Comparing Experimental and Theoretical Probability | Grade 7 Math | LevelKite

You can already use a probability to predict about how often something will happen. Now you will check predictions against real results and describe how the two compare.

A few words first. A trial is one run of a chance experiment, such as one flip of a coin. An outcome is one possible result of a trial. An event is an outcome, or a group of outcomes, that you are watching for, such as heads. An outcome in which the event happens is a favourable outcome.

For any event, there are two probabilities you can compare:

Theoretical probability is what a model predicts. A model is a description of the situation that gives each outcome a probability. You can reason it out without running a single trial.
Experimental probability is what happened. It is the event's relative frequency: the fraction of the trials in which the event occurred.

One comes from reasoning and the other from data, so they do not have to be the same number. Comparing them means finding both, then asking how far apart they are and why.

Two probabilities for the same event

Question	Theoretical probability	Experimental probability
Where does it come from?	A model, reasoned out without running any trial	The results of trials you have run
How do you find it?	For equally likely outcomes only: $\frac{\text{favourable outcomes}}{\text{total outcomes}}$	$\frac{\text{trials in which the event happened}}{\text{total trials}}$ , with at least 1 trial
Can its value change?	No. The model fixes it.	Yes. Running more trials can give a new value.
How far can you trust it?	As far as the model fits the real situation	It approximates the event's probability only over many trials

Before you compare, name the quantities. For the event you are studying:

$T$ is its theoretical probability and $E$ is its experimental probability. Both are numbers with no units. ( $E$ stands for experimental, not expected.)
$n$ is the number of trials, a whole number that is at least 1, so you never divide by zero.
$k$ is the number of those trials in which the event happened, so $0 \le k \le n$ and $E = \frac{k}{n}$ .

From your earlier work, the model's expected count of the event in $n$ trials is $T \times n$ . It is an approximate prediction, not a guarantee.

Key concept

Compare the model with the data

To compare the two probabilities of one event, write $T$ and $E$ in the same form (both decimals, or fractions with a common denominator) and subtract. The result is the gap.

Gap positive: the event happened more often than the model predicts.
Gap negative: it happened less often.
Gap zero: the two match.

A gap is normal, because chance makes results vary. Relative frequency approximates probability only over many trials, so judge every gap together with $n$ . Even over many trials, $E$ is usually close to $T$ and need not equal it. All of this assumes that $T$ fits the situation: favourable over total is a probability only when the outcomes are equally likely.

\text{gap} = E - T

Worked example

A fair coin, 50 flips

You flip a fair coin $50$ times and get heads $28$ times. Compare the experimental probability of heads with the theoretical probability.

Find the theoretical probability T. One flip has 2 outcomes, heads and tails. The coin is fair, so the two outcomes are equally likely. That condition is what lets you use favourable over total. Heads is 1 favourable outcome out of 2.
$T = \frac{1}{2} = 0.5$
Find the experimental probability E. It is the relative frequency: k = 28 flips gave heads out of n = 50 flips. Then multiply the numerator and the denominator by 2. Multiplying both by the same nonzero number keeps the fraction's value, and it gives hundredths, which you can read as a decimal.
$E = \frac{28}{50} = \frac{28 \times 2}{50 \times 2} = \frac{56}{100} = 0.56$
Both values lie from 0 to 1, as every probability must, and both are decimals, so you can subtract. Take experimental minus theoretical, in that order, so the sign shows which is larger. The gap is positive: heads came up more often than the model predicts.
$E - T = 0.56 - 0.50 = 0.06$
Say the gap in flips. The model's expected count is T times n, an approximate prediction and not a guarantee. The actual count k is 3 higher. The two views agree: n is positive, so E greater than T means k is greater than T times n. With only 50 flips, 3 extra heads is ordinary chance variation.
$T \times n = 0.5 \times 50 = 25 \qquad k - 25 = 28 - 25 = 3$

Answer

The experimental probability of heads is $0.56$ and the theoretical probability is $0.5$ . The experimental value is greater by $0.06$ , which is $3$ more heads than the expected count of about $25$ in $50$ flips.

Why is there a gap at all? A model cannot control single trials. Each flip is still decided by chance, so the number of heads can land above or below the expected count, and it often does. The useful question is how the gap behaves as the number of trials grows.

The table follows one class spinning a fair spinner with 4 equal sections, 1 of them blue. The sections are equal, so the 4 outcomes are equally likely, and for the event blue that gives $T = \frac{1}{4} = 0.25$ . The class keeps a running total, so each row includes all the spins counted in the rows above it.

Running totals for the event blue (theoretical probability 0.25)

Spins so far, n	Times on blue, k	Experimental probability, E	Gap, E minus T
$20$	$8$	$\frac{8}{20} = 0.40$	$0.40 - 0.25 = 0.15$
$100$	$29$	$\frac{29}{100} = 0.29$	$0.29 - 0.25 = 0.04$
$500$	$131$	$\frac{131}{500} = 0.262$	$0.262 - 0.25 = 0.012$
$2000$	$506$	$\frac{506}{2000} = 0.253$	$0.253 - 0.25 = 0.003$

Read down the last column: $0.15$ , $0.04$ , $0.012$ , $0.003$ . After 20 spins, $E$ was far from $T$ . After 2000 spins it was very close, though still not equal.

That is the usual pattern, and it has two limits. First, relative frequency approximates probability only over many trials, so a result from a few trials can sit far from $T$ with nothing wrong. Second, the pattern is a tendency, not a promise: a stretch of trials can make the gap wider before it narrows again. So whenever you compare $E$ with $T$ , state the number of trials too.

Worked example

A number cube: an event and its opposite

You roll a fair number cube with faces numbered $1$ to $6$ a total of $120$ times. A number greater than $4$ comes up $46$ times. Compare the experimental and theoretical probabilities of rolling a number greater than $4$ . Then do the same for not rolling one.

Find T for the event of rolling a number greater than 4. The cube is fair, so its 6 outcomes are equally likely. Each has probability one sixth, and the six probabilities add to 1, as a model's must. Equally likely outcomes allow favourable over total. The favourable outcomes are 5 and 6, which is 2 of the 6.
$T = \frac{2}{6} = \frac{1}{3}$
Find E from the data: k = 46 of the n = 120 rolls. Divide the numerator and the denominator by 2. Dividing both by the same nonzero number keeps the fraction's value.
$E = \frac{46}{120} = \frac{46 \div 2}{120 \div 2} = \frac{23}{60}$
Write T with the denominator 60 so that both fractions count parts of the same size. Then subtract experimental minus theoretical. The gap is positive, so the event happened more often than the model predicts.
$T = \frac{1 \times 20}{3 \times 20} = \frac{20}{60} \qquad E - T = \frac{23}{60} - \frac{20}{60} = \frac{3}{60} = \frac{1}{20} = 0.05$
Say the gap in rolls. The model's expected count is T times n, an approximate prediction and not a guarantee. The actual count is 6 rolls higher. As a check, 6 rolls out of 120 is one twentieth, the same gap as in the last step.
$T \times n = \frac{1}{3} \times 120 = 40 \qquad k - 40 = 46 - 40 = 6 \qquad \frac{6}{120} = \frac{1}{20}$
Now take the opposite event: not rolling a number greater than 4, which means rolling 1, 2, 3 or 4. The probability that an event does not happen is 1 minus the probability that it does. This holds for T. It holds for E too, because every roll either is greater than 4 or is not: 120 minus 46 leaves 74 rolls, and 74 out of 120 simplifies to 37 sixtieths.
$1 - T = 1 - \frac{1}{3} = \frac{2}{3} = \frac{40}{60} \qquad 1 - E = 1 - \frac{23}{60} = \frac{37}{60}$
Subtract experimental minus theoretical for this event. The gap has the same size and the opposite sign. That makes sense: the 6 extra rolls that were greater than 4 are 6 rolls missing from the other event.
$(1 - E) - (1 - T) = \frac{37}{60} - \frac{40}{60} = -\frac{3}{60} = -\frac{1}{20} = -0.05$

Answer

Rolling a number greater than $4$ has experimental probability $\frac{23}{60}$ , which is $\frac{1}{20}$ above its theoretical probability $\frac{1}{3}$ . Not rolling one has experimental probability $\frac{37}{60}$ , which is $\frac{1}{20}$ below its theoretical probability $\frac{2}{3}$ .

A common mistake: treating any gap as proof. Suppose you flip a coin 10 times and get 7 heads. It is tempting to say that $E = \frac{7}{10} = 0.7$ does not equal the fair-coin value $T = 0.5$ , so the coin must be unfair. The arithmetic is right, but the conclusion does not follow. With $n = 10$ , each flip moves $E$ by $\frac{1}{10} = 0.1$ , so the gap of $0.2$ is only 2 flips above the expected count of 5. A fair coin misses 5 heads by 2 or more in about one of every three sets of 10 flips. The same gap after 1000 flips would mean 700 heads against an expected count of 500, a difference of 200 flips. That would be strong evidence against a fair coin. A model describes what to expect over many trials. It never promises an exact match, least of all after a few.

Worked example

Two groups, one spinner

A fair spinner has $5$ equal sections: $2$ red and $3$ white. Group A spins it $25$ times and lands on red $13$ times. Group B spins it $400$ times and lands on red $148$ times. Whose experimental probability of red is closer to the theoretical probability? Is that what you would expect?

Find T. The 5 sections are equal, so the 5 outcomes are equally likely and favourable over total applies. Landing on red is favourable in 2 of the 5 sections.
$T = \frac{2}{5} = 0.4$
Find each group's experimental probability as k over n, written in hundredths. For Group A, multiply the numerator and the denominator by 4. For Group B, divide both by 4. Neither change alters the fraction's value. The small letter after each E names the group.
$E_A = \frac{13}{25} = \frac{52}{100} = 0.52 \qquad E_B = \frac{148}{400} = \frac{37}{100} = 0.37$
Subtract experimental minus theoretical for each group. Group A's gap is positive, so red came up more often than the model predicts. Group B's gap is negative, so red came up less often.
$E_A - T = 0.52 - 0.40 = 0.12 \qquad E_B - T = 0.37 - 0.40 = -0.03$
Closer means a smaller distance from T. Distance ignores direction, so compare the absolute values of the gaps, which are their distances from zero.
$|-0.03| = 0.03 \qquad |0.12| = 0.12 \qquad 0.03 < 0.12$
Check with counts, using the approximate expected count T times n. Group B is 12 spins from its expected count and Group A is only 3 spins from its own. B is still closer as a probability, because 12 out of 400 is 0.03 while 3 out of 25 is 0.12.
$0.4 \times 25 = 10 \qquad 13 - 10 = 3 \qquad 0.4 \times 400 = 160 \qquad 148 - 160 = -12$
Decide whether this is what you would expect. Relative frequency approximates probability only over many trials, so 400 spins would be expected to land closer than 25 spins. It was likely, not guaranteed. Group A's larger gap comes from just 3 spins, so it is no evidence that the spinner is unfair.

Answer

Group B is closer: its experimental probability $0.37$ is $0.03$ below the theoretical probability $0.4$ , while Group A's $0.52$ is $0.12$ above it. That is what you would expect, though it was not guaranteed, because Group B ran many more trials.

Takeaway. Theoretical probability is the model's prediction, and favourable over total gives it only when the outcomes are equally likely. Experimental probability is the relative frequency from real trials. To compare them, write both in the same form, subtract experimental minus theoretical, and report the gap along with the number of trials. Expect a gap, because predictions are approximate expected counts, not guarantees. Relative frequency approximates probability only over many trials, so a comparison tells you more as the number of trials grows.