Finite Probability Modeling Workshop: Cards and Dice

From Experiment Description to Probability Model

A card-and-dice question is not fully specified by naming the objects. We also need to know how cards are selected, whether draws are with or without replacement, whether a die is fair, and whether the parts of a combined experiment are independent. Those assumptions determine both the sample space and the probability assigned to each outcome.

For a finite experiment, the sample space $\Omega$ is the set of outcomes the model treats as possible. An event $A$ is a subset of $\Omega$, and its probability is the sum of the probabilities assigned to its outcomes. If every outcome has the same probability, counting favorable outcomes is enough. If probabilities differ, we must add their weights instead.

Definition: A finite probability model consists of a finite sample space $\Omega$ and a probability assignment $P$ on its events, with $P(\{\omega\})\geq 0$ for each $\omega\in\Omega$ and $\sum_{\omega\in\Omega}P(\{\omega\})=1$. For an event $A\subseteq\Omega$, $$ P(A)=\sum_{\omega\in A}P(\{\omega\}). $$ A uniform model assigns probability $1/|\Omega|$ to each outcome. A nonuniform model assigns different probabilities to at least two outcomes.

A useful modeling habit is to write the outcome in enough detail to tell apart all results relevant to the question. For a card and one die roll, an outcome can be an ordered pair $(c,d)$, where $c$ is the card and $d$ is the die face. The order matters because it records which result came from which part of the experiment.

Combining Independent Experiments

Suppose a card experiment has finite outcome set $C$ and probability assignment $p_C(c)$, and a die experiment has finite outcome set $D$ and assignment $p_D(d)$. If the two experiments are independent, the combined sample space is $C\times D$, and the joint probability of $(c,d)$ is the product of the component probabilities.

Theorem: For independent finite experiments with probability assignments $p_C$ and $p_D$, define $$ P(\{(c,d)\})=p_C(c)p_D(d),\qquad (c,d)\in C\times D. $$ This is a probability assignment on $C\times D$. For any $A\subseteq C$ and $B\subseteq D$, $$ P(A\times B)=P_C(A)P_D(B), $$ where $P_C(A)=\sum_{c\in A}p_C(c)$ and $P_D(B)=\sum_{d\in B}p_D(d)$.

Proof. Each product $p_C(c)p_D(d)$ is nonnegative. The sum of all joint probabilities factors into two finite sums:

\sum_{c\in C}\sum_{d\in D}p_C(c)p_D(d) = \left(\sum_{c\in C}p_C(c)\right) \left(\sum_{d\in D}p_D(d)\right) =1\cdot1=1.

Thus the assignment is normalized. Summing over $A\times B$ gives

P(A\times B) =\sum_{c\in A}\sum_{d\in B}p_C(c)p_D(d) =\left(\sum_{c\in A}p_C(c)\right) \left(\sum_{d\in B}p_D(d)\right) =P_C(A)P_D(B).

Both factorizations are valid because the sets are finite. This proves the claim. $\square$

Independence is an assumption about the experiment, not a consequence of writing outcomes as pairs. For instance, if the die result determines which card is selected, the product assignment need not describe the joint experiment.

Worked Example: A Uniform Card and a Fair Die

Draw one card uniformly from a standard 52-card deck and roll one fair six-sided die, independently. Find the probability that the card is red or the die shows six.

Build the model. The outcomes are the 312 ordered pairs $(c,d)$, with $c$ one of 52 cards and $d\in\{1,2,3,4,5,6\}$. Each pair has probability $(1/52)(1/6)=1/312$, so the combined model is uniform.

Let $R$ be the event that the card is red, and $S$ the event that the die shows six. There are $26\cdot6=156$ outcomes in $R$, and $52\cdot1=52$ in $S$. Their intersection has $26\cdot1=26$ outcomes: a red card together with a six.

Adding the two event counts without correcting for their overlap would count those 26 outcomes twice. The union therefore has $156+52-26=182$ outcomes, and

P(R\cup S)=\frac{182}{312}=\frac{7}{12}.

The same result follows from independence: $P(R)=1/2$, $P(S)=1/6$, and $P(R\cap S)=1/12$, so $1/2+1/6-1/12=7/12$. Here counting works because the stated assumptions make all 312 pairs equally likely.

Counting Under a Uniform Law

The shortcut “favorable outcomes divided by total outcomes” is justified only when the outcomes being counted are equally likely. The relevant count is the number of outcomes in the event, not the number of informal descriptions of the event.

Theorem: If $\Omega$ is a finite sample space and every outcome has probability $1/|\Omega|$, then every event $A\subseteq\Omega$ has probability $$ P(A)=\frac{|A|}{|\Omega|}. $$

Proof. By finite additivity, $P(A)$ is the sum of the point probabilities for outcomes in $A$. Each of the $|A|$ terms is $1/|\Omega|$. Hence

P(A)=\sum_{\omega\in A}\frac{1}{|\Omega|} =\frac{|A|}{|\Omega|}.

This argument also shows why the uniformity condition matters: replacing unequal point probabilities by equal contributions changes the sum. $\square$

Worked Example: A Biased Card Selection and a Loaded Die

A card is selected from a four-card set containing $A\spadesuit$, $K\heartsuit$, $7\diamondsuit$, and $2\clubsuit$. Their selection probabilities, in that order, are $1/2,1/4,1/8,1/8$. Independently, a die has face probabilities $1/2,1/4,1/8,1/16,1/32,1/32$ for faces 1 through 6. Find the probability of selecting a heart or rolling at least 5.

Check the component laws. The card weights sum to $1/2+1/4+1/8+1/8=1$. The die weights sum to $1/2+1/4+1/8+1/16+1/32+1/32=1$. The only heart is $K\heartsuit$, so its probability is $1/4$. The probability of rolling at least 5 is $1/32+1/32=1/16$.

Let $H$ denote selecting the heart and $T$ denote rolling 5 or 6. By independence, their intersection has probability $(1/4)(1/16)=1/64$. Adding the probabilities of $H$ and $T$ counts their intersection twice, so

P(H\cup T)=P(H)+P(T)-P(H\cap T) =\frac14+\frac1{16}-\frac1{64} =\frac{19}{64}.

As a check, the event consists of all outcomes with the heart, together with outcomes having a non-heart and die face 5 or 6. The first group has probability $1/4$. The non-heart cards have total probability $3/4$, so the second group has probability $(3/4)(1/16)=3/64$. Their disjoint sum is $1/4+3/64=19/64$.

For example, the particular outcome $(K\heartsuit,6)$ has probability $(1/4)(1/32)=1/128$, not $1/24$. There are 24 pairs in the sample space, but they are not equally likely.

Ordered Draws Without Replacement

A different experiment requires a different sample space. When two cards are drawn in sequence without replacement, the second card cannot equal the first, and $(c_1,c_2)$ records the order of the draws. If the deck is well shuffled, each ordered pair of distinct cards is equally likely. Combining this draw with an independent fair die roll keeps the product model uniform.

Worked Example: Two Cards and a Die Roll

Draw two cards in order without replacement from a well-shuffled standard deck, then independently roll a fair six-sided die. Find the probability that the first card is an ace, the second card is red, and the die shows six.

Sample space. There are 52 choices for the first card, 51 remaining choices for the second, and 6 die faces. Thus the sample space has $52\cdot51\cdot6=15{,}912$ outcomes. Each has probability $1/15{,}912$, under the stated shuffle and fairness assumptions.

Count favorable ordered pairs of cards. There are four possible first-card aces. If the first ace is red, 25 red cards remain; there are two such first aces, contributing $2\cdot25=50$ ordered pairs. If the first ace is black, all 26 red cards remain; there are two such aces, contributing $2\cdot26=52$ pairs. In total there are $50+52=102$ ordered card pairs meeting the card conditions.

For each such pair, exactly one of the six die faces is six. Therefore there are 102 favorable outcomes in the combined space, and

P(\text{first ace, second red, die six}) =\frac{102}{52\cdot51\cdot6} =\frac{102}{15{,}912} =\frac{17}{2{,}652}.

The calculation accounts for the dependence between the two card draws: after a red ace is drawn first, only 25 red cards remain, whereas after a black ace, all 26 remain. Treating the two draws as independent would ignore the removal of the first card.

Common Mistake: Counting Outcomes That Are Not Equally Likely

A tempting but incorrect argument in the biased four-card-and-die model is: “There are four cards and six faces, so there are 24 pairs; if 7 pairs meet the event, the probability is $7/24$.” The flaw is the assumption that every pair has probability $1/24$. The card and die weights differ, so some pairs are much more likely than others.

The correction is to sum the probabilities of the event’s outcomes, using the product of component weights for each pair. In that example, the union probability is $19/64$, as calculated from the event probabilities and their intersection. Counting is a valid shortcut only after the model has established equal point probabilities.

Key takeaway: State the outcome set and probability law before calculating. Count event outcomes only under a uniform model; otherwise, add the assigned probabilities of the outcomes in the event.

Check Your Understanding

For each question, identify the model assumptions that justify the calculation.

A fair die is rolled independently of a uniformly selected card. How many outcomes are in the combined sample space, and what is the probability of any one card-face pair?
For the four-card and loaded-die model, why is the probability of $(K\heartsuit,6)$ not the reciprocal of the number of pairs?
In a uniform finite sample space with 40 outcomes, an event contains 9 outcomes. What is its probability, and which assumption makes the count ratio valid?
Two cards are drawn in order without replacement. How many choices are available for the second card after the first has been observed?
Explain why independence of a card draw and a die roll allows the probability of a card event and a die event to be multiplied, but does not make two successive draws without replacement independent.