4.4 Discrete Distribution (Playing Card Experiment)

Step 1 — ask what each one is built from. The theoretical distribution is computed from the deck's structure: 13 diamonds in 52 cards, ten replaced draws. Nothing in that computation looks at data.

Step 2 — check which ones depend on data. The empirical distribution is a tally of what the cards actually did today, and the simulated distribution is a tally of what the random number generator actually did today. Both are measurements.

Answer: the empirical and simulated distributions would both change; the theoretical one would not. That is why the theoretical distribution is the standard the other two get compared against — not because it is more accurate about your class, but because it is the only one of the three that is not a sample. It is worth being clear about the direction of that comparison: you are not checking whether the formula got your class right, you are checking whether your class's ten-draw sample landed where a sample from that formula should land.

Step 1 — count the diamonds. A standard deck has 52 cards in four suits of 13 each, so 13 of the 52 cards are diamonds.

$$ P(\text{diamond}) = \frac{13}{52} = \frac{1}{4} = 0.25 $$

Step 2 — check the binomial conditions. There are ten draws, fixed in advance. Each draw is either a diamond (success) or not (failure). Because the card is replaced and the deck reshuffled, the draws are independent and the success probability stays at 0.25 for every one of them. All four conditions hold.

Step 3 — name the parameters. The number of trials is \(n = 10\) and the success probability is \(p = 0.25\).

Answer: the theoretical probability is \(0.25\), and \(X \sim B(10,\ 0.25)\). Notice how much the replacement step is carrying here. Without it, the second draw would face a 51-card deck whose composition depends on the first draw, \(p\) would drift from draw to draw, and \(X\) would be hypergeometric instead of binomial.

Step 1 — with replacement. The card goes back, so the second draw faces the same 52-card deck as the first and the two draws are independent.

$$ P(\text{both diamonds}) = \left(\frac{13}{52}\right)\left(\frac{13}{52}\right) = (0.25)(0.25) = 0.0625 $$

Step 2 — without replacement. The first diamond is gone, so the second draw faces 51 cards of which only 12 are diamonds.

$$ P(\text{both diamonds}) = \left(\frac{13}{52}\right)\left(\frac{12}{51}\right) = (0.25)(0.2353) \approx 0.0588 $$

Answer: 0.0625 with replacement and about 0.0588 without — and the binomial formula assumes the first one. The gap looks small, and on two draws it is. It is not small in principle: without replacement, the second draw's probability depends on the first draw's outcome, and dependence is the one thing the binomial model is not allowed to have. Replacing the card is not a fussy detail of lab procedure; it is the step that makes \(B(10,\ 0.25)\) the correct model rather than an approximation to one.

Step 1 — check the total. \(2 + 5 + 9 + 7 + 4 + 2 + 1 = 30\) repetitions, which matches the 30 groups she recorded.

Step 2 — take the relative frequency of 2.

$$ RF(x = 2) = \frac{9}{30} = 0.3000 $$

Step 3 — find the mean by weighting each value by how often it happened.

$$ \bar{x} = \frac{0(2) + 1(5) + 2(9) + 3(7) + 4(4) + 5(2) + 6(1)}{30} = \frac{0 + 5 + 18 + 21 + 16 + 10 + 6}{30} = \frac{76}{30} \approx 2.5333 $$

Answer: \(RF(x = 2) = 0.3000\) and \(\bar{x} \approx 2.5333\). That mean sits within a rounding error of the theoretical 2.5, which is what pooling 30 repetitions buys you. Any one of those 30 groups, looking only at its own ten draws, saw a whole number — 0, or 4, or 6 — and none of them saw 2.5 at all. The average is not a value the experiment can produce; it is a value the experiment's results cluster around.

Step 1 — get the sum of the squared values, weighted by frequency.

$$ \sum x^2 f = 0^2(2) + 1^2(5) + 2^2(9) + 3^2(7) + 4^2(4) + 5^2(2) + 6^2(1) = 0 + 5 + 36 + 63 + 64 + 50 + 36 = 254 $$

Step 2 — subtract the part explained by the mean. With \(n = 30\) and \(\bar{x} \approx 2.5333\),

$$ s^2 = \frac{\sum x^2 f - n\bar{x}^2}{n - 1} = \frac{254 - 30(2.5333)^2}{29} = \frac{254 - 192.5333}{29} = \frac{61.4667}{29} \approx 2.1195 $$

Step 3 — take the square root.

$$ s = \sqrt{2.1195} \approx 1.4559 $$

Answer: \(s \approx 1.4559\). Hold that beside the theoretical \(\sigma \approx 1.3693\) you will compute in the next section. The sample came out slightly more spread out than the model predicts, which is the ordinary behavior of 30 repetitions — the standard deviation of a sample is itself a measurement, and it wobbles just like the mean does. Divide by \(n - 1\), not \(n\); this is a sample, not the whole population of every ten-draw run that could ever happen.

Step 1 — count the arrangements. Two diamonds among ten draws can happen in \(\binom{10}{2} = \frac{10 \times 9}{2} = 45\) different orders.

Step 2 — take the probability of any one such order and multiply.

$$ P(X = 2) = 45 (0.25)^2 (0.75)^8 = 45 (0.0625)(0.1001) \approx 0.2816 $$

Step 3 — use the binomial shortcuts for the center and spread.

$$ \mu = np = 10(0.25) = 2.5 $$ $$ \sigma = \sqrt{npq} = \sqrt{10(0.25)(0.75)} = \sqrt{1.875} \approx 1.3693 $$

Answer: \(P(X = 2) \approx 0.2816\), \(\mu = 2.5\), and \(\sigma \approx 1.3693\). Compare that \(\mu\) to the \(\bar{x}\) your class computed and that \(\sigma\) to your \(s\). They are answering the same question — where is the center, how spread out is it — from two completely different directions, one from the formula and one from the cards.

Step 1 — apply the formula with \(x = 0\). There is exactly one way to get no diamonds at all, and \((0.25)^0 = 1\), so the whole expression collapses to the probability that all ten draws miss.

$$ P(X = 0) = \binom{10}{0}(0.25)^0(0.75)^{10} = (0.75)^{10} \approx 0.0563 $$

Step 2 — ask what the column covers. The values 0 through 10 are every number of diamonds ten draws could possibly produce. There is no twelfth outcome and no overlap between any two of them.

Answer: \(P(X = 0) \approx 0.0563\), and the column must add to exactly 1. That is the definition of a probability distribution function doing its job — §4.1's requirement that the probabilities of a discrete random variable are each between 0 and 1 and sum to 1. Use it as your check: if your eleven entries total 0.9987 or 1.0021 you have a rounding artifact and can move on, but if they total 0.87 you have dropped or mis-entered a row, and the histogram you build next will be wrong in a way that is hard to see by eye.

Step 1 — turn a probability into an expected count. Multiply the probability of the outcome by the number of repetitions.

$$ 30 \times 0.2816 = 8.448 $$

Step 2 — put it beside the observed count. The class observed 9 groups at \(x = 2\), against an expected 8.448.

Answer: about 8.4 groups were expected and 9 were observed — a difference of roughly half a group. Two things are worth noticing. The expected count is not a whole number even though the observed count has to be, which is the same "you cannot draw 2.5 diamonds" point from a different angle. And this is the arithmetic that turns "the graphs look similar" into something you can defend: instead of comparing two pictures by eye in the discussion questions, compare each bar's observed count against \(30 \times P(x)\) and say which bars are off and by how much.

Step 1 — translate each inequality into a list of values. \(1 < x < 4\) means \(x \in \{2, 3\}\). And \(x \geq 8\) means \(x \in \{8, 9, 10\}\).

Step 2 — add the probabilities for the first event.

$$ P(1 < x < 4) = P(2) + P(3) = 0.2816 + 0.2503 = 0.5319 $$

Step 3 — add the probabilities for the second event.

$$ P(x \geq 8) = P(8) + P(9) + P(10) = 0.0004 + 0.0000 + 0.0000 = 0.0004 $$

Answer: \(P(1 < x < 4) = 0.5319\) and \(P(x \geq 8) = 0.0004\). Those two numbers say something worth pausing on: more than half of all classes will land on 2 or 3 diamonds, while fewer than 1 class in 2,000 will land on 8 or more. If your class did get 8 or more, you have not broken statistics — you have met the tail, and that is a better lab result than the boring one.

Step 1 — take each count over the 30 repetitions.

$$ RF(x = 3) = \frac{7}{30} \approx 0.2333 \qquad RF(1 < x < 4) = \frac{9 + 7}{30} = \frac{16}{30} \approx 0.5333 $$

Step 2 — handle the empty event. No group in this class got 8, 9, or 10 diamonds, so the count is 0.

$$ RF(x \geq 8) = \frac{0}{30} = 0.0000 $$

Step 3 — line the three pairs up. Theoretical 0.2503 against empirical 0.2333; theoretical 0.5319 against empirical 0.5333; theoretical 0.0004 against empirical 0.0000.

Answer: \(RF(x = 3) \approx 0.2333\), \(RF(1 < x < 4) \approx 0.5333\), \(RF(x \geq 8) = 0.0000\) — close on the first two for her class, and dead-on for the middle one. The third pair is the interesting one, and it is where students most often write the wrong conclusion. An empirical relative frequency of 0 does not mean the event is impossible; it means 30 repetitions were nowhere near enough to see something that happens about four times in ten thousand. A rare event's relative frequency will read 0.0000 in almost every small sample, which is exactly why you need the theoretical column to know it is there at all.

Step 1 — identify what each table is made of. Table 4.4.1 holds frequencies and relative frequencies, which are counts of what actually happened. Table 4.4.2 holds \(P(x)\) values computed from \(n = 10\) and \(p = 0.25\).

Step 2 — see which one has data in it. Only Table 4.4.1 does. Table 4.4.2 never looks at the deck, so running the experiment more times cannot move a single entry in it.

Step 3 — say which direction Table 4.4.1 moves. With 500 repetitions instead of 30, each relative frequency is an average over many more trials, so the whole column settles closer to the matching \(P(x)\) — including \(RF(x \geq 8)\), which would finally start showing something other than 0.

Answer: Table 4.4.1 changes and Table 4.4.2 does not; the empirical relative frequencies tighten around the theoretical probabilities, and the empirical histogram takes on the theoretical one's shape. This is the whole meaning of "long-term probability" and it is the lab's real payoff. The 0.25 was never a promise about your ten draws — it was a promise about what the fraction of diamonds does as the number of draws grows. Ten draws cannot show you that. Five hundred can start to.