Probability & Expected Value

1.Equally likely outcomes and complementary counting

Most AMC probability problems are, at heart, counting problems. When every outcome of an experiment is equally likely, the probability of an event $E$ is $P(E)=\frac{|E|}{|S|}$, the number of favourable outcomes divided by the total number of outcomes in the sample space $S$. Nothing more is needed. The whole difficulty lies in describing $S$ so that its outcomes really are equally likely.

Why does the formula hold? Suppose each of the $N=|S|$ outcomes has the same probability $p$. The probabilities of all outcomes sum to $1$, so $Np=1$ and $p=\frac1N$. Additivity over disjoint outcomes then gives $P(E)=\sum_{\omega\in E}\frac1N=\frac{|E|}{N}$. The argument is one line, but it tells you what to check: the outcomes you list must be equally likely, not merely possible.

The most common error is listing outcomes that are not equally likely. The two dice can be listed as the sums $2,3,\dots,12$, but these eleven sums are not equally likely (a sum of $7$ has six ways to occur and a sum of $2$ has one), so dividing by $11$ is wrong. Always list outcomes at the level of the physical experiment: ordered pairs of faces, ordered strings of coin results, or ordered arrangements of distinct objects.

Complementary counting is the second essential tool. For an event described as 'at least one', count the complement 'none occurs' instead, because the complement is usually a single product: $P(\text{at least one})=1-P(\text{none})$. For $n$ fair coins, $P(\text{at least one head})=1-\left(\tfrac12\right)^n$. Check this on $n=1$: $1-\frac{1}{2}=\frac{1}{2}$, which is correct.

Ordered and unordered counting must agree. You may count sequences or count subsets, but if the numerator counts ordered pairs, the denominator must count ordered pairs too. A helpful check is to compute one probability two ways: when two distinct objects are drawn, the ordered and the unordered counts give the same fraction, because the same factor of $2$ appears on top and bottom.

Inclusion–exclusion repairs double counting. For the multiples of $4$ or $6$ in $\{1,\dots,100\}$, count $25+16$ and subtract the $8$ multiples of $12$ that were counted twice, giving $33$. This is the same idea as $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and it is the first formula to reach for whenever two conditions overlap.

On the AMC the counting is often hidden in a clever sample space: rolls of dice, flips of coins, or random permutations. A space of $6^4=1296$ rolls is still a finite list. By hand you exploit structure, such as symmetry or a recursion, rather than listing every case; the verification snippets in this topic do the full list by computer as a check.

• $P(E)=\frac{|E|}{|S|}$ when all outcomes are equally likely
• $P(\text{at least one})=1-P(\text{none})$
• $|A\cup B|=|A|+|B|-|A\cap B|$
• $P(\text{at least one head in } n \text{ fair flips})=1-\left(\tfrac12\right)^n$

• Treating outcomes that are not equally likely (such as the possible sums of two dice) as if they were.
• Mixing ordered and unordered counts, so the numerator and denominator count different objects.

2.Conditional probability

Conditional probability answers the question 'how likely is $A$, given that $B$ has happened?' We write $P(A\mid B)$. Learning that $B$ happened changes the universe: outcomes where $B$ fails are discarded, and we ask what fraction of the surviving outcomes also make $A$ true. The word 'given' is a signal to shrink the sample space, not to multiply by something.

Formally, $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, defined whenever $P(B)>0$. Rearranging gives the multiplication rule $P(A\cap B)=P(B)\,P(A\mid B)$, which is exactly the rule used along the branches of a probability tree. When outcomes are equally likely the formula reads $P(A\mid B)=\frac{|A\cap B|}{|B|}$: count inside $B$ only.

Why the formula is right. Restrict attention to $B$. Inside $B$ the outcomes are still equally likely, there are $|B|$ of them, and the ones in $A$ are exactly the $|A\cap B|$ elements of $A\cap B$. Dividing gives the ratio. Nothing beyond the definition of a uniform probability on the smaller universe is assumed.

Consider drawing two cards from a deck without replacement. Given that the first card is an ace, $P(\text{second is an ace}\mid\text{first is an ace})=\frac{3}{51}$, because three aces remain among $51$ cards. The denominator is not $52$: the given information changes both the numerator and the denominator. A student who writes $\frac{4}{52}$ has forgotten the conditioning.

Order matters. $P(A\mid B)$ and $P(B\mid A)$ are generally different: $P(\text{ace}\mid\text{spade})=\frac{1}{13}$ while $P(\text{spade}\mid\text{ace})=\frac{1}{4}$. Confusing the two is the source of the base-rate errors discussed under Bayes' theorem, and the cure is always to ask which event forms the new universe.

Conditioning on a sum of dice is a frequent AMC move. Given that two dice sum to $7$, only six outcomes remain, and the question becomes a count inside those six. Writing the restricted list explicitly takes seconds and removes almost every error, because you can see the double counting of $(1,6)$ and $(6,1)$ directly.

Conditioning can reveal hidden structure. Once we know the first flip is heads, the remaining flips form a fresh uniform universe, with the same rules as before. Recasting 'given' as a new universe is exactly what makes the recursive arguments later in this topic work.

• $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ for $P(B)>0$
• $P(A\cap B)=P(B)\,P(A\mid B)$, the multiplication rule along a tree branch
• With equally likely outcomes, $P(A\mid B)=\frac{|A\cap B|}{|B|}$, counting inside $B$ only
• $P(A\mid B)\neq P(B\mid A)$ in general

• Dividing by the size of the original sample space instead of the size of the conditioning event.
• Confusing $P(A\mid B)$ with $P(B\mid A)$.

3.Total probability and Bayes' theorem

Many problems describe a two-stage experiment: first a hidden cause is chosen (a bag, a coin, a disease status), then something is observed. Such problems ask for the overall probability of the observation, or for the probability of a cause given the observation. The two tools are the law of total probability and Bayes' theorem, and both are consequences of the definition of conditional probability.

Total probability. Suppose $B_1,\dots,B_k$ partition the sample space: they are pairwise disjoint and together cover everything. Then for any event $A$, $P(A)=\sum_i P(B_i)\,P(A\mid B_i)$. Proof: $A$ is the disjoint union of the pieces $A\cap B_i$, so $P(A)=\sum_i P(A\cap B_i)$, and each term equals $P(B_i)\,P(A\mid B_i)$ by the multiplication rule.

Bayes' theorem. Using the definition twice, $P(B_j\mid A)=\frac{P(A\cap B_j)}{P(A)}$, and then the total probability formula gives $P(B_j\mid A)=\frac{P(B_j)\,P(A\mid B_j)}{\sum_i P(B_i)\,P(A\mid B_i)}$. The prior $P(B_j)$ is updated by the likelihood $P(A\mid B_j)$ to give the posterior. The denominator is only a normalising constant: the numerators already tell you which cause is most likely.

Why the prior matters. A test with $95\%$ sensitivity sounds accurate, but if only $1\%$ of people have the condition, most positive results can be false positives. The posterior $P(\text{condition}\mid +)$ depends on the prior $\frac{1}{100}$ as well as the test's two error rates. The naive answer $\frac{19}{20}$ ignores the prior entirely, and that is the base-rate fallacy the AMC likes to test.

The tree method is almost always faster than the formula on a contest. Draw first-stage branches with priors, second-stage branches with likelihoods, and multiply along each path. The total probability of the observation is the sum of the paths that end in it, and the Bayes answer is the path through the cause of interest divided by that sum. There is no formula to memorise and no branch to confuse.

Sanity checks. Posteriors over a partition must sum to $1$; if they do not, a numerator was computed with the wrong prior. A plausibility check also catches many slips: the posterior should lie between the prior and $1$ when the observation favours that cause, and it should move in the expected direction when a likelihood grows.

• Total probability: $P(A)=\sum_i P(B_i)\,P(A\mid B_i)$ for a partition $\{B_i\}$
• Bayes: $P(B_j\mid A)=\frac{P(B_j)\,P(A\mid B_j)}{\sum_i P(B_i)\,P(A\mid B_i)}$
• Posteriors over a partition always sum to $1$

• Stopping at $P(A\mid B_j)$ without dividing by the total probability of $A$.
• Ignoring the prior, for example treating every bag as equally likely when the selection rule says otherwise.

4.Independence

Two events are independent when learning that one occurred tells you nothing about the other: $P(A\mid B)=P(A)$. Because $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, this is equivalent, when $P(B)>0$, to the product rule $P(A\cap B)=P(A)\,P(B)$. The product form is the one used in calculations; the conditional form explains why it is reasonable.

Independence is a property of the probability model, not of the words used to describe it. Flips of different coins and rolls of different dice are independent because the physical processes do not interact, so the probability of any combination of outcomes factors into a product. Draws without replacement from the same bag are dependent: the first draw changes the composition of what remains.

Independence passes to complements, and this is what makes many AMC calculations short. If $A$ and $B$ are independent, then $P(A\cap B^c)=P(A)-P(A\cap B)=P(A)-P(A)P(B)=P(A)\bigl(1-P(B)\bigr)=P(A)P(B^c)$, so $A$ and $B^c$ are independent as well. The same reasoning applies to $A^c$ and $B^c$.

Mutual independence is stronger than pairwise independence. Several events are mutually independent when the product rule holds for every subfamily, not merely for pairs. Three events can be pairwise independent without being mutually independent. On the AMC the events are almost always given as mutually independent (separate flips, separate players), and then you may multiply over any subfamily.

Do not confuse independence with disjointness. Two events with positive probability that are mutually exclusive satisfy $P(A\cap B)=0\neq P(A)P(B)$, so they are never independent. Knowing that $A$ happened tells you that $B$ did not: that is strong dependence, the opposite of independence.

Independence turns 'at least one' into a complement computation. For independent events $E_i$ with probabilities $p_i$, $P(\text{none occurs})=\prod_i(1-p_i)$, so $P(\text{at least one occurs})=1-\prod_i(1-p_i)$. For 'exactly one' you add the products over the choices of which single event occurs, as the worked examples show.

• $A$ and $B$ are independent exactly when $P(A\cap B)=P(A)\,P(B)$
• Independence passes to complements: if $A,B$ are independent, so are $A$ and $B^c$
• $P(\text{at least one of independent } E_i)=1-\prod_i\bigl(1-P(E_i)\bigr)$
• Mutually exclusive events with positive probability are never independent

• Assuming independence for draws without replacement, where the first draw changes the second.
• Multiplying the probabilities of mutually exclusive events, which are in fact strongly dependent.

5.Geometric probability

Geometric probability replaces counting by measure. If a point is chosen uniformly at random from a region $R$, every subregion of equal area is equally likely to contain it, and for a subregion $F\subseteq R$ we have $P(\text{point}\in F)=\frac{\text{area}(F)}{\text{area}(R)}$. On a segment the same idea uses length. This is the continuous analogue of $\frac{|E|}{|S|}$ from counting, and the ratio is dimensionless.

Why area is the right measure. Cut $R$ into many tiny cells of equal area. The point is uniform, so each cell has the same probability of containing it, and the probability of landing in $F$ is the number of cells inside $F$ divided by the total number of cells. Taking the cells smaller and smaller turns this count into the ratio of areas. Only uniformity matters; a non-uniform density would weight parts of the region differently.

A standard recipe. First draw the region and write its boundary as inequalities. Second, identify the favourable region by its boundary lines. Third, compute its area as a polygon or as the whole region minus corner pieces. Coordinates are usually simpler than geometric reasoning: write $|x-y|<1$ as the pair of linear inequalities $x-y<1$ and $y-x<1$, and then the region is a polygon cut out by lines.

Corner triangles are the most common shortcut. In a unit square, the event $|x-y|\le c$ with $0\le c\le 1$ excludes two right isosceles triangles, each with legs $1-c$. Each has area $\frac{(1-c)^2}{2}$, so the favourable area is $1-(1-c)^2$. For $c=\frac{1}{2}$ this gives $1-\frac{1}{4}=\frac{3}{4}$. The same pattern works in any square of side $s$ with the band of half-width $c$.

Piecewise linear boundaries reduce to polygons. Probabilities involving $y<f(x)$ for a function $f$ are areas under graphs divided by the area of the region. When $f$ is piecewise linear, split the domain at its corners; the resulting pieces are trapezoids and triangles, and the area is exact with fractions.

Two independent numbers. If $a$ and $b$ are chosen independently and uniformly from $[0,1]$, the point $(a,b)$ is uniform on the unit square. Problems about two such numbers are therefore area problems in a square, and conditions like $a<b$ split the square along a diagonal, which is why answers often come out as $\frac{1}{2}$ or $\frac{3}{4}$.

Sanity checks: the probability must lie between $0$ and $1$, and the favourable region must actually sit inside the sample region. Many wrong answers come from forgetting a corner or from a favourable region that extends outside the square.

• For a uniform point in $R$: $P(\text{point}\in F)=\frac{\text{area}(F)}{\text{area}(R)}$
• On a segment the analogous ratio uses lengths
• In the unit square, $|x-y|\le c$ has favourable area $1-(1-c)^2$
• Two independent uniform numbers on $[0,1]$ give a uniform point of the unit square

• Subtracting a corner triangle from the whole region but forgetting to remove it from the right place.
• Dividing a length by an area, or forgetting that the probability is a ratio of like quantities.

6.Expected value and linearity of expectation

The expected value of a random quantity $X$ taking values $x$ with probabilities $P(X=x)$ is $E[X]=\sum_x x\,P(X=x)$. It is the long-run average of $X$ over many independent repetitions, and it need not be a possible value: the expected roll of a fair die is $\frac{7}{2}$, which no single roll ever shows.

Linearity. For any two random quantities $X$ and $Y$ defined on the same sample space, $E[X+Y]=E[X]+E[Y]$, and $E[cX]=cE[X]$ for constants $c$. Proof: $E[X+Y]=\sum_\omega\bigl(X(\omega)+Y(\omega)\bigr)P(\omega)=\sum_\omega X(\omega)P(\omega)+\sum_\omega Y(\omega)P(\omega)$. The proof uses only the definition of expectation, so no independence is needed.

Why this is powerful. The expected value of a sum is the sum of the expected values even when the summands are heavily dependent. A count of successes can be decomposed into indicator variables, each with an easy probability, and then added. You never need the full distribution of the count, which is usually the hard part.

Indicators. For an event $A$, the indicator $I_A$ equals $1$ if $A$ occurs and $0$ otherwise, and $E[I_A]=P(A)$. Writing a count as $N=\sum_i I_{A_i}$ gives $E[N]=\sum_i P(A_i)$. This is the key move for 'expected number of' questions: choose the events so that each probability is simple to compute, even if the events overlap heavily.

Expectations of functions. For any function $g$, $E[g(X)]=\sum_x g(x)\,P(X=x)$. For example, the expected larger value of two dice is $\sum_k k\,P(\max=k)$ with $P(\max=k)=\frac{2k-1}{36}$, giving $\frac{161}{36}$. Note that $E[\max]\neq\max E$ and, more generally, $E[g(X)]\neq g(E[X])$ for nonlinear $g$, so linearity cannot be applied through a $\max$ or a product.

Fair games. A game is fair when the expected net gain is $0$, and a player is favoured when that expectation is positive. A common trap is averaging over the possible payoffs without weighting them by their probabilities; the weights are essential, and they must come from the actual experiment.

Waiting times. If each trial succeeds independently with probability $p$, the number of trials until the first success has expectation $\frac1p$. This is derived with the state method below: a single state with $E=1+(1-p)E$ gives $E=\frac1p$.

• $E[X]=\sum_x x\,P(X=x)$
• $E[X+Y]=E[X]+E[Y]$ always, even when $X$ and $Y$ are dependent
• $E[I_A]=P(A)$ for an indicator $I_A$
• The expected number of trials until the first success of probability $p$ is $\frac1p$

• Assuming $E[g(X)]=g(E[X])$ for a nonlinear $g$, such as the maximum of two dice.
• Forgetting to weight outcomes by their probabilities when averaging.

7.Symmetry arguments

A symmetry argument says that two situations are identical after relabelling, so they have the same probability or the same expected value. It is the cheapest tool in the kit. If you can see that a quantity does not depend on which of several objects you consider, you need not compute anything, and the same observation often replaces pages of casework.

Exchangeability. Random variables are exchangeable when every permutation of them has the same joint distribution. A uniformly random ordering of distinct objects is the standard example: each object is equally likely to occupy each position, so the probability that a particular object lands in position $k$ is $\frac1n$ for a row of $n$ objects.

Ranking arguments. Among $m+1$ specified distinct objects arranged in uniformly random order, each is equally likely to come first, so the probability that a given one precedes all the others is $\frac{1}{m+1}$. Only the relative order of those objects matters, and all $(m+1)!$ relative orders are equally likely. The same reasoning shows that the expected position of the largest of $k$ random distinct values is determined by symmetry rather than by computation.

Expectations from symmetry. Consider a uniformly random $3$-element subset of $\{1,\dots,9\}$. The reflection $x\mapsto 10-x$ maps 3-element subsets bijectively to 3-element subsets, so it preserves the uniform distribution. Under this map the middle element $m$ becomes $10-m$. Hence $E[m]=E[10-m]=10-E[m]$, forcing $E[m]=5$, with no need to list the $84$ subsets.

Symmetry plus linearity is the strongest combination. For a uniformly random permutation of $n$ distinct numbers, each of the $\binom n2$ unordered pairs is equally likely to appear in either order, so $P(\text{pair inverted})=\frac{1}{2}$. Linearity then gives $E[\text{number of inversions}]=\binom n2\cdot\frac{1}{2}$.

Adjacency. In a uniformly random line of $n$ people, the probability that two specified people stand next to each other is $\frac{2}{n}$. Count the ways: there are $n-1$ adjacent slot pairs, each with $2$ orders, out of $n(n-1)$ ordered placements of the two people, which gives $\frac{2(n-1)}{n(n-1)}=\frac2n$.

Warning. Symmetry must be exact. If the process treats the objects differently (a biased coin, a player who moves first, a position with a privilege), the symmetry is broken and the simple answer is wrong. Before invoking symmetry, state explicitly the bijection that exchanges the cases you want to identify.

• In a uniform random permutation of $n$ distinct objects, $P(\text{object }a\text{ is in position }k)=\frac1n$
• Among $m+1$ specified objects, each is equally likely to come first, so $P=\frac{1}{m+1}$
• $E[\text{number of inversions}]=\binom n2\cdot\frac{1}{2}$ for a uniform random permutation
• A bijection that preserves the random law forces equal expectations of the images

• Invoking symmetry when the process is biased or positional, such as a first-mover advantage.
• Counting unordered pairs as ordered pairs when computing inversions or adjacencies.

8.States and first-step analysis

Many repeated games stop when something happens: a player reaches a score, a pattern appears, a counter hits a boundary. To analyse them, describe the situation by a state, meaning the smallest piece of information that determines the future. In a pattern game the state is the longest suffix of the results so far that is also a prefix of the target pattern; in a score race it is the current score.

First-step analysis. Let $E_s$ be the expected number of further steps from state $s$ until the game stops. Conditioning on the first move from $s$ gives the equation $E_s=1+\sum_t P(s\to t)\,E_t$, where the sum runs over the states $t$ reachable in one step and $E_t=0$ for absorbing (stopping) states. The $1$ is the cost of the move just made, and the rest is the expected future from wherever the move lands. Probabilities of ending in a given absorbing state satisfy the same kind of equation with boundary values $0$ or $1$.

Why the equations determine the answer. Each undecided state gives one linear equation in the unknowns of its neighbours, so there are as many equations as undecided states. The system has a unique solution because the game ends with probability $1$, which makes the matrix $I-Q$ invertible, where $Q$ records transitions among undecided states. In practice, substitution is usually fastest on the AMC.

Example: waiting for two heads in a row. Let $E_0$ be the expected remaining time with no progress and $E_1$ the expected remaining time after one head. A tail returns to $0$ and a head moves to $1$ or ends the game, so $E_0=1+\frac{1}{2}E_0+\frac{1}{2}E_1$ and $E_1=1+\frac{1}{2}E_0$. Substituting the second equation into the first gives $\frac{1}{4}E_0=\frac{3}{2}$, so $E_0=6$ and $E_1=4$.

Geometric waiting times are the simplest case: one state that stays put with probability $1-p$, giving $E=1+(1-p)E$ and so $E=\frac1p$. Coupon-collector problems use a chain of states, one for each number of distinct items already collected, and the expected time spent in each state is the reciprocal of the success probability there.

Choosing states is the real skill. The state must capture everything that matters for the future and nothing that does not. If the future depends on the last two results, the state needs both; if only the last result matters, one suffices. Merge states that have the same future, which is often exactly what symmetry does, because it keeps the system small.

Checking. Substitute the solution back into every equation, and test plausibility. An expected time must be at least the minimum number of steps needed to finish, and it should grow as the stopping pattern becomes harder to reach.

• First-step equation: $E_s=1+\sum_t P(s\to t)\,E_t$, with $E=0$ at stopping states
• A geometric wait with success probability $p$ has expectation $\frac1p$
• Each undecided state gives one linear equation, and the system has a unique solution
• The state must capture everything about the future that matters

• Forgetting the $1$ for the step just taken, so the equations omit the cost of the move.
• Choosing a state that does not determine the future, so the equations look neat but are wrong.

9.Random walks and gambler's ruin

A random walk on the integers moves up or down by one at each step according to independent coin flips. It models a gambler who wins or loses one unit per round. Most AMC walk problems ask for the probability of reaching one boundary before another, the expected duration of the game, or the probability of returning to the start, and all three have clean answers.

Gambler's ruin with a fair coin. Let $x_k$ be the probability of reaching $N$ before $0$ starting from $k$, with $0\le k\le N$. First-step analysis gives $x_k=\frac{1}{2}x_{k-1}+\frac{1}{2}x_{k+1}$ for $0<k<N$, with boundary values $x_0=0$ and $x_N=1$. The sequence has equal increments, so it is linear: $x_k=\frac kN$. The probability of winning from $k$ is simply $\frac kN$.

Expected duration with a fair coin. Let $D_k$ be the expected number of steps until absorption from $k$. First-step analysis gives $D_k=1+\frac{1}{2}D_{k-1}+\frac{1}{2}D_{k+1}$ with $D_0=D_N=0$. The solution is $D_k=k(N-k)$. To check, expand $\frac{1}{2}\bigl[(k-1)(N-k+1)+(k+1)(N-k-1)\bigr]=k(N-k)-1$, so the right side of the recursion matches $k(N-k)$.

Biased walk. With up-probability $p\neq\frac{1}{2}$ and $q=1-p$, set $r=\frac qp$. The recursion $px_{k+1}+qx_{k-1}=x_k$ has characteristic roots $1$ and $r$, so the probability of hitting $N$ before $0$ from $k$ is $\frac{1-r^k}{1-r^N}$. For $p=\frac{1}{2}$ this formula is indeterminate, and the linear answer $\frac kN$ takes over as its limit.

Harmonic functions. The fair equation $x_k=\frac{1}{2}(x_{k-1}+x_{k+1})$ says that the value at each point is the average of its neighbours. Such functions are called harmonic, and bounded harmonic functions on a segment are linear. This viewpoint explains why the answer is linear in $k$, and it is the reason the optional-stopping argument gives the same result quickly.

Returning to the origin. For the simple symmetric walk, $P(S_{2n}=0)=\binom{2n}{n}/4^n$. A classical companion identity says that the walk avoids $0$ at all the times $1,2,\dots,2n$ with the same probability, namely $\binom{2n}{n}/4^n$. For $n=1$ both sides equal $\frac{1}{2}$, which is a quick sanity check.

Practice. For each walk problem, identify the boundary values, write the first-step equation, solve the recurrence, and check that the boundary conditions hold. For a long game, verify the formula on a small case: with $N=2$ and $k=1$, the walk ends after exactly one step, and $k(N-k)=1$ agrees.

• Fair gambler's ruin: $P(\text{reach }N\text{ before }0\text{ from }k)=\frac kN$
• Fair expected duration on $\{0,\dots,N\}$ from $k$ is $k(N-k)$
• Biased walk with $r=\frac qp\neq1$: $P=\frac{1-r^k}{1-r^N}$
• For the simple symmetric walk, $P(S_1,\dots,S_{2n}\neq0)=\frac{\binom{2n}{n}}{4^n}$

• Using the fair formula $\frac kN$ for a biased walk, which needs the ratio $r$.
• Forgetting the boundary values $x_0=0$ and $x_N=1$, which are needed to pin down the solution of the recurrence.

Example 1: Two socks from a drawer

A drawer holds $4$ black socks and $2$ white socks. Two socks are taken out at random, one after the other, without replacement. What is the probability that both socks are black?

Example 2: Even product, then a sum

Two fair six-sided dice are rolled. Given that the product of the two numbers is even, what is the probability that their sum is $7$?

Example 3: Which coin was it?

A box contains two coins. Coin $P$ is fair. Coin $Q$ has heads on both faces. One coin is chosen uniformly at random from the box and flipped twice, and both flips show heads. What is the probability that the chosen coin is $Q$?

Example 4: Two sixes in a row

A fair six-sided die is rolled repeatedly. What is the expected number of rolls until a $6$ appears on two consecutive rolls for the first time?

Example 5: Descents in a random permutation

A permutation $\sigma$ of $1,2,3,4,5,6$ is chosen uniformly at random from all $6!$ orderings. A descent is a position $i$ with $1\le i\le5$ and $\sigma(i)>\sigma(i+1)$. What is the expected number of descents?

Example 6: Four doors, one host

A prize is hidden uniformly at random behind one of four closed doors. You pick one door. The host then opens one of the other doors that does not hide the prize, chosen uniformly at random among those available. You may keep your door or switch to one of the two remaining closed doors, chosen uniformly at random. What is the probability that switching wins?