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Probability with Cards

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Four problems: Probability of drawing 3 aces; Probability of drawing 5 cards of the same suit; Dividing 52 cards among 4 people; Probability of 4 people getting four of a kind (with only 4 card hands)

Four problems: Probability of drawing 3 aces; Probability of drawing 5 cards of the same suit; Dividing 52 cards among 4 people; Probability of 4 people getting four of a kind (with only 4 card hands)

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  • Slide 3 has an error. It has the correct answer, but the fractions needs to be over 52,51,50.
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  • 1. Probability With Cards
  • 2. Drawing Aces
    • We have a standard deck of 52 playing cards. We are drawing 3 cards without replacement.
      • What is the probability that all three are aces?
      • What is the probability that none of the three cards are aces?
    • Method 1
    • We have 52 cards, and 4 of them are aces. Then the chance of drawing the first card as an ace is
    • We now have 51 cards left, and 3 of them are aces. The conditional probability of drawing a second ace given that the first card was an ace is then
    • Finally, we have 50 cards left, and 2 of them are aces. The conditional probability of drawing a third ace given that the first two cards were also aces is then
    • How do we combine these individual probabilities? We know that to find the probability that all three cards a re aces we must draw the first card as an ace AND draw the second card as an ace, AND draw the third card as an ace. Those ANDs indicate the fact that these tasks are all necessary to complete the full process – they are dependent on each other. This means that we multiply the individual probabilities:
  • 3. Drawing Aces
    • Method 1 continued
    • The probability that none of the cards are aces is found similarly. We begin with 52 cards and there are 48 cards which are not aces, so the probability of drawing one card which is not an ace is
    • Now there are 51 cards left, 47 of which are not aces, so the conditional probability of choosing a non-ace when the first card was not an ace is
    • Finally there are 50 cards left, 46 of which are not aces, so the conditional probability of choosing a non-ace when the first two cards were both non-aces is
    • Similarly to the previous section we multiply the individual probabilities in order to find the probability that none of the three cards will be aces:
  • 4. Drawing Aces
    • Method 2
    • For this method I assume an understanding of combinations, including factorials. Here are refreshers on combinations and factorials .
    • To find the probability of drawing 3 aces we can count the number of ways in which we could draw three aces, and then divide by the total number of possible draws of three cards.
    • The total number of ways to draw three aces: We have 4 aces and we need to choose three, and since we don’t care what order the aces are chosen, the number of ways is
    • The total number of possible draws of three cards: We have 52 cards and we need to choose 3, so similarly to the previous step we have
    • The probability of choosing three aces is then:
    • We can see that both methods yielded the same results.
  • 5. Drawing Aces
    • Method 2 continued
    • To find the probability of choosing three cards, none of which are aces we follow the same pattern as the previous section. To count the number of ways to choose 3 non-aces we must choose 3 cards out of 48 non-aces, and we don’t care what order they are in, so we have
    • To count the total number of ways to select three cards we must choose 3 out of 52 total cards. Similarly to the previous section we have
    • Then the probability of choosing 3 cards, none of which are aces is then
    • Again, we see that this method yielded the same answer as method 1
  • 6. Flush
    • Using a standard deck of 52 playing cards, what is the probability of drawing 5 cards of the same suit – called a flush – without replacing cards in the deck? Click on the pictures of a speaker to hear further explanation.
    • Method 1
    • There are four different suits: hearts, diamonds, clubs, and spades. Let’s start by finding the probability of drawing 5 heats, and then we can extend our result to include all four suits
    • First card: We know there are 13 hearts out of a total of 52 cards in a standard deck, so the probability of drawing one heart is
    • Second card: We already removed one card, and it was a heart. There are now only 51 total cards remaining in the deck, and only 12 of those are hearts. The conditional probability of drawing a heart now, given that the first card was also a heart, is
    • Third card: After drawing the first two cards as hearts, there are 11 hearts left out of 50 total cards. The conditional probability of drawing a heart, given that two hearts have already been removed from the deck is then
  • 7. Flush
    • First card:
    • Second card:
    • Third card:
    • Fourth card: Similarly to the previous three cards, there are only 10 hearts left out of a total of 49 cards, so the probability of drawing another heart is
    • Fifth card: Finally, after drawing 4 hearts, there are now 9 hearts left out of a total of 48 total cards. The conditional probability of drawing a heart given that 4 hearts have already been drawn is then
  • 8. Flush
    • Combining the individual probabilities: Each of the 5 drawing steps is required to complete the task. That is we’re drawing the first card AND then the second card AND then the third, etc. We can’t draw the first card OR the second card – we have to do both. I find this AND/OR distinction helpful. If we do task 1 AND then task 2 then we generally multiply the individual probabilities; if we do task 1 OR we do task 2 then we add the individual probabilities. In this case we are doing all 5 steps, and each is dependent on the last, so we multiply the probabilities to get the total probability of drawing 5 hearts from a standard deck of 52 cards: we’re almost done…
    • Now we need to consider all four suits. Since there are an equal number of cards of each suit in a standard deck, the probability is the same for each individual suit. How do we combine these? Well, we can either draw five hearts OR five spades OR five clubs OR five diamonds. These are all mutually exclusive events – they can’t occur together. As I described in the previous step, that means we add the individual probabilities:
    • which is equal to
  • 9. Flush
    • Method 2
    • Another way to start this problem would be to assume that we don’t care what card we choose first, and then figure out the chance of choosing four more of the same suit.
    • First card: We don’t care what card we choose first – we’ll take any card in the deck. Another way to say that is we have 52 possible “correct” choices out of 52 total cards. One way to write the probability is which is equal to 1
    • Second card: We have removed one card. That first card belongs to one of the four suits – we don’t care which one, we just want to know the probability of drawing a second card of the same suit. Well, since we removed one from the deck already, there are 12 left of that particular suit and there are 51 total cards. Then the probability of choosing a second card that is the same suit as the first is:
    • Third, fourth and fifth cards: just like in method 1 we want to choose more cards of the same suit , so we get probabilities of , , and .
    • Combining the probabilities: This is exactly like method 1 in that we multiply the individual probabilities because they are all happening in order (think AND). So we get:
    • We get the same answer as method 1! Note that in method 1, 4 multiplied by 13 divided by 52 is equal to 1, so we can easily rewrite the solution from method 1 to look like the solution from method 2:
    • Method 2
    • Another way to start this problem would be to assume that we don’t care what card we choose first, and then figure out the chance of choosing four more of the same suit.
    • First card: We don’t care what card we choose first – we’ll take any card in the deck. Another way to say that is we have 52 possible “correct” choices out of 52 total cards. One way to write the probability is which is equal to 1
    • Second card: We have removed one card. That first card belongs to one of the four suits – we don’t care which one, we just want to know the probability of drawing a second card of the same suit. Well, since we removed one from the deck already, there are 12 left of that particular suit and there are 51 total cards. Then the probability of choosing a second card that is the same suit as the first is:
    • Third, fourth and fifth cards: just like in method 1 we want to choose more cards of the same suit , so we get probabilities of , , and .
    • Combining the probabilities: This is exactly like method 1 in that we multiply the individual probabilities because they are all happening in order (think AND). So we get:
    • We get the same answer as method 1! Note that in method 1, 4 multiplied by 13 divided by 52 is equal to 1, so we can easily rewrite the solution from method 1 to look like the solution from method 2:
  • 10. Flush
    • Method 3
    • Another way to compute this would be to count the number possible combinations of 5 cards of the same suit and divide by the total number of possible combinations of five cards. For this method I assume an understanding of combinations, including factorials. Here are refreshers on combinations and factorials .
    • Total number of combinations of 5 cards of the same suit: First we choose one of four suits – that is 4 choose 1, AND then we choose 5 of the 13 cards in that suit. So we have:
    • Total number of combinations of 5 cards. That is simply choosing any 5 out of 52 total cards: .
    • Now we divide as stated in the first step. Instead of simply plugging this in to a calculator, I will manipulate the result so that we can see it is the same as methods 1 and 2.
  • 11. Dealing Cards
    • What is the total number of ways to divide up 52 playing cards among 4 people?
    • Method 1
    • We assume that the deck of cards is shuffled perfectly so that the order of the cards is random. I will try to convince you that the order in which we deal does not matter.
    • Cards are usually dealt one at a time beginning with the person to the left of the dealer and continuing clockwise. We’ll call the first to receive a card person A, and the other people B, C, and D in the order that they are dealt cards.
    • We deal first to person A, and there are 52 possible cards of which we give them 1. AND then we deal to B, and there are 51 cards left, of which we give them 1. And then C, and then D, and then A, etc. until all of the cards are dealt.
    • However, we don’t care which order each person receives their cards – for example it doesn’t matter whether A gets the 2 of hearts as their first card or as their second card. We can visualize this by saying the letters “A,” “B,” “C,” “D” when we deal a card to that person. We then have a string of 52 letters: ABCDABCDABCD…ABCD.
    • We know that each person gets 13 cards, so this string of letters has 13 As, 13 Bs, 13 Cs, and 13 Ds. Each A is indistinguishable from the others, meaning that you could switch the As around and the list of letters would look the same. Not only that, but we could switch the order of all of the letters around. This represents the fact since the cards are in a random order to begin with , it doesn’t matter how we deal the cards. We could deal 5 cards to person A, then 2 to person D, then 12 to person C, etc.: AAAADDCCCCCCCCCCCC… as long as each person ends up with 13 cards.
  • 12. Dealing Cards
    • The result of this is that dividing 52 cards among 4 people is the same as counting the number of distinct strings of 13 As, 13 Bs, 13 Cs, and 13 Ds. This is found by counting the total number of strings of 52 distinct letters and dividing by the number of strings we overcounted. As we noted before, 52! gives us the number of strings possible with 52 distinct letters. In any one of those particular strings, however, there are 13 As which are not distinct from each other. There are 13! ways to arrange those As in every string without changing the position of any of the Bs, Cs, or Ds. Similarly there are 13! Arrangements of the Bs, and of the Cs, and of the Ds. Since all of these arrangements are possible in each of the 52! strings, we have overcounted 13!*13!*13!*13! extra strings.
    • Therefore, the number of strings of 13 As, 13 Bs, 13 Cs, and 13 Ds is and as mentioned in an earlier step, this is also equal to the number of ways to divide 52 cards among 4 people. This number is huge, approximately
  • 13. Dealing Cards
    • Method 2
    • This method is more of a corollary to the previous method in that it helps to accept that the order in which the cards are dealt does not matter if the deck is in a random order.
    • Let’s assume that the deck is shuffled perfectly again, that is, the order of the cards is random. Then we can choose 13 out of the 52 cards for one person AND then 13 out of the remaining 39 for another person AND then 13 out of the remaining 26 for another person AND finally 13 out of the remaining 13 for the final person.
    • Those “AND”s were to help us realize that we should be multiplying, since we are carrying out each step and they are dependent. So we have:
  • 14. Four of a Kind
    • We have a standard deck of 52 cards which is in a random order, and we deal 4 cards to each of 4 players. What is the probability that each player receives four of a kind (four of the same number)?
    • We can find this result by dividing the number of hands in which each player receives four of a kind by the total number of possible ways to deal 4 cards to 4 people from a standard deck of 52 cards.
    • Finding the total number of ways to deal 4 cards to 4 players is similar to the problem about dealing cards. Let’s call our players A, B, C, and D. When we deal a card to each player, we label it with their letter, and we are left with 36 cards which we can label X. We then have a string of letters which includes 4 As, 4 Bs, 4 Cs, and 4 Ds, and 36 Xs. Then, just like the Dealing Cards problem, we have 52! total possibilities for strings of 52 distinct letters divided by 4!*4!*4!*4!*36! which is the number of times overcounted. This is also the total number of ways to deal 4 cards to 4 people.
    • The total number of possible hands in which all four players receive four of a kind is much smaller than the total number of possible hands. There are 13 different card values to choose from, each one containing 4 cards of the various suits: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.
  • 15. Four of a Kind
      • Method 1
      • Player A must have 4 of a kind of one of those 13 values, mathematically that is 13 choose 1 AND then 4 choose 4. Of the 13 types of cards we choose 1, and then of the 4 cards of that type we choose 4.
      • Then player B must have four of a kind of one of the 12 remaining types of card. Similarly to the previous step this is
      • Players C and D follow similarly with and , respectively.
      • We know that each of these steps is dependent on the previous, (and remember the AND guide), so the number of hands in which all four players receive four of a kind is:
    • Method 2
      • We choose 4 of the 13 values. Then we choose all 4 of each type. But in card games we care which player has which cards, so we multiply by the 4! different ways we could arrange those hands among the 4 players.
  • 16. Four of a Kind
    • Then the probability of each player receiving four of a kind is:
    • which is a very low probability, as
    • we would expect