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PLAYING IT SMART by ALAN KRIGMAN How Many Sevens Should You Expect in Six Throws of the Dice? Every craps player knows that the chance a throw of the dice will yield a seven is one out of six, which is 1/6 or 16.7 percent. Most craps players know why. The dice can land 36 ways, six of which total seven. And six out of 36 reduces to one out of six. What does this say, though, about the results you can expect in six random hurls of the hexahedrons? Some solid citizens think it says nothing, since anything can happen in six throws. Others take the probability literally. They assume, for instance, that after five other totals in a row, a seven is "due." If not with absolute certainty, then surely with more than the usual chance. The laws of probability give a single seven as most likely in six tosses, but not staggeringly more so than other possibilities. To illustrate, picture a series of six throws, labeling outcomes with the letter A when the total is seven and B when it's anything else. The probability of each A is accordingly 1/6 (16.7 percent) and that of each B the complementary 5/6 (83.3 percent). The chance of getting BBBBBB, no sevens, is 5/6 multiplied by itself six times, which is 33.5 percent. The chance you'll see a single seven in the order ABBBBB is the product of 1/6 (for the A), and 5/6 multiplied by itself five times (for the Bs), which equals 6.7 percent. But there are six ways to get a single seven: ABBBBB, BABBBB, BBABBB, BBBABB, BBBBAB, or BBBBBA. Each has a 6.7 percent probability. So the chance the series will include a single seven is six ways times 6.7 percent, or 40.2 percent. Assume you want to know the chance of exactly two sevens. This might be AABBBB, ABABBB, ABBABB, etc. For the probability of any one such combination, multiply 1/6 times 1/6 (for the As) and 5/6 by itself four times (for the Bs), then multiply these two products together. The result is somewhat over 1.3 percent. But two As and four Bs can be arranged 15 ways. The math mavens have a formula that tells them it's 15, but you can find it by "brute force," writing down all the possibilities. So, to get the net chance of two sevens in six throws, multiply 1.3 percent by 15. This works out to be 19.5 percent. Without the errors introduced by rounding off, it would actually be closer to 20.1 percent. The accompanying table shows how many ways various numbers of sevens and non-sevens can occur in six throws, along with the chance of any particular sequence and the overall probability associated with the numbers in question. An example may help you interpret the table. A sequence of four sevens and two non-sevens can be formed 15 ways. The probability of any particular sequence of these sevens and non-sevens is 0.0536 percent. The chance of six throws having these results in any order is 0.8038 percent. Probabilities associated with six random throws of the dice
The table shows that of the alternate individual sequences, six non-sevens is the most likely, at nearly 33.5 percent. Any particular set of one seven and five non-sevens is next, at about 6.7 percent. Ignoring order of occurrence, one seven in six tries has the greatest prospects at close to 40.2 percent. The theoretical probability, in this instance one out of six, will always be the likeliest to occur. However, as the figures indicate, this probability may be far from overwhelming; it's only moderately higher than those of zero or two sevens. The same exercise for 36 throws would show the most likely result to be six sevens and 30 non-sevens at 17.59 percent. Compare this with 17.02 and 15.07 percent for five and seven sevens, respectively. The six sevens are more probable than the other alternatives, but only marginally. Oh, by the way, if you try to work out the combinations of six sevens and 30 non-sevens by brute force, plan on spending a while. There are 1,947,792 of 'em. Calling to mind this verse by the poet, Sumner A Ingmark: Intuition's a good tool but use it with discretion, (c) 2005, ICON/Information Concepts Inc |
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