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Understanding Chances and Probabilities in Random Experiments

Learn the meaning of honesty in coins, dice, and cards and how probabilities are calculated in equiprobable spaces. Explore examples of coin tossing and rolling dice.

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Understanding Chances and Probabilities in Random Experiments

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  1. 15 Chances, Probabilities, and Odds 15.1 Random Experiments and Sample Spaces 15.2 Counting Outcomes in Sample Spaces 15.3 Permutations and Combinations 15.4 Probability Spaces 15.5 Equiprobable Spaces 15.6 Odds

  2. What Does Honesty Mean? What does honesty mean when applied to coins, dice, or decks of cards? It essentially means that all individual outcomes in the sample space are equallyprobable. Thus, an honest coin is one in which H and T have the same probabilityof coming up, and an honest die is one in which each of the numbers 1 through 6is equally likely to be rolled. A probability space in which each simple event hasan equal probability is called an equiprobable space.

  3. Equiprobable Spaces In equiprobable spaces, calculating probabilities of events becomes simply amatter of counting. First, we need to find N, the size of the sample space. Each individual outcome in the sample space will have probability equal to 1/N (they allhave the same probability, and the sum of the probabilities must equal 1). To findthe probability of the event E we then find k, the number of outcomes in E. Sinceeach of these outcomes has probability 1/N, the probability of E is then k/N.

  4. PROBABILITIES IN EQUIPROBABLE SPACES If k denotes the size of an event E and N denotes the size of the sample spaceS, then in an equiprobable space

  5. Example 15.19 Honest Coin Tossing Suppose that a coin is tossed threetimes, and we have been assured that the coin is an honest coin. If this is true, theneach of the eight possible outcomes in the sample space has probability 1/8. (Recall that there areN = 8 outcomes in the sample space.)The probability of any event E is given by the number of outcomes in E divided by 8.

  6. Example 15.19 Honest Coin Tossing This table shows each of the events with their respective probabilities.

  7. Example 15.20 Rolling a Pair of Honest Dice Imagine that you are playing some game that involves rolling a pair of honest dice, and the only thing thatmatters is the total of the two numbers rolled. As we saw in Example 15.4 thesample space in this situation isS = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12},where theoutcomes are the possible totals that could be rolled.

  8. Example 15.20 Rolling a Pair of Honest Dice This sample space has N = 11possible outcomes, but the outcomes are not equally likely, so it would bewrong to assign to each outcome the probability 1/11. In fact, most people withsome experience rolling a pair of dice know that the likelihood of rolling a 7 ismuch higher than that of rolling a 12.

  9. Example 15.20 Rolling a Pair of Honest Dice How can we find the exact probabilities for the various totals that one mightroll? We can answer this question by considering the sample space discussed inExample 15.5, where every one of the N = 36 possible rolls is listed as a separateoutcome. Because the dice are honest, each ofthese 36 possible outcomes is equally likely to occur, so the probability of each is1/36.Now we have something!

  10. Example 15.20 Rolling a Pair of Honest Dice Table 15-5 (next slide) shows the probability of rolling a 2, 3, 4,. . . , 12. In each case the numerator represents the number of ways that particular total can be rolled.For example, the event “roll a 7” consists of six distinct possible rolls: Thus, the probability of “rolling a 7” is6/36 = 1/6. The other probabilities are computed in a similar manner.

  11. Example 15.21 Rolling a Pair of Honest Dice: Part 2 Once again, we are rolling a pair of honest dice. We are now going to find theprobability of the event E: “at least one of the dice comes up an ace.”(In dice jargon, a is called an ace.) This is a slightly more sophisticated counting question than the ones in Example 15.20. We will show three different ways to answer the question.

  12. Example 15.21 Rolling a Pair of Honest Dice: Part 2 Tallying.We can just write down all the individual outcomes in the event Eand tally their number. This approach gives and Pr(E) = 11/36.There is not much to it, but in a larger sample space itwill take a lot of time and effort to list every individual outcome in theevent.

  13. Example 15.21 Rolling a Pair of Honest Dice: Part 2 Complementary Event.Behind this approach is the germ of a really importantidea. Imagine that you are playing a game, and you win if at least one of thetwo numbers comes up an ace (that’s event E). Otherwise you lose (call thatevent F). The two events E and F are called complementary events, and E iscalled the complement of F(and vice versa).

  14. Example 15.21 Rolling a Pair of Honest Dice: Part 2 The key idea is that the probabilities of complementary events add up to1. Thus, when you want Pr(E) butit’s actually easier to find Pr(F) then you first do that. Once you have Pr(F), you are in business: Pr(E) = 1 – Pr(F) Here we can find by a direct application of the multiplication principle.

  15. Example 15.21 Rolling a Pair of Honest Dice: Part 2 There are 5 possibilities for the first die (it can’t be an ace but it can beanything else) and likewise there are five possibilities for the second die.Thismeans that there are 25 different outcomes in the event F, and thus Pr(F) = 25/36. It follows that Pr(E) = 1 – (25/36) = 11/36.

  16. Example 15.21 Rolling a Pair of Honest Dice: Part 2 Independent Events. In this approach, we look at each die separately and, asusual, pretend that one die is white and the other one is red. We will let F1 denote the event “the white die does not come up an ace” and F2 denote theevent “the red die does notcome up an ace.”

  17. Example 15.21 Rolling a Pair of Honest Dice: Part 2 Clearly, Pr(F1) = 5/6 and Pr(F2) = 5/6 andNow comes a critical observation: The probability that bothevents F1 and F2 happen can be found by multiplying their respective probabilities. This means thatPr(F) = (5/6)  (5/6) = 25/36.We can now find Pr(E)exactly as before: Pr(E) = 1 – Pr(F) = 1 – (25/36) = 11/36.

  18. Independent Events Of the three approaches used in Example 15.20, the last approach appears tobe the most convoluted, but, in fact, it is the one with the most promise. It is basedon the concept of independence of events. Two events are said to be independentevents if the occurrence of one event does not affect the probability of the occurrence of the other.

  19. Independent Events When events E and F are independent, the probability thatboth occur is the product of their respective probabilities; in other words, Pr(E and F) = Pr(E) • Pr(F). This is called the multiplication principle for independent events.

  20. Example 15.22 Rolling a Pair of Honest Dice: Part 3 Imagine a game in which you roll an honest die four times. If at least one of yourrolls comes up an ace, you are a winner. Let E denote the event “you win” and Fdenote the event “you lose.” We will find Pr(E) by first finding Pr(F) using thesame ideas we developed in Example 15.21. We will let F1, F2, F3, and F4 denote the events “first roll is not an ace,” “second roll is not an ace,” “third roll is not an ace,” and “fourth roll is not an ace,”respectively.

  21. Example 15.22 Rolling a Pair of Honest Dice: Part 3 Then Pr(F1) = 5/6, Pr(F2) = 5/6 Pr(F3) = 5/6, Pr(F4) = 5/6 Now we use the multiplication principle for independent events: Pr(F) = 5/6  5/6  5/6  5/6 = (5/6)4 ≈ 0.482 Finally, we find Pr(E) Pr(E) = 1 – Pr(F) ≈ 0.518

  22. Example 15.23 Five-Card Poker Hands: Part 2 One of the best hands in five-card draw poker is a four-of-a-kind (four cardsof the same value plus a fifth card called a kicker). Among all four-of-a-kindhands, the very best is the one consisting of four aces plus a kicker. We will letF denote the event of drawing a hand with four aces plus a kicker in five-carddraw poker. (The F stands for “Fabulously lucky.”) Our goal in this example isto find Pr(F).

  23. Example 15.23 Five-Card Poker Hands: Part 2 The size of our sample space is N = 52C5 = 2,598,960. Of these roughly 2.6 million possible hands, there are only 48 hands in F: four ofthe five cards are the aces; the kicker can be any one of the other 48 cards in thedeck. Thus Pr(F) = 48/2,598,960 ≈ 0.0000185, roughly 1 in 50,000.

  24. Example 15.24 The Sucker’s Bet Imagine that a friend offers you the following bet: Toss an honest coin 10 times. Ifthe tosses come out in an even split (5 Heads and 5 Tails), you win and yourfriend buys the pizza; otherwise you lose (and you buy the pizza). Sounds like areasonable offer–let’s check it out.We are going to let E denote the event “5 Heads and 5 Tails are tossed,” andwe will compute Pr(E).

  25. Example 15.24 The Sucker’s Bet We already found (Example 15.7) that there are N = 1024equally likely outcomes when a coin is tossed 10 times and that each ofthese outcomes can be described by a string of 10 Hs and Ts.We now need to figure out how many of the 1024 strings of Hs and Ts have 5Hs and 5 Ts. There are of course, quite a few: HHHHHTTTTT, HTHTHTHTHT,TTHHHTTTHH, and so on.

  26. Example 15.24 The Sucker’s Bet Trying to make a list of all of them is not a very practical idea. The trick is to think about these strings as 10 slots each taken up by anH or a T, but once you determine which slots have Hs you automatically knowwhich slots have Ts. Thus, each string of 5 Hs and 5 Ts is determined by the choiceof the 5 slots for the Hs. These are unordered choices of 5 out of 10 slots, and thenumber of such choices is 10C5= 252.

  27. Example 15.24 The Sucker’s Bet Now we are in business: Pr(E) = 252/1024 ≈ 0.246 The moral of this example is that your friend is a shark. When you toss anhonest coin 10 times, about 25% of the time you are going to end up with an evensplit, and the other 75% of the time the split is uneven.

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