USING LOTTERIES IN TEACHING A CHANCE COURSE

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lottery sites encourage this by making this historical data available. Sometimes the argument is simply that, when a particular number has not turned up as often as would be expected, then this number is more likely to come up in the future. This is often called the "gambler's fallacy" and all too many people believe it. The fact that it is not true is the basis for many beautiful applications of chance processes called Martingales. Paulson remarks that he particularly enjoys discussing the following system. Consider, the six winning numbers in the Powerball Lottery. If they occur randomly their sum should be approximately normally distributed with mean 6(1+45)/2 = 138 and standard deviation approximately 32. Thus, sets of six numbers whose sum is more than 64 away from the mean 138, are not likely to occur as winning sets and should be avoided. It is better to pick six numbers whose sum is near 138. We leave the reader to ponder this last system and with the following advice which paraphrases the advice of our teacher Joe Doob. Play the lottery once. Then wait until there has been a drawing without a winner and play again. Then wait until there have been two drawings without a winner and play again. Continue in this manner. You will enjoy playing and not lose too much. APPENDIX THE BIRTHDAY PROBLEM Note: We wrote this when we thought the birthday problem would help settle useful computations relating to the lottery. We did not use it as much as we thought we would so we leave it as an appendix. The birthday problem asks: what is the necessary number of people in a room to make it a favorable bet that two people have the same birthday? The surprising answer is 23. To show this, we compute the probability that there is no duplication of birthdays in a room with 23 people. Since there are 365 possibilities for each person's birthday, our familiar counting principle shows that there are 36523 possible assignments of birthdays for the 23 people. How many of these assignments give all different birthdays? For this to happen, the first person can have any of 365 possible birthdays; but, for each of these, the second person must have one of 364 possible birthdays; and then the third person one of 363; etc. Hence the probability of no match is 26

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