21-110: Problem Solving in Recreational Mathematics

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1 -0: Problem Solving in Recreational Mathematics Homework assignment solutions Problem. An urn contains five red balls and three yellow balls. Two balls are drawn from the urn at random, without replacement. (a) In this scenario, what is the experiment? What is the sample space? (b) What is the probability that the first ball drawn is red? (c) What is the probability that at least one of the two balls drawn is red? (d) What is the (conditional) probability that the second ball drawn is red, given that the first ball drawn is red? Solution. (a) The experiment is the drawing of two balls from the urn without replacement. The sample space is the set of possible outcomes, of which there are four: drawing two red balls; drawing two yellow balls; drawing a red ball first, and then a yellow ball; and drawing a yellow ball first, and then a red ball. One way to denote the sample space is in set notation, abbreviating the colors red and yellow: sample space = {RR, YY, RY, YR}. Note that these four outcomes are not equally likely. We can also represent the experiment and the possible outcomes in a probability tree diagram, as shown below. Note in particular the probabilities given for the second ball. For example, if the first ball is red, then four out of the remaining seven balls are red, so the probability that the second ball is red is / (and the probability that it is yellow is /). On the other hand, if the first ball is yellow, then five out of the remaining seven balls are red, so the probability that the second ball is red is / (and the probability that it is yellow is /). The probabilities of each of the four outcomes can be computed by multiplying the probabilities along the branches leading to the outcomes. First ball Second ball Outcome Probability R R RR = Y RY = Y R YR = Y YY = (b) Before the first draw, five out of the eight balls in the urn are red. Each ball is equally likely to be drawn. So the probability that the first ball drawn is red is /. (c) The event that at least one of the two balls is red contains three outcomes: RR, RY, and YR. Since we know the probabilities of all of these outcomes, we can find the probability of this event by adding the probabilities of the individual outcomes. So the probability that at least one of the two balls is red is / + / + / = /. Alternatively, we can observe that the event that at least one of the two balls is red is the complement of the event that both balls are yellow. The probability that both balls are yellow is /, so the probability that it is not the case that both balls are yellow (i.e., the probability that at least one ball is red) is / = /. Page

3 of the die there are, or,, possible outcomes (because the order of the rolls is important). These outcomes are all equally likely. How many outcomes are in the event A? The outcomes in the event A are those that look like,,,,,?, where the question mark can be replaced by any number from through. So the cardinality of A is. Likewise, the outcomes in the event B are those that look like?,,,,,, where the question mark can be replaced with any number from through. So the cardinality of B is also. How many outcomes are in both A and B? In other words, what is the cardinality of A B? There is only one such outcome, namely,,,,,,. So A B =. By the principle of inclusion exclusion, we have A B = A + B A B = + =. Alternatively, we could have found A B by listing all of the elements of A B explicitly: A B = {,,,,,;,,,,,;,,,,,;,,,,,;,,,,,;,,,,,;,,,,,;,,,,,;,,,,,;,,,,,;,,,,, }. So, using S to denote the sample space, we have Pr(A B) = A B S =,. Therefore, the probability that a is rolled at least five times in a row is /,, or about two hundredths of one percent. Problem. (From The Colossal Book of Short Puzzles and Problems by Martin Gardner.) A secretary types four letters to four people and addresses the four envelopes. If she inserts the letters at random, each in a different envelope, what is the probability that exactly three letters will go into the right envelopes? Solution. If three letters go into the right envelopes, then the fourth letter must go into the fourth envelope, which must be the right one! So the event that exactly three letters go into the right envelopes contains zero outcomes, which means that the probability of this event is zero. Problem. (From The Colossal Book of Short Puzzles and Problems by Martin Gardner.) Bill, a student in mathematics, and his friend John, an English major, usually spun a coin on the bar to see who would pay for each round of beer. One evening Bill said: Since I ve won the last three spins, let me give you a break on the next one. You spin two pennies and I ll spin one. If you have more heads than I have, you win. If you don t, I win. Gee, thanks, said John. On previous rounds, when one coin was spun, John s probability of winning was, of course, /. What are his chances under the new arrangement? Solution. Let s make a table of the possible outcomes. Remember, John wins if he gets more heads than Bill. If there is a tie, Bill wins. Bill John Winner H H H John H H T Bill H T H Bill H T T Bill T H H John T H T John T T H John T T T Bill Page

4 The eight outcomes in the table above are equally likely. John wins in four of the eight cases. So the probability that John wins under the new arrangement is /, or /, the same as it was before. [Gardner points out that John s probability of winning remains / whenever he has one more coin than Bill. For example, if John has coins and Bill has 0, John s probability of winning is still /.] Problem. has three fair six-sided dice, colored green, red, and blue, but these dice are not numbered in the standard way. Instead, they are numbered like this: Green: Red: Blue: In other words, the green die has three faces labeled and three labeled ; the red die has five faces labeled and one labeled ; and the blue die has five faces labeled and one labeled. proposes to that they play a little game. Here s how it will work, explains. In each round, we will each choose one die and discard the third one. Then we will simultaneously roll our chosen dice. Whoever rolls the higher number will win a dollar from the other player. (Note that there can never be a tie.) And since I m such a nice person, I will let you choose your die first, before I choose mine. (a) Suppose chooses the red die and chooses the green die. Show that the probability that wins is greater than /. (b) After using the red die for a while and losing more often than not to s green die, begins to suspect that the odds are not in his favor. So he chooses the blue die instead, and chooses the red die. Show that the probability that wins is still greater than /. (c) continues to lose when he plays the blue die against s red die. He reasons, The green die is better than the red die, and the red die is better than the blue die. So, clearly, I should choose the green die, because it must be the best of all. He takes the green die, and takes the blue die. Show that the probability that wins is still greater than /. (d) s dice are an example of nontransitive dice. How is this dice game similar to the game of Rock, Paper, Scissors? Solution. (a) We draw a probability tree diagram showing s and s rolls when has the red die and has the green die. Winner Probability = = = = To find the overall probability that wins, we add the probabilities of each outcome in which wins. So the probability that wins is /+/+/ = /, which is greater than / (it is about %). Page

5 (b) When has the blue die and has the red die, the probability tree diagram looks like the following. Winner Probability = = = = wins with probability /, which is greater than / (it is about 9%). (c) When has the green die and has the blue die, the probability tree diagram looks like the following. Winner Probability = = = = wins with probability / + / + / = /, which is greater than / (it is about %). (d) In the game of Rock, Paper, Scissors, each object wins over one of the other two objects but loses to the other one: rock beats scissors but loses to paper, paper beats rock but loses to scissors, and scissors beats paper but loses to rock. In s dice game, each die wins over one of the other two dice (with probability greater than /) but loses to the other one (with probability greater than /): green wins over red but loses to blue, red wins over blue but loses to green, and blue wins over green but loses to red. [True story: Warren Buffett once challenged Bill Gates to play this game, offering to let Gates choose his die first. Gates was intrigued, examined the dice, figured out the trick, and then demanded that Buffett choose first.] Problem. Ten mathematicians are captured by pirates. The pirate captain tells them, I have an unusual custom for dealing with prisoners on my ship. Tomorrow at dawn I will put all ten of you in a line, blindfolded and facing the same direction. I will assign each of you either a black hat or a white hat, depending on the outcome of the flip of a fair coin. (Since there are ten of you, I will flip the coin ten times, once for each of you.) After you have each been given a hat, I will remove your blindfolds. Each of you will then be able to see the colors of the hats of all the people standing in front of you in the line, but you will not be able to see your own hat, nor will you be able to see the hats of the people behind you. Page

6 Then I will begin my questioning. I will start with the person at the back of the line (who can see all nine of the other hats). I will ask him what color his hat is. He must announce his guess, either black or white, in a flat, monotone voice, but loudly enough for all the rest of you to hear. If his guess is correct, I will allow him to go free. If his guess is incorrect, however, he will be killed. Next, I will move to the second-to-last person in line (who can see eight other hats), and I will repeat the procedure, moving up the line, until I have dealt with each of you in turn. By the way, once the hats have been distributed, you will not be allowed to pass any information among yourselves except your guesses. Well, I suppose it s getting pretty late, and we have a big day tomorrow. Sleep well, and I ll see you in the morning! That night the mathematicians get together and decide that they must devise a strategy to maximize the expected number of them who will go free. What strategy should they adopt? Solution. The surprising answer is that nine of the ten mathematicians can be saved with certainty! Let s number the mathematicians through 0, where mathematician is at the front of the line (and can therefore see no hats) and mathematician 0 is at the back of the line (and can see all nine of the other hats). The key to the best strategy is the following: Mathematician 0, who is the first to guess, counts the number of white hats he sees in front of him. If he sees an even number of white hats, he guesses white ; if he sees an odd number of white hats, he guesses black. Since all of the remaining mathematicians can hear his guess, they will know whether there is an even number or an odd number of white hats among them. When it is time for mathematician 9 to guess, he can see the number of white hats in front of him, and he knows (from mathematician 0 s guess) whether there is an even number or an odd number of white hats among mathematicians through 9. This information is enough to allow mathematician 9 to deduce the color of his hat with certainty. For instance, if he sees an even number of white hats in front of him and he knows that mathematician 0 also saw an even number of white hats, then his own hat must be black. On the other hand, if he sees an even number of white hats in front of him and mathematician 0 saw an odd number of white hats, then his own hat must be white. So mathematician 9 can be assured of guessing his own hat color correctly. Mathematician can use similar reasoning to deduce the color of his hat. When it is his turn to guess, he can see the seven hats in front of him, he knows whether mathematician 0 saw an even number or an odd number of white hats, and he also knows the color of mathematician 9 s hat (from mathematician 9 s guess, which was correct). This is enough information to allow him to determine his own hat color, so he is also assured of guessing his own hat color correctly. In general, when it is time for mathematician in the middle of the line to guess, the information available to him includes the colors of all the hats in front of him (which he can see directly); whether mathematician 0 saw an even number or an odd number of white hats (from mathematician 0 s guess); and the colors of all the hats behind him, except the hat of mathematician 0 (from the guesses of the other mathematicians). From this information he can work out what color his own hat must be, and so he will be able to guess correctly (simultaneously passing the information about the color of his own hat to the people in front of him, who will need it to determine the colors of their hats). Therefore, if the mathematicians follow this strategy, mathematicians through 9 will be saved with certainty. Unfortunately, mathematician 0 will only guess correctly half the time (by sheer luck), but, as he has no possible source of information about the color of his own hat, this is the best that can be done. In other words, with probability / it will be the case that nine mathematicians go free, and with probability / it will be the case that all ten mathematicians go free. The expected number of survivors is therefore (9) + (0) = 9., which is really quite impressive. Page

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