[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, released April 2000]
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 establish the mathematical likelihood of a
particular outcome and thus to advise oneself 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.
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.
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.
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).
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).
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).
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).
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.
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 (those interested in reading about the construction of such a table should
consult Appendix A to this module).
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 |
||||||||
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).
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.
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 on the next page (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?
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 |
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.
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, the theoretical calculations
come closer and closer to what we actually observe in experiments as we
increase the number of trials. 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, 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.
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.
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.
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 at the end of this module (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.
Here are some things to think about before,
during, and after the experiment.
What is the most reasonable number of heads to
expect on any one toss of ten coins?
Why?
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?
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 |
M.
Answers to Self-Test Exercises
1. For 5 percent probability: p = .05; even odds: p = .5; 16 cases out of 20: p
= .8; 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); p
= .04 means 4 cases out of 100; p = .023
means 23 cases out of 1000; p = .001
means 1 case in 1000; p = .0345 means
345 cases in 10000; p = .0006 means 6
cases in 10000.
1.
The
probability of drawing a king is 4 in 52 (p
= .077), a king of hearts is 1 in 52 (p
= .019; for drawing a king or a queen is 8 our of 52 (p = .154).
2.
The
probability of an odd number on the die after one throw is 3 out 6 (p = .5); for getting a number greater
than 3 is 3 out of 6 (p = .5); for
getting a 1 or a 2 is 2 out of 6 (p =
.333).
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).
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.
(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 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 million).
Thus the occurrence of such a hand is very 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 wheel) bets of sticking with the
same number and continually increasing the bet. No matter what has happened in previous rolls 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 roll are any different. [Back to Text]
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