Math Steven Noble. November 24th. Steven Noble Math 3790

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1 Math 3790 Steven Noble November 24th

2 The Rules of Craps In the game of craps you roll two dice then, if the total is 7 or 11, you win, if the total is 2, 3, or 12, you lose, In the other cases (when the sum is 4, 5, 6, 8, 9 or 10), your sum is referred to as your point. You get the dice again. Now you keep rolling the dice until the sum is either 7, in which case you lose, or the sum is equal to your point, in which case you win.

3 Probability of winning Let us calculate the probability of winning at the game of craps.

4 Probability of winning Let us calculate the probability of winning at the game of craps. The probability of rolling a 7 with two dice is 6 36, since there are 36 possible outcomes for the two dice, and there are six desired outcomes, i.e., those that add up to seven, namely (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). Similarly, the probability of rolling an 11 with two dice is 2 36, since the only ways to roll 11 are (5, 6) and (6, 5).

5 Probability of winning Let us calculate the probability of winning at the game of craps. The probability of rolling a 7 with two dice is 6 36, since there are 36 possible outcomes for the two dice, and there are six desired outcomes, i.e., those that add up to seven, namely (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). Similarly, the probability of rolling an 11 with two dice is 2 36, since the only ways to roll 11 are (5, 6) and (6, 5). Therefore, the probability of winning on the first roll is = 8 36 = 2 9.

6 What if you don t win right away? Now, let s examine what happens when we roll a 4, 5, 6, 8, 9 or 10. Let s consider each case separately. First, consider the case when the point is 4.

7 What if you don t win right away? Now, let s examine what happens when we roll a 4, 5, 6, 8, 9 or 10. Let s consider each case separately. First, consider the case when the point is 4. We continue to roll the dice until the sum is either 4 (in which case we win), or roll a 7 (in which case we lose). We know that the game does not end until either of these two scenarios occur, so we want to determine the probability that the sum is 4 given that either the sum 4 or 7 has occurred. To elaborate, to find out the probability of winning the game when the point is 4, this simply the probability that we roll 4 before we roll 7. So, this is the same as saying, what is the probability that we roll a 4 given that we roll either a 4 or a 7.

8 Conditional Probability This brings us back to conditional probability. We say that P(A B) represents the probability that event A occurred given that event B occurred.

9 Conditional Probability This brings us back to conditional probability. We say that P(A B) represents the probability that event A occurred given that event B occurred. There is a nice formula for conditional probability: P(A B) = P(A B). P(B) (Here P(A B) is the probability that both A and B will occur.)

10 Returning to our example So in our example, let A be the event that the sum of the dice is 4, and let B be the event that the sum of the dice is either 4 or 7. We wish to find P(A B). Well, A B is simply A, namely the event that the sum is 4. Hence P(A B) = P(A) = 3 36, and P(B) = = 1 4, since there are three ways to roll a 3 with two dice, and six ways to roll a 7.

11 Returning to our example So in our example, let A be the event that the sum of the dice is 4, and let B be the event that the sum of the dice is either 4 or 7. We wish to find P(A B). Well, A B is simply A, namely the event that the sum is 4. Hence P(A B) = P(A) = 3 36, and P(B) = = 1 4, since there are three ways to roll a 3 with two dice, and six ways to roll a 7. Hence, P(Roll 4 Roll 4 or 7) = 3/36 9/36 = 1 3.

12 The rest of the cases Similarly, we find that P(Roll 5 Roll 5 or 7) = 4/36 10/36 = 2 5. P(Roll 6 Roll 6 or 7) = 5/36 11/36 = P(Roll 8 Roll 8 or 7) = 5/36 11/36 = P(Roll 9 Roll 9 or 7) = 4/36 10/36 = 2 5. P(Roll 10 Roll 10 or 7) = 3/36 9/36 = 1 3.

13 Putting it together Hence, the probability of winning when our first roll is 4 is P(Initial Roll Is 4) P(Roll 4 Roll 4 or 7) = = 1 36.

14 Putting it together Hence, the probability of winning when our first roll is 4 is P(Initial Roll Is 4) P(Roll 4 Roll 4 or 7) = = Similarly, we have P(Initial Roll Is 5) P(Roll 5 Roll 5 or 7) = = 2 45, P(Initial Roll Is 6) P(Roll 6 Roll 6 or 7) = 5 36 P(Initial Roll Is 8) P(Roll 8 Roll 8 or 7) = = , 5 11 = , P(Initial Roll Is 9) P(Roll 9 Roll 9 or 7) = = 2 45, P(Initial Roll Is 10) P(Roll 10 Roll 10 or 7) = = 1 36.

15 Ok, now we ll really put it all together So all of these fractions above represent the probabilities of winning when the first roll is 4, 5, 6, 8, 9 and 10. Notice the symmetry of the numbers. So this is saying that the probability of winning by rolling 10 on your first roll is only 1 36, which is about three percent. To find the probability of winning at the game, we just add all these probabilities: = Since is approximately 49.3 percent, the probability of winning at this game is just under fifty percent.

16 The value of playing Let us say the casino lets you bet money on the Pass Line, or on the Don t Pass Line. If you bet 5 of the Pass Line, you win 5 dollars if the roller wins the game. If you bet 5 on the Don t Pass Line, you win 5 dollars if the roller loses the game. Since we ve just shown the probability of winning the game is less than fifty percent, it is statistically better to always bet on the Don t Pass Line - in other words, always bet that the roller will lose.

17 The value of playing Let us say the casino lets you bet money on the Pass Line, or on the Don t Pass Line. If you bet 5 of the Pass Line, you win 5 dollars if the roller wins the game. If you bet 5 on the Don t Pass Line, you win 5 dollars if the roller loses the game. Since we ve just shown the probability of winning the game is less than fifty percent, it is statistically better to always bet on the Don t Pass Line - in other words, always bet that the roller will lose. To find out the expected value of a decision you add up each each possible outcome times the probability of the outcome. So if you bet on Pass Line the expected value is (5 dollars) ( ) + ( 5 dollars) ( ) 251 = and the expected value of betting on Don t Pass Line is ( ) ( ) ( 5 dollars) + (5 dollars) =

18 Another Value So if you were to bet on the Don t Pass Line one thousand times, statistically you would expect to win about 70 dollars. But what if instead of winning 5 dollars you only win 4 dollars when you bet 5 dollars on Don t Pass Line. The the expected value is ( 5 dollars) ( ) + (4 dollars) ( ) = So no matter whether you choose to bet on Pass Line or Don t Pass Line you still have a negative expected value. However there is a third option of not playing that has an expected value of 0.

19 Monty Hall A (female) contestant on a game show is shown three doors by the (male) host. Behind one of these there is a car, and behind each of the other two there is a goat. She chooses one of the doors, hoping of course to get the car. Before the door is opened, the host (who knows what s behind each door) opens one of the remaining two doors to reveal a goat. At this point he offers her the chance to switch her choice if she so wishes. Should she switch, and if so, how does that change the probability of winning a car?

20 Monty Hall Solution Let us say that A is the event that the car is behind door A (the door she chooses); B is the event that the car is behind door B; and C is the event that the car is behind door C. Finally we will say that b is the event that the host opens door B.

21 Monty Hall Solution Let us say that A is the event that the car is behind door A (the door she chooses); B is the event that the car is behind door B; and C is the event that the car is behind door C. Finally we will say that b is the event that the host opens door B. What we want to know is P(A b) and P(C b). For this we enlist another formula. If A 1, A 2,..., A n are all the possible events that can happen, and X is some other event, then P(A 1 X ) = P(X A 1) P(X A 1 ) + P(X A 2) P(X A n ).

22 Applying this theorem So here our A i s are A, B, C and X is b. If the car is behind door A then there is a 0.5 chance that he ll open door C and a 0.5 chance that he ll open door B (this is because there are two scenarios and we don t have any extra knowledge to distinguish between the two). So P(b A) = 0.5. If the car is behind door B then the host must open door C so P(b B) = 0. And if the car is behind door C then the host must open door B so P(b C) = 1.

23 Applying this theorem So here our A i s are A, B, C and X is b. If the car is behind door A then there is a 0.5 chance that he ll open door C and a 0.5 chance that he ll open door B (this is because there are two scenarios and we don t have any extra knowledge to distinguish between the two). So P(b A) = 0.5. If the car is behind door B then the host must open door C so P(b B) = 0. And if the car is behind door C then the host must open door B so P(b C) = 1. Plugging into the formula we get that P(A b) = 1/2 1 = 1/3, P(C b) = 3/2 3/2 = 2/3.

24 Applying this theorem So here our A i s are A, B, C and X is b. If the car is behind door A then there is a 0.5 chance that he ll open door C and a 0.5 chance that he ll open door B (this is because there are two scenarios and we don t have any extra knowledge to distinguish between the two). So P(b A) = 0.5. If the car is behind door B then the host must open door C so P(b B) = 0. And if the car is behind door C then the host must open door B so P(b C) = 1. Plugging into the formula we get that P(A b) = 1/2 1 = 1/3, P(C b) = 3/2 3/2 = 2/3. So seeing as the car is more likely to be behind door C than A it makes more sense to switch than to stay where you are.

25 Sample problem 1 You are given one true or false question on a test. If you get the question right, your final mark is increased by one percent, but if you get the question wrong, your final mark is decreased by one percent. You can guess, or you can copy from your neighbour, but you know from experience that you you neighbour answers incorrectly 20% of the time. Besides, there is a 10% chance that the teacher would catch you cheating, and in this case, she would deduct five percent from your final mark. Moral issues aside what should you do: copy from your neighbour, or randomly guess on the question?

26 Sample problem 2 Two players alternately shoot themselves with a six-shooter, only one chamber of which contains a bullet. (This is called Russian Roulette). You have the first shot, so you decide the rules. Either both players take turns shooting the next chamber, or both players randomly spin the chamber before shooting. Naturally, you want to maximize your chances of surviving. Should you spin first and shoot, or shoot without spinning?

27 Sample problem 3 Alison, Bernon, and Chantel play the following game. They take turns (in the order A, B, C, A, B, C,...) rolling one die. Alison wins if she rolls 1, 2, or 3 on her turn. Bernon wins if he rolls 4 or 5 on his turn. Chantel wins if she rolls 6 on her turn. They keep repeating this until there is a winner. What is the probability that Alison wins the game? For example if Alison rolls a 5 then it is Bernon s turn. Say her rolls a 6. Then it s Chantel s turn. Say she rolls 6. Then Chantel wins.

Math Steven Noble. November 22nd. Steven Noble Math 3790

Math Steven Noble. November 22nd. Steven Noble Math 3790 Math 3790 Steven Noble November 22nd Basic ideas of combinations and permutations Simple Addition. If there are a varieties of soup and b varieties of salad then there are a + b possible ways to order