while a 10 can be obtained by rolling:

47 Chapter 4 Probability One of the original motivations behind counting was the beginning of the taming of uncertainty that occurred in the 16th an...
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Chapter 4 Probability One of the original motivations behind counting was the beginning of the taming of

uncertainty that occurred in the 16th and early part of the 17th century. Why it took so long to even develop to the extent it did in those early centuries is indeed interesting, but not for us to speculate (gambling is very old indeed.) What is relevant to us is that by 1600 it was reasonably clear in many people's minds what some aspects of probability were about. But let's progress by example, a historical one. Galileo himself was posed this question, and as usual he analyzed it correctly. Example 1. Suppose we are going to play the following game (those early years were mostly concerned with gambling questions (as mentioned above, gambling is old definitely thousands of years old.): we roll 3 dice, if a 9 shows up, I pay you $1, if a 10 shows up, you pay me $1, if anything else shows up, we roll again. Naturally you are mistrustful since I am proposing the game, but how do you know I am not the idiot by offering it to you, or better, that it may be a fair game and you are just missing the opportunity to have fun. Of course, if you are just going to play one hand, no calculation is really necessary and you are just going to make your decision based on your mood, who makes the offer, etc. But suppose you intend to do this for three hours every Saturday for the next three years (we live in the age of individual preference.) At first thought it seems like a reasonable game: one can obtain a 9 by rolling:

while a 10 can be obtained by rolling:

At first thought it seems like a fair and reasonable game. Both numbers can be rolled in six different ways, as the two lists of possibilities indicate. But what is the logic behind this attempt? It has something to do with the number of ways of doing something and if one thing has more ways of occurring than another, then it is more likely to occur. After all nobody would play the previous game if the competition were between rolling a 3 and rolling a 10 since intuitively one feels that a 3 is much rarer than a 10. Although there is common sense behind this, it is not quite correct. It needs to be improved upon. The basic principle that we are going to use for our probability calculations is

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suppose an activity or experiment is to be performed, and we have equally feasible outcomes, then the probability for a given event to occur is the number of outcomes that give the desired event divided by the total number of outcomes.1 But extra emphasis needs to be made on the premise of the principle: one must first reduce the outcomes of the activity to equally feasible outcomes2. Then you can start looking at the probability of the event that you are interested in. Going back to the game in question: what is the activity in this example? Rolling 3 dice. What are the outcomes? It seems acceptable to say that the outcomes are, in addition, to the two lists above:

3 4 5 6 7 8 11 12 13 14 15 16 17 18 and thus there would be a total of 56 outcomes, so the probability of a and a 10 would have the same probability, so the game is seemingly fair.

1

9 would be 6 , 56

This is the first enunciated principle in the theory of probability, and the simplest one. As simple as it is, it was not stated clearly until the 16th century by the inimitable Cardano, great scholar and scoundrel. 2 What are equally feasible outcomes can be in itself a polemic. How do you know a coin is fair? But we will be naive about the subtleties of statistical analysis, and only insist that, from what we know, we can honestly claim that the outcomes we are taking are equally feasible.

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However, if we apply the same reasoning, then the probability of a

3 is

1 , and a 56

4 has

the same probability. So if we keep rolling the three dice for a long time, the number of 3's occurring should roughly be the same as the number of 4's. It does not take much experimentation to perhaps start doubting our premise, and maybe we should question why did we label those outcomes as equally feasible? So let's rethink a bit. Is a as equally feasible as a

?

Suppose we had a yellow die, a white die and a blue die. Then, to roll a 3, we would have to have

, we need to show a

do it by:

, or

in each die, but to get a

, or

,

4, we can

, there are three ways since

any of the three dice could show a and the other two need to show a . It seems like we have some more choices in the latter situation. Three times as many, actually. A way out of the quagmire is to take for our outcomes the 216 different ways there are to roll three dice if one of them is yellow, other one white and the third one blue. We get the 216 from: 6 6 6 . Nobody can argue on the equal feasibility of these 216 outcomes. So we start now from there. How many ways can we roll a 3? As before, only one way, so the probability is 1 , not 1 . But how many ways can we roll a 4? Three ways as we 216

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saw above, so the probability of a

4 is

often than 3's. Let's go back to the we roll a which the

3 . So 216

4's should occur about three times more

9 and the 10 of our game. Of the 216 ways, how many ways can ? Easily, we have three decisions, which die shows the

and which the

:

,

3 2 1 6 ways:

By identical reasoning, there are 6 ways to roll a

. But what about a

? Here our tree of options has only two stages, since once we have decided which die shows the

, the other two dice must show a

and we have nothing left

to decide (or equivalently, we only have one option, once we have placed the there are only 3 ways to roll a

:

), so

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Identically, there are 3 ways to roll a

, and 6 ways to roll a

Finally, there is only 1 way: show

to roll a

.

(all 3 dice have to

's.) So how many ways can we roll a 9? Totaling our options we obtain:

6 6 3 3 6 1

25 ways to roll a 9. Putting it in a table, together with the

similar calculations for 10.

Roll of 9

# of Ways

Hence the probability of a

9

# of Ways

6

6

6

3

3

6

3

6

6

3

1

3

25

Total

Roll of 10

Total

is 25 while the probability of a 216

27

10

is 27 . So on the 216

average, after 216 rolls of the dice, Person A would have lost 25 times, but would have won 27 times, and the rest would have been draws. So on the average, after 216 rolls of the dice, you would have won 25 times, I would have won 27 times, and the rest would have been draws. So your net outcome would be a loss of $2. Now whether you play or not is your decision: after all you could consider $2 cheap entertainment (millions of people go to Vegas). One of the common errors made in the past by mathematicians (including some of the best like Leibniz, D'Alembert and others) is the one of presuming equally feasible outcomes to an experiment without further analysis. In this course we will not have the chance to get too subtle into this subject, but remember to always be careful to set up the outcomes to the experiment before you start asking about the event that you are interested in, and try to analyze your outcomes so that they seem, as best as you can tell, equally feasible. Example 2. Suppose that a happily married couple has two children. How likely is it that they will have one of each sex? D'Alembert incorrectly analyzed this by saying there were three outcomes to the experiment: Two Boys, Two Girls and One-Of-Each. So the

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probability of One-Of-Each is 1 . Actually, if we assume that a boy being born is as 3

likely as a girl being born in any given birth3, then there are four equally feasible outcomes: BB, BG, GB and GG. Of those, 2 give us children of both sexes, so the probability is 2 1 . Again this estimate conforms to reality much better. 4

2

But to more relevant matters: Example 3. Suppose you come to take a test totally unprepared. The test consists of 10 True-False questions, each of which you will answer at random (but you will answer them all, since there is no penalty for guessing.) How likely is it that you will achieve a passing score of 70% or better? First, what is the experiment? Answering the exam. How many ways can you do this? By the second counting principle, 210 1024 (10 decisions, 2 choices for each). What is the event we are pursuing? To obtain at least 7 correct answers. Let's partition this set into: exactly 7 correct, exactly 8 correct, exactly 9 correct and exactly 10 correct. In how many ways can you answer the exam so that you have exactly 10 correct? 1 way. How about 9 correct? Build your tree, the first stage is to decide which question you are going to answer incorrectly, for this stage you have 10 choices. After you have done that, there are no options left since the question you are to answer incorrectly has to be answered that way while all the others have to be answered correctly. So there is a total of 10

10 1

ways of getting 9 correctly. What about 8?

There we have the decision of which two questions out of the 10 we are to answer incorrectly, after that we have no options left, so the answer is similar reasoning, we get

10 3

10 2

45 . Finally, by

120 ways of getting exactly 7 correct. So the number of

ways of getting at least 7 is: 1+10+45+120=176. Hence the probability of getting a passing score in the exam is 176 , which is approximately 17%. Not bad for total 1024

ignorance! Observe that The probability of an event is always a number between 0 and 1, inclusive. Example 4. This is a slight, but very important variation of Example 3. Suppose you come to take a test (totally unprepared as usual), but that the test is Multiple Choice, with ten questions and each question having three choices, only one of them correct. What is the probability that you score at least a 70% on the test? How many ways can we answer the exam? Easy, 310 59049 . How many ways can we get all correct: 1 way. Nine 3

As it happens, this is not quite correct. One realization came very early—as soon as statistical tables of birth were gathered in the 1660's: more boys are born than girls—approximately in an 18-to-17 ratio (girls are more likely to survive, so not so many need to be produced.) The other complication is that a given couple, because of the chemistry, has a certain small factor of repeating the sex of previous children (this is small).

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correct?

10 9

19 21

20 . The surprising ingredient here might be the 2, which is

coming from the 2 ways we can answer a question incorrectly. Reviewing the three factors we get:

10 9

as the number of ways of choosing the questions that we are going

to answer correctly, 19 as the number of ways of answer those questions correctly, 21 as the number of ways of answering the remaining questions incorrectly. How many ways can we get an 80%? 10 7

17 23

10 8

18 22 180 . Finally, the number of ways of getting 70% is

960 , so the probability of passing the exam is: 1 20 180 960

1161

59049

59059

.0196 ,

much smaller than in the True-False exam. Keep the model of the multiple-choice exam in mind. Many situations can be modeled using the same kind of thinking. There is no reason, for example, why there can be only one way to answer the question correctly. Hence our counting becomes a little bit more complicated, but totally manageable. Let's next play poker. Example 5. A typical deck of cards consists of 52 cards in 4 suits (spades, hearts, diamonds and clubs) and 13 denominations

(Ace, 2,3,4,5,6,7,8,9,10, Jack, Queen, King). A poker hand consists of 5 cards out of the 52. There are special hands, combinations of either denominations or suits or both, that are ranked higher than others. The rankings from best to worse are as follows: Name of the Hand Royal Flush

Description of the hand Example 10,J,Q,K,A of the same suit.

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Straight Flush

5 cards in sequence in the same suit, not royal.

4-of-a-Kind

All four cards of the same denomination.

Full House

3 cards in the same denomination, 2 in another.

Flush

5 cards in the same suit, but not in sequence.

Straight

5 cards in sequence, but not in the same suit.

3-of-a-Kind

3 cards in the same denomination, other 2 different

Two Pairs

2 cards in one denomination, 2 cards in another denomination, fifth card in yet another.

Pair

2 cards in one denomination, nothing else.

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Bust

None of the above.

What is the probability of a Royal Flush? There are at least two ways to view our experiment: one way is I am dealt 5 cards out of a deck of 52 (as in 5-card draw), another is I am dealt one card at a time until I have 5 (as in 5-card stud). Should the probabilities be different? Of course not. But what may happen is that it is easier to do a problem one way than the other. With the Royal Flush there is no hassle. With the first approach, there are

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2,598,960 equally feasible outcomes, of those 4 of them give a royal

flush (the only option we have is what suit the flush is to be in), so the probability is 4 1 0.000001539 , not very likely indeed. In the other approach we have 2598960

649740

52 51 50 49 48 311875 , ,200 equally feasible outcomes. How many of them give us a royal flush? The first card has to be any of the 20 possible cards (any 10 or J or Q or K or A), the second card has to be in the same suit, so we have only 4 choices, for the third card we have 3 choices, for the fourth one, 2 choices and for the fifth one, 1 choice, so by the second counting principle, we have a total of 20 4 3 2 1 480 ways, so 480 1 0.000001539 , which is as expected. the probability is 311875200

649740

Let's try to count Straight Flushes. In the first approach, how many decisions do we have to make? The suit of the flush (4 choices for this decision) and the type of straight (9 choices for this decision—just count them in your fingers by deciding which denomination is the lowest), so in total we have 36 options, so the probability is 36 2598960

0.000013852 (you should not hold your breath until you get one of these either). What about the second approach? The first card can be anything, but what about the second card. We are in trouble. The number of options for the second level of the tree depended on which branch of the first level you are in (for example, if the first one is a king, then the second one can only be a 9,10,J,Q,A: 5 options, while if the first one is an 8, then the second one can be a 4,5,6,7,9,10,J,Q: 8 options.) We certainly don't want to start drawing trees. As it turns out this tree has 4,320 terminal nodes! (We might let a machine do this, but certainly not by hand.) This is an important lesson. If you are in trouble counting the outcomes for an event, then by moving laterally and changing the set up maybe the trouble can be avoided. In reality, the second approach only worsens as we go down the list of hands. So we will stay with the first approach.

For 4-of-a-Kind: we have to decide the denomination (13 ways), and then the odd card (48 ways), so we have 13 48 624 options all together, giving us a probability of 0.00024.

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For a Full House: in order to build a full house we need to decide which 3-of-a-kind we are going to have (13 options), which suit those 3 cards are going to have,

4

4 , which

3

pair (12 options) and the suits for the pair,

4

6 options, so the total is

2

13 4 12 6

3,744 , so the probability is 0.001440576.

For a Flush: we have to decide the suit (4 options) and which 5 cards out of the 13 in the suit, which gives

13

1,287 options, so we have 5148 ways, but these include the 40

5

hands that are straights (36 straight flushes + 4 royal), hence the number is 5108, and the probability is 0.0019654. Note that if we had missed the subtlety of the 40 hands we had counted before, the answer wouldn't be that much different: 0.0019807, with a difference of 0.000015. For a Straight: we have to decide the type of straight (10 options), and then decide the suit for each of the cards (4 options for each), as before, we don't worry about the straight flushes, we will just subtract them. So in total we have 10*4*4*4*4*4=10,240 ways from which we subtract the 40 straight flushes, to give 10,200, so the probability is 0.0039246. For 3-of-a-Kind: choose the denomination (13 ways), the suits for these 3 cards ways), the two other denominations

12

4

4

3

66 ways), and the arbitrary suits for the two

2

new cards: 4 2 16 . Total (things are getting better).

13 4 66 16 54,912 . The probability is, thus, 0.021128

Now we will do one of the subtlest ones: Two-Pair. One trap that is commonly fallen into is as follows: choose the denomination for the first pair—we have 13 ways of doing this, then choose the suits for this pair: 4 6 ways. Then we have 12 ways of choosing 2

the second pair, and again 6 ways of choosing its suits. Now all we have to do is choose the remaining card out of a possible 48, giving us a grand total of 13 6 12 6 48 269,568 . There are two most foul errors in this discussion. The latter is the easiest one to catch. The 48 is wrong. You are not controlling full houses! Hence it should be one card out of the remaining 44, giving an adjusted count of 247,104. But one error remains that is most subtle and very common (and tempting—can you detect it?) Remember the distinction between a committee and an executive board. Can we tell the difference between the first pair and the second pair? We certainly have counted them as if we could, and that is definitely wrong. Instead we have counted every hand twice and the real count should be 123,552. Just to make you totally comfortable with this let's count them another way. Let's start by choosing the two denominations for the two pairs: 13

78 ways. Then we have 6 ways of choosing the suits for one of the pairs and

2

another 6 of choosing them for the other one, and then we have as before 44 ways of

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choosing the extra card: 78 6 6 44 123,552 . For a simple Pair, we have first to decide the denomination (13 ways), then the suits for the pair,

4

6 . Then we must have three other denominations, 12

2

3

arbitrary suits for those denominations, 4 4 4 13 6 220 64 1,098,240 . Finally, for a Bust, we must have 5 denominations,

220 , and then the

64 , which gives us the total of

13

1,287 , and a suit for each,

5

4 5 1,024 . But have included the Royal Flushes, the Straight Flushes, the Flushes and the Straights, so we have 1287 1024 4 36 5108 10200 1,302,540 . But wait, you say, I was stupid to have done this calculation since we could have found the number of bust hands by subtracting all the previous hands from the total number of hands. However, as we will see in a latter Name of the hand Number of ways Probability chapter, it is important to have Royal Flush 0.000002 4 redundancy, especially in a Straight Flush 0.000014 36 sophisticated calculation like this 4-of-a-Kind 0.000240 624 one. And the fact that, as the table Full House 0.001441 3,744 below shows us, our numbers add Flush 0.001965 5,108 0.003925 up to the correct total assures us Straight 10,200 0.021128 54,912 that we have not possibly made 3-of-a-Kind 0.047539 Two Pair 123,552 only one error. Pair Bust TOTAL

1,098,240 1,302,540 2,598,960

0.422569 0.501177 1.000000

Example 6. Dominoes. In one version of the game of dominoes there are two players, each of which draws seven tiles. For those unaware of the tiles, each tile consists of two entries (not necessarily distinct) from the numbers from 0 to 6 (inclusive). A tile with both numbers being the same is called a double. So there are 7 tiles with high mark being a 6, 6 with the high mark being a 5, etcetera. So there are T7 28 tiles. The player with the highest double starts the game by laying down that domino. If neither player has a double, the tiles are reshuffled and the game starts again. We will pose some questions about the game of dominoes.  What is the probability the game will start with the double 6? Since there are 14 tiles out of 28 being chosen, the probability that the double 6 is 27 27! 13 13!14! 14 1 . chosen is simply 28! 28 28 2 14 14!14! 

What is the probability that Tony (one of the two players) leads the double 6? Clearly, it is half of the previous answer, so it is 14 .

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What is the probability the game will start by leading the double 4? To start with a double 4, it must have happened that neither the double 6 nor the double 5 were chosen among the 14 tiles, and that the double 4 was indeed 25 25! 13 7 13!12! 14 14 13 12.96% . chosen. So the probability is 28! 28 28 27 26 54 14 14!14!



What is the probability that the tiles will have to be reshuffled? By similar reasoning to the previous, we have the probability to be given by 21 21! 14 14!7! 21!14! 0.28% . 28! 28 28!7! 14 14!14!

Example 7. We next look at another historical example. It involves two of the best mathematicians of the 17th century, both French: Fermat and Pascal (both of whom we have encountered before). Pascal was proposed the following problem: Two parties, of equal ability, will play a game until one of them has won six hands. Each of them has placed 32 coins in the pot to be collected by the winner. For some unexpected reason they have to stop when one of them has won 5 games and the other 3. How should the 64 coins be divided? You may think of this problem as that of flipping a coin until you get a total of 6 heads or 6 tails. The problem had been around for a long time, and several proposed answers had been given, including 2:1 and 5:3. Pascal corresponded with Fermat on it, and they both solved it correctly, but in very different ways.

Game 1

Game 2

Game 3

Winner

A A A A B B B B

A A B B A A B B

A B A B A B A B

A A A A A A A B

First we look at Fermat's solution: let's call the two players A and B. Then if they were to play 3 more hands (or flip the coin three more times), they would have decided for sure on the winner, for then either A would have at least won 1 or B would have won the needed 3. Of the possible, equally feasible, 8 outcomes to these 3 future games, 7 of them make A the 7 winner, while one of them makes B the winner, so the probability of A winning is so 8 the stakes should be divided 7-to-1, or 56 coins for A and 8 for B. Note that Fermat was very careful to make sure he had equally feasible outcomes to his claim. As a matter of fact, a contemporary of them, Roberval, complained that the outcomes of 3 the future would be: A, BA , BBA , or BBB . So the probability of A winning is , so the 4

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stakes should be divided 3-to-1. The problem, of course, is that Roberval had no reason to presume his outcomes to be equally feasible. And indeed they are not: A has probability 1 1 1 , BA has probability , and both BBA and BBB have probability , and the answer is 8 2 4 7 correctly . 8 A word one encounters often in the outside world is odds. In general, the odds for an event are the probability of the event occurring divided by the probability of it not occurring. Equivalently, it is how the stakes should be divided. So in the previous situation, the odds for A to win are 7-to-1. The odds for B to win are 1-to-7. If you want to bet on B, for every $1 you contribute to the pot, your opponent should contribute $7. 5-4

5-5 A=32

6-4 A=64

On to Pascal's solution. He solved the problem recursively! He argued as follows. If the players had each won 5 games, nobody would disagree with splitting the stakes 32-32. What would happen if A had won 5 and B had won only 4? Well, if they played one more hand they would either be 6-4 or 55. In one situation, A would receive all 64 coins while in the other, A would receive 32. Thus, 5-3 we should average the two, and

64 32 48 coins, and B would receive 16. So we 2 understand 5-4. How about 5-3? With one more hand, they 5-4 would be at either 6-3 or 5-4. In one situation, A gets 64, while in the other, 48 averaging we get 56. The same A=48 answer as Fermat! give A,

6-3 A=64

Example 8. The next idea we introduce in this section is fundamental: averages. Let's look at the following game (once more, not atypical of the 17th century): Since one has a chance in six of rolling a 1 with a die, one has an even chance to roll at least one 1 when one rolls 3 dice. Hence I propose the following game to you. You will roll 3 dice. If three 1's show up you win a wonderful $5, if only two 1's, you win $2, while if only one 1 is rolled, you will still win $1. If, unfortunately, on the other hand, no 1's show up you pay me only $1. Naturally, you are suspicious of my proposition, but it is much better to pin point the reasons for your suspicions. What we need is to compute your expected value when you play this game this is equivalent to what your average performance is going to be. Of course, if you are going to play this game just once, then it does not matter what you opt to do, but as a long-range strategist, you need to compute.

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Outcome

Probability 1

$

$5

216 15

The computation is just common sense. You are basically asking: suppose I played the game so many times, what would happen?

$2

In any one roll, you can win either 5, 2 or 1 dollars, or you can lose 1. Let's say you played 216 times (Why this number?) Then on the $1 216 average, three 1's would show up once, while 125 $1 two 1's would show up 15 times ( 15 3 5 ), 216 the 3 is the number of options of which two dice are going to show the two 1's while the 5 is what the other dice is going to show). How many times does one 1 show up? Choose which die shows the 1 (3 options), and then choose what the other two dice show ( 5 5 ) for a total of 75 times. Finally, no 1's will show 125 times (which is 5 5 5 216 1 15 75 ). So what are your winnings? 5 1 2 15 1 75 1 125 15 or equivalently, your expectation is 216 75

5

1 216

2

15 216

1

75 216

1

125 216

15 . 216

So on the average you will lose $15 in 216 rolls, or approximately 7¢ a roll. Naturally, you would rather not play a game when your expectation is negative, unless you will have so much fun you are willing to pay the fee. Example 9. Dominoes Revisited. A tile in Dominoes has the value of the sum of the two numbers in it. Thus the highest valued tile is the double 6 with the value of 12. What is the average value of a tile? There are 28 tiles, and the sum of all the tiles is 8 6 5 4 3 2 1 0 8 21 , so the average value of a domino piece is 6.By a famous theorem which states simply that the average of a sum is the sum of the averages, the average total of points a domino hand of 7 tiles should have 7 6 42 .

We end the section with a brief discussion of conditional probability. The next example points out the subtleties inherent in probability—but just because it is subtle you shouldn't mistrust it, instead you should take the opportunity to fine tune your brain. This is a famous problem that has amused, confused and bedazzled many people. Example 10. We are going to play a game where I am going to give you a choice of three doors. Behind one is an extra point for this class, behind the other two, nothing. You pick a door, and all-knowing I, before showing you what is behind your chosen door, open another door which has nothing behind it. I then give you a choice of either retaining your door or switching to the only other unopened door. On the average, suppose you were doing this every day there is class. What should be your standard operating procedure?

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Before we discuss the problem as stated, let us solve a simpler problem. Suppose I had just given you a door to choose out of 3 doors. Then nobody would argue that you have 1 3

of a chance of guessing the correct door. Is that agreed on? Now, let's convolute the problem by my showing you a door without a price. One easily arrived, yet wrong, conclusion is that it really does not matter if you have a standard procedure. This wrongful reasoning goes as follows. Originally each door had a 1 chance 3

of having the one point behind it. One of them has been eliminated, so now each door has only 1 of having the point behind it, hence it is really the same if you switch as if you 2

don't switch. Isn't this absolutely reasonable? I will try to convince you that it is not. But first let me get a little philosophical and point out that what probability tries to accomplish is to measure uncertainty, and that the only uncertainty in this problem is strictly from your point of view. (After all, I know everything.) Hence you have to try to use every bit of information available to you (sort of squeeze blood out of rocks). What piece of information you have not weighed in the argument in the previous paragraph? The fact that although under no circumstances I would show you what is behind your door, I had perhaps a choice of which door to show you and that I indeed chose the door I chose. More directly put, it is correct to say that at the beginning every door has 1 of a chance of being rich. But what is crucial is that the new 3

information could not have affected the probability behind your door, but instead has affected the probabilities of the other two, one going to 0 and the other to 2 . And that it 3

indeed it behooves you to switch doors all the time! To further convince you of this we will run an experiment (the ultimate test of any discussion). We are going to play the game 50 times in three different ways. In one strategy you never change doors, in another you always change doors, and in yet another you will change or not change at random. I will in return also show you a door at random every time I have a choice. In order to be absolutely fair about this experiment, I let the computer choose several random vectors of size 50. One for where the point is going to be placed, another for which door you pick, a third for which door I show you, and finally for the last strategy, for whether you change or not. I will label the 3 doors 1,2 and 3. Here is the data: Game Door With Door You Your Your Your Door I Door You Random Outcome Point Choose Outcome Show You Switch To Outcome 1 0 1 1 1 3 2 1 switch 2 1 0 1 2 2 1 3 keep

61 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Totals

3 2 2 3 2 1 3 3 1 3 1 2 3 1 2 2 2 1 2 3 2 3 2 1 1 1 2 1 2 2 1 2 1 1 2 3 2 3 3 1 3 1 1 2 1 2 3 1

2 1 1 1 1 1 1 3 1 3 2 2 1 1 1 1 3 2 2 2 1 1 2 2 2 2 2 3 1 2 3 1 3 1 1 2 1 3 3 3 2 2 3 3 2 3 1 3

0 0 0 0 0 1 0 1 1 1 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 14

1 3 3 2 3 2 2 1 3 1 3 1 2 3 3 3 1 3 1 1 3 2 3 3 3 3 3 2 3 1 2 3 2 3 3 1 3 1 2 2 1 3 2 1 3 1 2 2

3 2 2 3 2 3 3 2 2 2 1 3 3 2 2 2 2 1 3 3 2 3 1 1 1 1 1 1 2 3 1 2 1 2 2 3 2 2 1 1 3 1 1 2 1 2 3 1

1 1 1 1 1 0 1 0 0 0 1 0 1 0 1 1 1 1 0 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1 1 1 1 36

keep keep keep switch keep switch switch switch switch switch switch switch keep keep switch keep switch keep switch switch switch switch keep keep keep keep keep switch keep switch switch keep keep switch switch keep keep switch switch switch keep keep keep switch keep keep switch keep

0 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 0 1 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 19

which clearly indicates that switching is a much better strategy. We also understand why it is a much better strategy. In the always-keep strategy, you win only when you were correct to start with, that is, when you were correct in your original choice before I showed another door, which is about one third of the time—as we all

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know, and just what the data points out. The last example dealt with the fundamental notion of conditional probability. If A and B are two events, one defines the probability of A given B , in symbols P( A | B) , by P( A B ) . P( A | B ) P( B ) Namely, we restrict our set of possibilities to those in which B has occurred, and thus our denominator while in the numerator we put the situations when both events occur. Example 11. The situation is simple. You are to visit a potential customer who is known to have two children. You are speculating whether the customer has two boys. Knowing nothing else, you know that the likelihood that you are right is 14 (as worked out in Example 2 above). You arrive at the house and you see a boy playing in the backyard. You ask the customer who the boy is and the customer replies either He is my oldest child. He is one of my children. The subtleties in measuring information are reflected in the difference between the two statements. One would hardly think there is a measurable difference between them. But let us see what we can conclude from each. Let A be the event that both children are boys, B be the even that the oldest child is a boy, and C the event that one of the children is a boy. Then, if B is given, then the likelihood that A will occur is the same as the likelihood that the second child is a boy, which is simply 12 . But given C , then out of the four possibilities for two children only one is ruled out, that of two girls, so our denominator is 3, and clearly the numerator is only one, and so in summary we get the surprising fact that P( A) 14 , P( A | B ) 12 , but P( A | C ) 13 .

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