I'll Give You a Definite Maybe
A Handbook on Probability, Statistics, and Excel

[This e-text has been prepared by Ian Johnston of Malaspina University-College, Nanaimo, for the use of Liberal Studies students. The text is in the public domain and may be used by anyone, in whole or in part, without permission and without charge, provided the source is acknowledged, released April 2000.  This document was prepared in April 2000, and minor editorial and formatting changes were made in November 2004]

For comments, questions, corrections, improvements, and what not contact Ian Johnston

Section One: Basic Concepts of Probability

[This module was prepared, in part, using material written by of Dr. Rod Church of Malaspina University-College]

A. Introductory Note

The term Probability, as the name suggests, refers to a calculation, a mathematical estimate concerning the possibility of a particular outcome in a situation where there are a number of different possible outcomes. If, for example, we know that in a certain situation (e.g., a roll of the dice) there is a determined number of possible outcomes (e.g., six possible faces of the die), we can calculate the probability of a particular outcome or of a particular sequence of outcomes if the event is repeated.

For example, with a normal die in a fair roll, there is a 1 in 6 chance of any particular face of the die showing on the top (since there are six surfaces on the cube); in a fair toss of a normal coin there is a 1 in 2 chance of a head appearing on top, a 1 in 2 chance of a tail appearing on top, and a 2 in 2 chance of a head or a tail appearing on top.

It is important to stress at the outset that probability is a theoretical calculation of the likelihood of a particular outcome. Thus, it is based, not upon observation of actual events (e.g., listing the results from several tosses of the coin) but on calculations which are based on certain assumptions (e.g., that the coin or die toss is fair, that there are no hidden factors, like cheating, influencing the outcome).

Because the probability of an outcome is a theoretical calculation, knowing the probability does not mean that we can always predict a particular outcome with certainty (please understand this very important point). Often the result of a particular event or series of events will not match the theoretical probabilities. For example, in a sequence of ten coin tosses, theory tells us that the most probable result will be 5 outcomes with heads on top and 5 outcomes with tails on top. But that calculation does not mean that, if we observe an actual 10-toss sequence, we will always get that most probable result.

Checking probabilities (that is, the theoretical calculations) against the observed outcomes of a real event is, as we shall see, an essential part of the study of probability and statistics (and a central exercise in this module). But always keep in mind this opening point: the probability of an event is a theoretical calculation; the observed frequency with which an event actually occurs is something different: the first is based upon mathematical analysis of an event (independent of observation); the second is based upon an actual experiment.

Probability theory has a number of extremely important uses. A simple but powerful application is its ability to assist us in establishing the mathematical likelihood of a particular outcome and thus to advise ourselves of the prudence of a certain action (e.g., betting). For instance, to take a very simple example, in a fair toss of a coin, theory indicates that the probability of the coin landing with heads on top is 50 percent, with tails on top 50 percent, and with either heads of tails on top 100 percent. In a bet on the result of a coin toss, therefore, one would be unwise to wager, say, ten dollars that a head would appear against your opponent's bet of two dollars that a tail would appear.

In the course of this module, we will become more familiar with a number of important applications of probability theory.

B. Expressing Probability Mathematically

We can express the probability of an event mathematically in a number of ways, but the most common is as a decimal between 1 and 0. Thus, a probability of 1 (p = 1) is certainty that the event will occur. A probability of 0 (p = 0) is certainty that event will not occur. Thus, all probabilities for an event are expressed as a decimal between 0 and 1.

The probability that a single fair coin toss will produce a head on top is expressed as follows: p = .5. This we might also express as meaning that the probability is 5 out of 10 or 50 percent. The decimal notation is the most common and will be standard throughout the rest of this workbook.(1)

In the same way, if p = .75, that means that the outcome will theoretically occur 75 times out of 100; if p = .234, that means that the outcome will theoretically occur 234 times out of 1000; if p = .01345, that means the outcome will theoretically occur in 1345 cases out of 100,000.

It will be clear that any value for p greater than .5 means that the result is theoretically more likely to occur than not; a value for p less than .5 means that the result is more likely not to occur than it is to occur. The smaller the value of p, the less likely that outcome.

C. Self-Test on Probability Notation

1. Translate the following statements of probability into the decimal notation discussed above: 5 percent, even odds, 16 cases out of 20, 3 chances in 1000.

2. Express the following values of p as a numerical expression stating the number of probable outcomes out of the total number of possible outcomes (e.g., p = .45 is a probability of 45 out of 100): .6; .04; .023; .001; .01; .0345; .0006.

Please make sure you are thoroughly comfortable with this notation before moving on. Note that you can check the answers to the above questions at the end of this chapter.

One point to note. In the decimal notation of probability, the sum total of all the probabilities for all possible outcomes of a single event should add up to 1 (certainty). For example, in a single coin toss, the probability of the outcome heads is .5, and the probability of the outcomes tails is .5. These are the only two possible outcomes, and they total 1 (certainty). In the roll of a six-sided die, the probability for any particular number to come up is 1 in 6 or .1667. There are only six possibilities. The total of all the probabilities is therefore .1667 x 6, or 1 (certainty).

D. Probability Theory and Observation

One of the major applications of probability is to check the theoretical possibility of a particular outcome (calculated mathematically) with the observed result of an event. If we find that the observed frequency of a particular occurrence departs widely from theoretical probabilities (i.e., the result is occurring more frequently than we would expect from our calculations), sometimes we have good reason to investigate the event more carefully.

For example, in a fair toss of a coin, we theoretically calculate p as .5 for a head to appear on any single toss. But what is the probability that someone will throw the coin four times in a row and have the outcome heads each time? We can calculate the probability by multiplying the probabilities together. Since there are four tosses and the probability of a head is .5 on each of these, the probability of a sequence of four heads is as follows:

p = .5 x .5 x .5 x .5 = .0625

In other words, in a sequence of four consecutive coin tosses, probability theory indicates that in 625 cases out of 10,000 (or 1 outcome in 16) four heads will come up in a row (2)Thus, if you were making a bet in a number of different trials of four consecutive coin tosses that four heads would appear in a row, you could theoretically expect to win 1 out of every 16 trials and lose 15 out of 16 trials.

Now, this figure of 1 in 16 (p = .0625) does not claim that in every 16 trials of a four-coin-toss sequence, one result will always be four heads. The theory states that, on average, in a sequence of four-toss trials, the most likely outcome is that there will be four heads appearing in 1 out of every 16 trials. But in any particular series of four-toss trials, it is quite possible for more than sixteen four-toss trials to happen before the first four-head result appears or for a four-head result to occur before sixteen trials are completed or more than once in one sixteen four-toss sequence. As we shall see, the more times one repeats the four-toss sequence, the closer the observed result will come to the theoretically calculated result.

Thus, given that one four-head sequence every sixteen throws is the most likely (but not the inevitable) outcome of a series of four-toss sequences, if you were betting on the outcome of many four-toss sequences (that there would be four consecutive heads appearing), for the game to be truly fair, you should put up \$1 against your opponent's \$15 for each four-coin sequence (i.e., if four heads appear you win \$15; if they do not your opponent wins \$1). With a great many four-toss trials, you and your opponent should come out more or less equal.

Hence, if you are betting \$1 that four heads will appear in a four-toss sequence of coins and if your opponent is unwilling to put up \$15, you should not undertake the bet (the probabilities are against you). If she is willing to put up more than \$15, then you should take the bet (the probabilities are in your favour).

E. An Introductory Note about Significance

When the probability of an event (theoretically calculated) is very low and yet the event occurs, sometimes we are entitled to be suspicious or alarmed at the result. Such moments are an invitation to investigate the event to see if there are hidden factors at work (e.g., loaded dice, cheating, trick coins, some hitherto unknown cause). When do such moments occur? At what point is the probability so unlikely (i.e., so low) that I should investigate further?

Well, let us first, as an example, consider the probability of someone's tossing a coin with ten consecutive heads appearing. Using the same way of calculating the probability as we did above (multiplying together the probabilities for a single event), we can calculate the theoretical probability of ten consecutive heads appearing as follows:

p = .5 x .5 x .5 x .5 x .5 x .5 x .5 x .5 x .5 x .5 = .000977

This rounded off result indicates that the probability for ten consecutive heads in a series of ten fair coin tosses is 977 out of 1,000,000. Obviously the probability here is much smaller than for the four-head sequence (p = .0625), and we would be right to be much more surprised at such a result. But do we have the right to be suspicious? In other words, how improbable does an event have to get before we begin to think that there is something strange going on?

Scientists and social scientists have set conventional standards for how much an observed result has to differ from the theoretical expectation in order for the result to be considered significant (i.e., worthy of notice or of further investigation, because it is not likely to have occurred by chance). That level is p = .05 (or 5 results in 100). If there is more than a probability of .05 (i.e., if p is greater than .05) that an event could have occurred by chance, scientists do not usually pay much attention to it. If p is less than .05, however, then the result is considered significant and warrants analysis, because something other than chance may be affecting the outcome.

Thus, for example, someone's tossing a coin and getting four consecutive heads (where p = .0625) would not prompt any enquiry; the probability is low, but not so low that we need to start wondering about the fairness of the toss (remember the cut off point is p = .05). However, someone's tossing a coin and getting ten heads in a row (p = .000977 or, rounded off, .001) is significant. The probability is so low (1 chance in 1000) that we have some grounds for thinking the result might not be produced by chance and that it is time to check the coin.

The justification for this convention of a probability of .05 as the dividing line between significant and non-significant results is not mathematical but methodological and practical. Given the constraints, economic for example, upon investigation, it has been found to pay, in terms of success in explaining the world, to concentrate research in areas where such a degree of significance is obtainable. In some tests, as we shall see, where the procedure requires a much higher level of certainty, the cut-off line is sometimes reduced to .01 (or 1 chance in 100). In general, however, the figure of .05 probability is the accepted level below which results become significant and above which we can still ascribe the results to chance.

Note the very important point that a low probability (e.g., p = .001) does not indicate impossibility. It may well be that the ten-head coin-toss sequence, although very very improbable, is fair (i.e., produced by chance). The very low figure for p is a warning that something may be (and probably is) interfering with the result, but the figure does not prove such interference (3).   These basic principles indicate one important aspect of probability, its use as a diagnostic tool, a means of recognizing that in a natural event something may be interfering with the results (e.g., cheating or some hidden factor). Significant discrepancies between theoretical calculations of probable outcomes and the observed results of actual outcomes are often the first clue that something needs investigating. And from this initial observation, the investigators often discover something they had not known about before (e.g., a new disease, like Legionnaire's Disease or AIDS, or a new natural phenomenon).

F. The Fundamental Basis of Probability Theory

The basis upon which the mathematical treatment of probability rests is as follows: we assume that the phenomenon we wish to investigate can be described in terms of a given number of different possible but equally probable outcomes. For instance, in a simple coin toss there are two equally probable outcomes, heads or tails. In one roll of a die, there are six equally probable outcomes, represented by the six sides of the die, each with a number of dots from 1 to 6. In drawing a card from a full and well shuffled normal deck with no jokers, we know that there are four equally probable outcomes for the suit of the drawn card (hearts, diamonds, spades, or clubs), there are two possible outcomes for the colour of the card (red or black), and there are thirteen possible outcomes for the value of the card. Naturally we are assuming that in all these trials the procedure is fair (i.e., no cheating).

Multiple outcomes (e.g., so many heads in consecutive coin tosses or so many sixes in consecutive rolls of a single dice) can be calculated from the probability of a single event (as we did above by multiplying the frequency of repeated events by the probability of the single outcome in one event). Events which cannot be described in terms of equally probable outcomes are beyond the scope of probability theory discussed here.

The probability of any specific outcome is given by the number which is equal to the number of possibilities which involve that outcome divided by the total number of possibilities. For example, in a simple coin toss, there is only one possibility that a head will appear, and the total number of possible outcomes is two. Thus, the probability that a head will appear is 1 divided by 2 or .5.

In a roll of a single die, there is one chance that a six will appear, and there are six possible outcomes. Therefore the probability of a six appearing on a single roll of a normal die is 1 divided by 6 (p = .1667). If I draw a card from a normal full deck, there is only one chance that a heart will be on the card, and there are four possibilities for different suits. Therefore the probability of drawing a heart from a full deck is 1 divided by 4 (p = .25).

What is the probability that on a single roll of a die, the result will be an even number? Well, there are three possible even outcomes (the numbers 2, 4, and 6) and there are 6 possible outcomes (1, 2, 3, 4, 5, 6). Thus, the probability that on a single roll of a die the result will be an even number of dots is 3 divided by 6: p = .5.

More complicated outcomes (e.g., four heads in a row in consecutive coin tosses, three sixes in a row in consecutive rolls of the die, four number 12 results in four consecutive spins of the roulette wheel, and so on) are more complicated, but we can work out the possible outcomes for such events and calculate the probabilities involved (as we did above for sequences of four and ten coin tosses).

G. Self-Test on Simple Probabilities

Try the following problems. To inspect the answers, see the section at the end of this chapter.

1. What is the probability (expressed as a decimal) that from a well-shuffled normal and full deck of cards you will draw a king, a king of hearts, a king or a queen?

2. A fair six-sided die is to be thrown once. What is the probability (expressed as a decimal) of getting an odd number, a number greater than 3, a 1 or a 2?

3. Try your hand at this multiple probability problem. A box contains four apples, two of which are rotten. Two apples are selected at random one after the other. What is the probability that the two apples selected are both rotten? What is the probability that only one of the two selected is rotten? One suggested method: consider all the different possible outcomes of the two picks, calculate the probability of each of these outcomes, and then inspect the outcomes which answer the questions asked.

H. The Binomial Distribution

Many events have only two possible results: success or failure. In tossing a coin, there are only two possible results—heads and tails—each with a probability of .5. Success in a coin toss is calling the correct result. In answering multiple choice tests where there are four possible answers for each question, there are again only two possible results for each question, a correct answer or an incorrect one. If we are trusting entirely to chance in answering such a test (i.e., guessing), we have a one in four chance of getting the answer for any question correct (i.e., on each question for success p = .25, for failure p = .75).Sorting out the probabilities for a particular outcome requires us to figure out all the possible outcomes and to calculate the distribution for all those possible outcomes. And these calculations will vary depending on the number of possible outcomes. Thus, there are many possible probabilities for different events (e.g., the distribution of outcomes—heads or tails—for a four coin-toss sequence will be different from the distribution of outcomes—heads or tails—for a ten coin-toss sequence; the distribution for the results of a repeated event with a .5 probability will be different from a similar sequence with a .1667 probability, and so on).

Constructing such a mathematical calculation for a particular series of outcomes is called calculating the binomial distribution. Constructing a binomial distribution for a particular sequence of events is not very difficult, but it is rather time consuming. Fortunately we have tables of binomial distributions so that we do not have to go through the calculations here.

However, in this section we are concerned with being above to read and use a simple binomial distribution table, so that we can judge the probabilities of simple events with only two possible outcomes (success or failure). The simple table below provides an example of two binomial distributions. An explanation of the table follows immediately after it.

 Table 1: Selected Binomial Distributions p = .5 n = 10 n = 20 r ind com r ind com 0 .001 1- 0 0+ 1- 1 .010 .999 1 0+ 1- 2 .044 .989 2 0+ 1- 3 .117 .945 3 .001 1- 4 .205 .828 4 .005 .999 5 .246 .623 5 .015 .994 6 .205 .377 6 .037 .979 7 .117 .172 7 .074 .942 8 .044 .055 8 .120 .868 9 .010 .011 9 .160 .748 10 .001 .001 10 .176 .588 11 .160 .412 12 .120 .252 13 .074 .132 14 .037 .058 15 .015 .021 16 .005 .006 17 .001 .001 18 0+ 0+ 19 0+ 0+ 20 0+ 0+ Note that the entry 1- means a probability larger than .9995 but less than 1 (about as certain as one can be); the entry 0+ means a probability less than .0005 but greater than 0 (extremely unlikely).

This table indicates two different binomial distributions. The one on the left side indicates the probabilities of success in a series of 10 coin tosses; the one on the right side indicates the probabilities of success in a series of 20 coin tosses; this distinction is indicated by the label n = 10 on the left and n = 20 on the right. For both tables the probability of success in any one toss is .5 (hence heading p = .5 below the title).

The first column on each side has the heading r. This indicates the number of successes in the sequence. So, for example, on the left side, the values go from 0 to 10, indicating the possible outcomes of 0 heads in 10 tosses of the coin to 10 consecutive heads in 10 tosses. On the right side, the r values go from 0 to 20, indicating the possible outcomes from 0 heads in 20 tosses to 20 heads in 20 tosses.

The ind (= individual) column in the middle of each side of the table gives the probability of one's achieving that exact r value in a full sequence of tosses. So on the left side, for example, the probability of getting exactly 5 heads in a sequence of 10 tosses is .246 (the ind value on the same line as the r value of 5). In the right side of the table, the probability of getting exactly 5 heads in a sequence of 20 tosses .015.

The cum (= cumulative) column gives the probability of r or more successes in the total number of trials (or at least r successes). So, for example, on the left side of the table, the chances of getting 6 or more (or at least 6) heads in 10 consecutive tosses of a coin is .377; in the right side of the table, the probability of getting 5 or more (or at least 5) heads in 20 tosses of the coin is .994.

Look on the right side of the table at the r value 4. As you would expect, the probability of getting exactly four heads in a 20-toss sequence (the ind value) is very low (.005), but the chances of getting at least 4 heads in a 20-toss sequence (the cum value) is very high (.999, close to certainty).

I. Self-Test on a Binomial Table

Use Table 1 (on the previous page) to determine the answers to the following questions:

1. What is the probability of getting exactly six heads in a random toss of ten coins? What is the probability of getting exactly six tails?

2. What is the probability of getting at least 70 percent heads in ten consecutive tosses of a coin (i.e., of getting at least 7 heads in 10 consecutive tosses)? What is the probability of getting at least 70 percent success in a sequence of 20 tosses (i.e., 14 or more heads in 20 consecutive tosses)? In which coin tossing sequence is it easier for one to score a 70 percent or higher success rate?

Use Table 2 (given below) to deal with the following problems.

3. A friend of yours says that she is willing to bet that she can roll six or more sixes in 24 consecutive rolls of a single die (i.e., a six-sided crap die). She then puts \$20 on the table as her stake. Using the binomial distribution table given immediately below (a distribution for p = .1667), find out how much should you put on the table to make sure the bet is truly fair? You really like this friend, and you do not wish to take advantage of her. But you do not wish to be a sucker. As a student in Liberal Studies you are very anxious to do the mathematically right thing. So how much should you bet?

4. The friend mentioned in the above question, impressed with your honesty, agrees to accompany you on a trip to Reno. On your first evening there, at the dice table, she gives you this advice: "Look, the probability that a six will be thrown on one roll of one die is 1/6 (p = .1667). Therefore, the probability that we will get at least one six in three rolls of the die is 3 x 1/6 or ½ (p = .5). Since we have an even chance, let's make the bet with all your student loan money." You feel you are on a lucky streak, and the thought of even odds in the casino is tempting. But would you accept this reasoning? If not, explain why (please confine your reasons to the mathematical variety). Note that a good way to proceed here is to list all the possible outcomes, to calculate the probability of each outcome, and then to inspect and add up the ones which will bring you success in the bet

 Table 2: Binomial Distribution for a Six-Sided Die p = .1667 n = 24 r ind cum 0 .013 1- 1 .060 .987 2 .139 .927 3 .204 .788 4 .214 .584 5 .171 .371 6 .108 .200 7 .056 .091 8 .024 .035 9 .008 .012 10 .003 .003 11 .001 .001 The other ind and cum values from r =12 to r = 24 are 0

J. Some Observations on Binomial Distributions

The binomial distribution, as illustrated in Tables 1 and 2 above, is appropriate only when the events in the sequence are independent of each other (i.e., when the outcome of one event does not affect the outcome of another). For example, in any coin toss the probability of a head is always .5; it is not affected by the result of any outcome before it. In the roll of a single die, the probability of a 6 showing as the outcome is always the same (p = .1667), no matter what the results of previous rolls might have been. In other words in sequences of coin tosses or die rolls, the outcome of any single event is independent of the outcome of any other event (4)

This independence does not always apply in probability situations. For example, when one draws a single card from a full, well-shuffled deck, the probability of the outcome producing a red card is .5, because 26 of the 52 cards are red (26 divided by 52 equals .5). However, if I do draw a red card and do not return it to the deck, then the probability of a second draw producing a second red card is no longer .5. Since one red card has been removed from the deck on the first draw, there are now only 51 cards in the deck, 26 black cards and 25 red ones. Therefore, the probability of drawing a red card on the second draw is now 25/51 (p = .4902); whereas, the chance of drawing a black card on the second draw with one red card removed is now 26/51 (p = .5098).

In this example, the probabilities in the second draw have been affected by the result of the first draw (one card less in the deck, one red card less in the deck). What this means is that the probability of a red card in the second draw is not independent of the first draw. Thus, the sort of binomial distribution which we have been considering in the coin toss or roll of the die (where the probability remains the same in each toss or roll) would not be appropriate.

In any card game where the deck is not full and shuffled before each event, the probabilities are constantly changing as cards are played and removed from the deck. This is the case, most famously, with blackjack, where after every hand the played cards are normally collected and placed on the bottom and a new hand dealt from the remaining cards. A careful player who keeps close track of the cards which have been played and not returned to the deck and who understands probabilities very well can calculate when the probabilities of a favourable outcome shift in her favour (i.e., when p for success becomes greater than .5). This is the basis for a number of well-known and successful legal systems for beating the casino dealer, since this is the only casino game in which the probabilities occasionally favour the player.

Here, for example, is a simple illustration. Suppose in playing blackjack at the casino (in a game between you and the dealer only), you keep careful track of all the cards which have been played and put on the bottom of the deck. Suppose, as a result of this careful observation, you know that there are only five cards left unplayed and, further, that these five are 3 eights and 2 sevens. In such a situation you pawn everything you have, sell your car, bet everything, and stand pat with the two cards dealt to you. Since in routine casino blackjack the dealer cannot stand pat with 16 or under (and you know that from the remaining cards he must have 16, 15, or 14), it is certain that the dealer will have to draw and that his hand will be over 21. In other words, the probability for your success here is 1 (certainty).

That is a very simple example. Sophisticated mathematicians have developed systems of keeping track of the shifting probabilities in blackjack, betting low when the probability is less than .5 and high when the probability for success shifts to above .5. Over time, given the fairness of the game, such players will win (and some have won enormous amounts). Casinos routinely cope with such gamblers, when they recognize them, by shuffling the deck before every hand, thus restoring the favourable probabilities to the dealer in every hand. At this point, the sophisticated gambler with a system moves on down the strip to another casino.

K. Bernoulli's Theorem

Let's return to Question 2 in the Self-Test in Section I for a moment, in which we were asked to figure out from the binomial distribution table whether it was easier to get a 70 percent success rate in a 10-toss sequence of coin flips or in a 20-toss sequence. We noted then that getting the 70 percent success rate was much more probable in the 10-toss sequence than in the 20-toss sequence (a probability of .172 for 7 or more in a 10-toss sequence compared with a probability of .058 in a 20-toss sequence).

If we were to expand this study of a 70 percent success rate with various coin-toss sequences, we would find that the greater the number of tosses in the sequence, the harder it is to obtain a 70 percent success rate. Or alternatively put, the greater the number of coin tosses, the less probable become outcomes which deviate from the mid point.

When you think about this point, it is obvious enough. In a normal sequence of coin tosses we expect from probability theory that half the outcomes will be heads and half will be tails. We would not be surprised, however, if in a relatively small number of coin tosses the number of heads exceeded the number of tails (e.g., in a four-toss sequence, an outcome of 3 heads and 1 tails would not amaze us). However, in a 1000-toss sequence, a result of 750 heads and 250 tails should really surprise us, because in such a long sequence of tosses we would expect that chance will bring the total number of heads closer to the expected 50 percent.

This principle is known as Bernoulli's Theorem, which states that the observed frequencies (i.e., the observed results) should approximate more closely the theoretical probabilities as the number of trials (or events) increases. In other words, as we increase the number of observed trials of an event, the recorded results of those events come closer and closer to the theoretical calculations (the mathematical probabilities). In a small number of trials, the observed results may not be too close to what we expect from our calculations; but as we increase the number of trials, the results should get closer and closer to what we expect to happen by chance.

The theorem is a very important concept to grasp because it stresses that a particular percentage success rate (e.g., 70 percent success in coin tosses) may or may not be significant; what is crucial is the number of trials or repetitions of the event. The greater the number of trials the more significant any departures from the theoretical probabilities become.

Here is an important and frequent application of Bernoulli's Theorem. Suppose your Uncle Joe claims to have powers of ESP (extra sensory perception) and backs up his claim by a statistical "proof" of a 70 percent success rate in "influencing" through special psychic powers the outcome of a coin toss. How should you treat this claim? Is Uncle Joe a liar, a weirdo, a bore, a psychic, or what? Well, that all depends upon the number of trials he has carried out to obtain his 70 percent figure. To score 70 percent or higher in a 10-trial run is not particularly significant (getting 7 heads out of 10 is not all that uncommon). However, if your uncle obtains a 70 percent success rate or higher on a 20-trial run, the result is getting more interesting (not yet statistically significant, but close). If Uncle Joe can score 70 percent or higher on a 100-trial run, ask him if you can be his agent—you may be onto something spectacular.

The point is that getting a result 20 percent away from the average of 50 percent (what we would expect in a series of coin tosses as the most probable result) becomes less and less probable as the number of trials (coin tosses) increases. To achieve such a departure from the average in a large number of trials would be very remarkable indeed and invite investigation (into the fairness of the procedure, the nature of the cards, or the possibility of hidden powers, like ESP).

Many of the claims to psychic powers fall into this category (i.e., the report of a statistically significant departure from expected probabilities). The response of sceptics is frequently a request for a replication of the sequence of events under controlled conditions which eliminate as much as possible any possible hidden variables (e.g., secret signals, various forms of cheating, and so on) and a sufficient number of trials to eliminate luck.

L. Experiment A: An Investigation of Bernoulli's Theorem

The rest of this module describes an experiment to illustrate and test Bernoulli's Theorem, to see whether our observed results get closer and closer to the calculated probabilities as we increase the number of trials.

Equipment

For this experiment you need to collect together the following items:

-ten coins, preferably all the same size and in good condition;
-a can or cup in which to shake the coins, which must be able to tumble freely in the container when you hold your hand over the opening and shake the coins;
-a tally sheet to keep track of the results (number of heads) of head throw (the tally sheet in given below)

Procedure

Find a quiet place at a table onto which you can spill the coins without bothering your family, friends, neighbours, or the cat. Then go through the following steps:

1. Carefully shake out the ten coins so that they are well mixed and then spill them out on a table or similar surface. Count the number of heads, and enter the number in the appropriate place on the tally sheet on the next page (using a vertical stroke in the appropriate box for each head; for example, if on the first trial you get six heads, then put a vertical line on the six line in the table, under the Trials 1-25 column; then on the second trial, if you get three heads, then put a line in the three box in the same column, and so on).

2. Collect the data for the first 25 tosses (each with 10 coins) in the first column, so that at the end of the 25 trials there is a record of how many tosses produced no heads, one head, two heads, three heads, and so on. Then do the same for the next 25 trials, entering the results in the second column. Enter the data for the third group of 25 tosses in the third column and for the fourth group of 25 trials in the fourth column. Now you have a record of the number of heads in 100 tosses of 10 coins.

3. Calculate the average number of heads for each of the 25 toss sequences, and enter that figure in the bottom row (labelled Mean) under the appropriate column. Calculate the average by adding up the numbers in the column (to get the total number of heads) and dividing by 25 (i.e., the number of trials). Note that you add up the number of heads in any one square by multiplying the number of vertical strokes times the number which the square represents. For example, if you have four vertical strokes in the 7 box (indicating that on 4 occasions there were 7 heads in the 10 coin result), then the number of heads for that box is 4 x 7 or 28.

Thus, you will have four figures for the average number of heads in each of the four 25-toss sequences.

4. Add up the totals for the first two 25-trial sequences, and enter them in the appropriate boxes in Column 5 (headed Total 1-50). Then add up the totals for the first three 25-trial sequences, and enter them in the appropriate boxes in Column 6 (headed Total 1-75). Then do the same for all the trials, entering the results in the boxes under the heading Total 1-100.

5. Then calculate the average number of heads (the Mean) for each column, once again adding up the total number of heads and dividing by the number of trials. Thus, you should produce an average number of heads for the first 50 trials, the first 75 trials, and for the total 100 trials.

Some Questions to Consider

Here are some things to think about before, during, and after the experiment.

1. What is the most reasonable number of heads to expect on any one toss of ten coins? Why?

2. The Law of Large Numbers (Bernoulli's Theorem) says that the averages one obtains by experiment and observation (as in our coin-flipping exercise) should get closer to the theoretical average as we increase the number of trials (in this case, coin tosses). If the law is clearly operating in your experiment, then the average number of heads on your tally sheet should be closer to the theoretical average (5.0) for all 100 trials than it is for most or all of the separate studies of 25 trials. Is this the case? Is the Mean under the Total 1-100 column closer to 5.0 than the Mean figures under Trials 1-25, Trials 26-50, Trials 51-75, Trials 76-100?

3. What about the first 50 trials? Is the Mean at the foot of the column Trials 1-50 closer to 5.0 than the Mean for the Trials 1-25, 26-50, 51-75, 76-100? What about the Mean figure for the Total 1-75? Is that closer to 5.0 than the other Means for 25 trials?

In the class, we will add your results to those from the other similar experiments conducted by other students, so that we can calculate the average for, say, 1000 trials. If Bernoulli's Theorem is operating in our experiment, we would expect the average from 1000 trials to be closer to the theoretical average than the average for 25 or 100 trials.

We should note, however, that it is possible (by chance) for any one of the means from a smaller number of trials to produce the exact average.

 TALLY SHEET FOR 100 TOSSES OF 10 COINS No. of Heads Trials 1-25 Trials 26-50 Trials 51-75 Trials 76-100 Total 1-50 Total 1-75 Total 1-100 p None .001 One .010 Two .044 Three .117 Four .205 Five .246 Six .205 Seven .117 Eight .044 Nine .010 Ten .001 Mean

Answers to Exercise C, Self-Test on Probability Notation

1. For 5 percent probability, p =.05; even odds: p = .5; 16 case out of 20: p = .8; for 3 chances in 1000: p = .003.

2. Values of p are as follows: p = .6 means 6 cases out of 10 (or 60 out of 100 or 60 percent); p = .04 means 4 cases out of 100; p = .023 means 23 cases out of 1000; p = .001 means 1 case in 1000; p = .01 means 1 case in 100;  p = .0345 means 345 cases out of 10,000; p = .0006 means 6 cases in 10,000.

Answers to Exercise G: Self-Test on Simple Probabilities

1. The probability of drawing a king is 4 out of 52 (p = .077), a king of hearts is 1 in 52 (p = .019); for drawing a king or a queen is 8 out of 52 (p = .154).

2. In one throw of a six-sided die, the probability of an odd number is .5; the probability of a number greater than 3 is .5; the probability of a 1 or a 2 is .33.

3. If you are picking two apples from a box with two rotten and two good apples, then the only possible outcomes are as follows (with R indicating a rotten apple and G indicating a good apple): RR, RG, GR, and GG. These are the only possible outcomes of two consecutive picks.

On the first pick, the chances of a rotten apple are 2 out of 4 (p = .5). The chances of picking a good apple on the first pick are also .5.

On the second pick, the probabilities will depend upon the results of the first pick. Thus, if the first apple picked was rotten, then the chances of getting a second rotten apple are 1 in 3 (p = .33)—since there are only 3 apples left and two of them are good. If the first pick was a rotten apple, then the chances of picking a good apple on the second pick are 2 out of 3 (p = .66). If the first pick was a good apple, then the probabilities for the second are reversed: p = .33 for a second good apple; p = .66 for a rotten apple.

We can thus list all the possible outcomes of two consecutive picks and by multiplying together the probability of each pick, calculate the probability for each outcome, as follows:

RR: .5 x .33 = .17
RG: .5 x .66 = .33
GG: .5 x .33 = .17
GR: .5 x .66 = .33

Thus, the chance of picking two rotten apples on two consecutive picks is .17 (or 17 times out of 100, or 17 percent); the chance of picking one rotten and one good apple is the sum of the probabilities for RG and GR, or .66 percent. Notice in the list above of all the possible outcomes, the total of all of the probabilities is equal to 1 (certainty)

Answers to Section I: Self-Test on a Binomial Table

1. The probability of getting exactly six heads: p = .205. the probability for exactly six tails is the same

2. For at least 70 percent heads in 10 consecutive tosses of a coin (i.e. getting 7 or more in 10 tosses),  p = .172. In a 20-toss sequence (n = 20), the probability of at least 70 percent success (i.e., 14 or more):  p = .058. Hence, the probability is higher for the 10-toss sequence.

3.  For the game to be entirely fair, you should put up \$80. The probability for her success (6 or more sixes in 24 consecutive rolls of a single die) is .200. The probability that you will win is therefore .800. Since the probability that you will win is four times the probability that she will win, a fair bet on your part is \$80 for her \$20.

4. You should not undertake the bet, especially with your student loan money, because the probabilities are greater that you will lose than those that you will win. The following is an explanation of the mathematical reasoning.

So far as the frequency of sixes is concerned, there are eight possible outcomes of a three-roll sequence for one die. We can represent these as follows (letting S indicate a result of six, and N represent a result of non-six—that is, any other number):

SSS, SSN, SNN, NNN, SNS, NSN, NNS, NSS

These are clearly the only possible outcomes for three rolls of a single die. On any one throw the probability of getting a six result is 1 divided by 6 or .1667, and the probabilities for non-six are 5 out of 6, or .8333. Therefore we can calculate the different probabilities for these eight possibilities by multiplying together the probabilities, as follows:

SSS: .1667 x .1667 x .1667 = .0045
SSN: .1677 x .1667 x .8333 = .0232
SNN: .1667 x .8333 x .8333 = .1158
NNN: .8333 x .8333 x .8333 = .5786
SNS: .1667 x .8333 x .1667 = .0232
NSN: .8333 x .1667 x .8333 = .1158
NNS: .8333 x .8333 x .1667 = .1158
NSS: .8333 x .1667 x .1667 = .0232

The total of all these probabilities is 1 (certainty). Now, the bet is that there will be at least one six in three throws. Seven of the eight possible outcomes involve at least one six. So to get the probabilities of at least one six, we simply add up the probabilities of all those outcomes. The total, from the above list, is .4215.

Thus, although there is only one outcome which does not involve at least one six (NNN), the probability of that outcome is .5786, more than half. So any bet that one six will appear in three throws of a die has the odds against it.

Notes to Section One

(1) Note that the symbol for probability is p and that it is written in italics. This convention (using italics) is common with such mathematical symbols taken from the alphabet, and we will be meeting more of them in the course of this module. [Back to Text]

(2) We will not concern ourselves here with the mathematical theory which establishes that the probability of a multiple outcome is derived by multiplying together the probabilities of a single event. This is an important theoretical concept first developed by an eighteenth-century English mathematician, Abraham de Moivre. [Back to Text]

(3) For example, the probability of one person's receiving a perfect bridge hand, that is, a hand with all thirteen cards of the same suit, is approximately .0000000000015 (or 15 times in 100 trillion). Thus, the occurrence of such a hand is very improbable. But given the millions of bridge rounds dealt every month, it is not surprising that now and then one hears of someone's receiving a perfect hand. [Back to Text]

(4) This point should remind us of the common gambler's folly in dice (or roulette ) games of sticking with the same number and continually increasing the bet. No matter what has happened in previous rolls of the dice or spins of the wheel, the probability for success remains the same on any single roll or spin. The fact that a 6, for example, has failed to appear in many previous rolls does not mean that the odds of a 6 on a particular new roll are any different. [Back to Text]