the combinations. This is just one of the 487,635 possible outcomes. So the probability of 3, 15, 46, 49 winning is just equal to-- well, this is just one of the outcomes out of 487,635. So that right there is your probability of winning. This is one outcome out of all the potential outcomes or combinations when you take 60 and you choose four from that.