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For $x>0$, reflect a Brownian path after its first hitting of $x$. The Strong Markov property and symmetry preserve its law and exchange the events $\{S_1\ge x,B_1<x\}$ and $\{B_1>x\}$. The terminal equality event has probability zero, so the Brownian reflection principle givesContinuity of this distribution also gives no atom at zero. ThereforeA useful global bound follows from the maximum of the standard normal distribution density:

Monotonicity of $f$ gives, for each $n\ge 1$,Hence ${\sum }_{n\ge 1}f(2^{-n})<\infty$; the additional $n=0$ term is finite by continuity at one. Also $f(t)\to 0$ as $t\downarrow 0$, since a positive limiting lower bound would make the integral diverge.

By Brownian scaling, $S_{2^{-n-1}}$ has law $2^{-(n+1)/2}{S}_1$. Apply the small-value bound for the Brownian running maximum to obtainSumming yields the requested finite probability sum. The argument needs no independence of these nested-time events.

At level $n$, the $N_n$ dyadic cells fully contained in $J$ are disjoint. Their occupied indicators are independent by the preceding part, and every cell has length $2^{-n}$, so each has success probabilityTheir sum therefore satisfiesIf there are no fully contained cells, $N_n=0$ and this means the deterministic zero count. The binomial law follows directly from independence of occupation indicators, without assuming that the full counts $M(D_k^n)$ are already Poisson.

Write $w_{xy}=w_{yx}>0$ for an edge conductance and $w_x={\sum }_{y\sim x}{w}_{xy}$. Take $s\ne t$, as implicit in a unit source-to-sink electrical network. An oriented current flow is antisymmetric, $i_{yx}=-i_{xy}$. Kirchhoff's first law is conservation of current: ${\sum }_y{i}_{xy}=0$ at every vertex other than the source and sink, with the divergences there equal to the prescribed total flow and its negative. Kirchhoff's second law says that the sum of oriented voltage drops around any closed cycle is zero. Ohm's law isThus the resistance of an edge is $1/w_{xy}$, and the cycle law reads ${\sum }_{j=0}^{m-1}{i}_{x_j{x}_{j+1}}/w_{x_j{x}_{j+1}}=0$ for $x_m=x_0$.

Let $\tau =\inf \{n\ge 0:X_n=t\}$. Its expectation is finite. Indeed, from each state there is a path to $t$ of length at most $|V|-1$ with positive probability; finiteness gives a common lower bound $a>0$ on reaching $t$ within that many steps. The Markov property then gives $\mathbb{P}(\tau >k(|V|-1))\le (1-a{)}^k$, proving the assertion. Define the expected occupation count before absorptionConditional expectation at every pre-stopping visit to $x$ gives ${\mathbb{E}}_s{N}_{xy}=G(s,x)p_{xy}$ for the number $N_{xy}$ of directed traversals. Hence the expected signed count isThis proves Ohm's law with voltage $v_x=G(s,x)/w_x$, grounded at $v_t=0$. Summing its voltage drops along a cycle telescopes, proving Kirchhoff's second law.

For Kirchhoff's first law, count entrances and exits on each stopped path. Their net difference at $x$ is the initial indicator minus the final indicator. Taking expectations,Thus the source emits exactly one unit, the sink receives one unit, and all other vertices conserve current.

We also establish the uniqueness required for the conclusion. The cycle law implies that edge voltage drops are differences of vertex potentials: define a potential by summing drops along a path to $t$, and cycle sums make this independent of the path. If two unit flows obey both laws, their difference has the form $j_{xy}=w_{xy}(h_x-h_y)$ with zero divergence at every vertex. ThereforeStrict positivity of the conductances and connectedness force $h$ to be constant, so $j=0$. We conclude the electrical current from stopped random-walk traversals identity

Let $o(\omega )$ be the number of open edges, $k(\omega )$ the number of connected components in the open subgraph, counting isolated vertices, and $m=|E|$. Summing the joint weight over spins requires a constant spin on every open component. There are exactly $q^{k(\omega )}$ choices. Thus the edge marginal isThis is the random-cluster measure, and the joint normalizing constant is $Z_{\mathrm{RC}}$.

Summing instead over edge configurations factors independently over edges. For a fixed spin configuration the resulting weight issince $1-p=e^{-\beta }$. The factor $e^{-\beta m}$ is independent of the spins. Hence the spin marginal is exactly the ferromagnetic Potts measure in the conventionIts partition function satisfies $Z_{\mathrm{RC}}=e^{-\beta m}{Z}_{\mathrm{P}}$.

Conditionally on $\sigma$, edge states are independent: an edge with different endpoint spins is closed with probability one, while one with equal endpoint spins is open with probability $p$. Conditionally on $\omega$, color its $k(\omega )$ open components independently and uniformly from $\{1,\dots ,q\}$. Equivalently, every compatible spin configuration has conditional probability $q^{-k(\omega )}$, and all incompatible configurations have probability zero. These are the two conditional laws of the Edwards-Sokal coupling.

If $x,y$ belong to the same open component, their colors agree surely; otherwise the two independent component colors agree with probability $1/q$. ThereforeRearranging proves the spin-connectivity identity for the Potts model:

Phase transitions refer to infinite-volume limits, not a single finite graph. The identity shows that connectivity is exactly the excess spin agreement above its independent-color baseline. More directly, use a wired random-cluster boundary condition and fix the corresponding boundary spin to color $1$. Its component has color $1$, while every finite component has a uniform color. In the compatible infinite-volume coupling,Consequently normalized spontaneous magnetization isThe onset of a wired infinite cluster and the onset of spontaneous magnetization in the Potts model are the same transition under $p=1-e^{-\beta }$, giving ${\beta }_c(q)=-\log (1-p_c(q))$. This also transfers any jump or continuity of that order parameter between the two descriptions; it makes no assumption about behavior exactly at the critical point.

On the coordinatewise ordered configuration space, ${\mu }_1$ stochastically dominates ${\mu }_2$ when $\int fd{\mu }_1\ge \int fd{\mu }_2$ for every increasing real function $f$. Equivalently, ${\mu }_1(A)\ge {\mu }_2(A)$ for every increasing event. The Holley condition is the sufficient cross-lattice inequalityHere join and meet are coordinatewise maximum and minimum. This condition implies ${\mu }_1{\ge }_{\mathrm{st}}{\mu }_2$ by the Holley inequality; it is not asserted to be necessary. It can be checked on unnormalized weights, since their normalization constants cancel.

Use the weightsFor the upper bound, take ${\mu }_1={\varphi }_{p,1}$ and ${\mu }_2={\varphi }_{p,q}$. Since open-edge count is modular, the Bernoulli factors cancel in the Holley ratio, leavingThe last inequality uses $q\ge 1$ and the fact that removing edges cannot reduce the number of components. Thus ${\varphi }_{p,q}{\le }_{\mathrm{st}}{\varphi }_{p,1}$.

For the lower bound, let $p^{\prime }=p/[p+q(1-p)]$. Its odds satisfySet $a=o(\omega \vee \eta )-o(\omega )=o(\eta )-o(\omega \wedge \eta )$, the number of edges added to $\omega$ in the union. For ${\mu }_1={\varphi }_{p,q}$ and ${\mu }_2={\varphi }_{p^{\prime },1}$, the corresponding ratio isAdding one edge either merges two components or leaves the count unchanged. Therefore adding $a$ edges reduces $k$ by at most $a$, proving the exponent nonnegative. This also proves the hinted monotonicity of $k(\omega )+o(\omega )$ rather than assuming it. The Holley condition now proves both Bernoulli bounds for the random-cluster model:

The same argument holds with wired boundary conditions: identify the wired boundary vertices first and count components on that quotient graph. Taking finite-volume wired limits preserves every increasing cylinder-event inequality. Approximating $\{0\leftrightarrow \infty \}$ by the decreasing events $\{0\leftrightarrow \partial [-n,n{]}^d\}$ consequently givesLet $a=p_c(1)$. For $p<a$, the upper bound makes the wired percolation probability zero, so $p_c(q)\ge a$. If $a<1$ and $p^{\prime }>a$, the lower bound makes it positive. Solving this strict inequality,Taking infima provesFor $a=1$, the lower bound and the trivial upper bound $p_c(q)\le 1$ give the same conclusion. No assumption about percolation at the critical parameter is required.

Let $m=|V|\ge 1$ and put $H(\sigma )={\sum }_{\{x,y\}\in E}{\sigma }_x{\sigma }_y$, counting each edge once. If only the spin at $x$ changes from $-1$ to $+1$, the change in $H$ is $2N(x)$, soThe given opposite transition probabilities have the same ratio. More explicitly,For pairs differing at more than one site both sides vanish, and for equal configurations balance is automatic. This proves detailed balance, hence the chain is reversible with stationary law $\pi$. All single-spin changes have positive probability, so the finite chain is irreducible; it also has positive holding probability and is aperiodic.

Its random-map version is a heat-bath Markov chain. Choose a vertex $I$ uniformly and a uniform $U\in [0,1]$, independently, and set its spin to $+1$ whenand to $-1$ otherwise, leaving all other spins unchanged. This also specifies the holding probabilities omitted from the off-diagonal formulas. The conditional Ising law at $I$ is exactly this Bernoulli choice.

Order spin configurations coordinatewise with $-1<+1$. Since both the neighbour sum and $h$ are increasing, every map defined by a fixed $(I,U)$ preserves this order. For monotone coupling from the past, generate one two-sided independent sequence of such maps $F_n$, $n\in \mathbb{Z}$, and, for a horizon $T$, apply $F_{-T},\dots ,F_{-1}$ to the all-minus and all-plus states at time $-T$. If their time-zero images agree, every initial state has that same image by monotonicity. Return it. Otherwise enlarge $T$, for example by doubling it, retaining all maps already generated at existing negative times. Any earlier starting configuration is also forced to the same output once a horizon coalesces, so the result is consistent under further backward extension.

To prove termination, let $D={\max }_x\mathrm{deg}(x)$ andList the vertices $v_1,\dots ,v_m$. A block of $m$ updates choosing them in that order and having every uniform less than $a$ sets every vertex to $+1$, regardless of the starting configuration. Each choice has probability $1/m$, and every uniform condition has probability $a$. Thus a block is a universal reset with probabilityIndependent disjoint blocks in the past giveOnce such a block occurs, later maps preserve agreement. Therefore the backward search terminates in finite time almost surely, including the doubling implementation.

Finally we prove that the common output $Y$ has law $\pi$. For a deterministic horizon $T$, independently start another chain at time $-T$ with initial law $\pi$ and use the same maps. Its time-zero state $Z_0^{(T)}$ has law $\pi$ by stationarity. On the event that this deterministic horizon coalesces, $Z_0^{(T)}=Y$, so for every event $B\subseteq \Sigma$,Letting deterministic $T\to \infty$ provesThis argument justifies exact sampling without treating a random, data-dependent horizon as a deterministic stationary run.

For a $d$-dimensional normed vector space, with unit ball $K$ and John ellipsoid $\mathcal{E}$, its volume ratio isIt is at least $1$ and invariant under invertible linear coordinate changes.

For ${\ell }_1^n$, the unit ball has volume $2^n/n!$: its intersection with each orthant is the standard simplex of volume $1/n!$. Its John ellipsoid is $n^{-1/2}{B}_2^n$. To verify maximum volume, let $A{B}_2^n$ be any centered inscribed ellipsoid. For each sign vector $\epsilon \in \{-1,1{\}}^n$, containment gives $|A^T\epsilon |\le 1$. Averaging over independent Rademacher random variables yields $\mathrm{tr}(A{A}^T)\le 1$. The arithmetic-geometric mean inequality for its squared singular values gives $|\det A|\le n^{-n/2}$. The scalar matrix $n^{-1/2}I$ attains this bound because $\Vert x{\Vert }_1\le \sqrt{n}|x|$. Centered ellipsoids suffice: if $a+A{B}_2^n$ is contained in the symmetric convex unit ball, averaging it with $-a+A{B}_2^n$ gives the centered copy with the same volume.

Using the stated Euclidean ball volume and $n=2k$,since $(2k)!/k!={\prod }_{j=k+1}^{2k}j\ge k^k$. Therefore $\mathrm{vr}({\ell }_1^{2k})\le \sqrt{8/\pi }$ uniformly in $k$.

The subsequent assertion for an arbitrary $2k$-dimensional space is false as printed. Here is a counterexample independent of any asymptotic section theorem. Take $E={\ell }_{\infty }^{2k}$. Its John ellipsoid is $B_2^{2k}$: an inscribed $A{B}_2^{2k}$ has every row of $A$ of Euclidean length at most $1$, so the Hadamard inequality gives $|\det A|\le 1$, attained by $A=I$. Let $F$ be any $k$-dimensional subspace and $f_1,\dots ,f_k$ an orthonormal basis of it. The random unit vectorhas $|x|=1$. Since ${\sum }_j{f}_j(i{)}^2\le 1$ for each coordinate $i$, the moment-generating function bound for independent Rademacher random variables givesChoose $r=\sqrt{2\log (8k)/k}$; the latter probability is at most $1/2$, so some unit vector in $F$ has $\Vert x{\Vert }_{\infty }\le r$. The proposed section inequality would therefore forceThus even one half-dimensional section cannot have the printed universal constant.

The valid orthogonal splitting with bounded volume ratio is that $L$ may depend on $V=\mathrm{vr}(E)$, and $L=32V^2$ suffices. This supplies a dimension-independent constant for the preceding ${\ell }_1^{2k}$ case, or whenever $V$ is bounded uniformly. We prove the complete qualified assertion, including its epsilon-net ingredient.

In John position, $\Vert x{\Vert }_E\le |x|$. Polar integration of the radial function $\rho (u)=1/\Vert u{\Vert }_E$ giveswhere $\sigma$ is normalized spherical measure.

A maximal $\epsilon$-separated subset of $S^{k-1}$ is an epsilon-net. Balls of radius $\epsilon /2$ about its points have disjoint interiors and lie in the ball of radius $1+\epsilon /2$. Comparing volumes bounds its cardinality by $(1+2/\epsilon {)}^k$. Finite existence follows from this packing bound, and maximality proves the net property without assuming a separate covering theorem.

Choose two orthogonal $k$-dimensional subspaces $F_0,F_0^{\perp }$ and one such net on each unit sphere. Rotate both by a common uniformly distributed orthogonal matrix $U$. This distribution can be obtained from Question 1 applied to the compact orthogonal group with its Frobenius metric; left and right multiplication are transitive isometries. Consequently $Uv$ is uniformly distributed on $S^{2k-1}$ for every fixed unit vector $v$. The union bound shows that the probability that some rotated net point has norm less than $2\epsilon$ is at mostSet $\epsilon =(32V^2{)}^{-1}$. Because $V\ge 1$, the bracket is at most $1/4+1/256$, and twice its $k$th power is less than $1$ for every $k\ge 1$. Some rotation therefore makes every point of both nets have norm at least $2\epsilon$.

The triangle inequality and $\Vert x{\Vert }_E\le |x|$ imply that $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric. Every unit vector of $U{F}_0$ or $U{F}_0^{\perp }$ is within $\epsilon$ of a good net point and thus has norm at least $\epsilon$. By homogeneity, on both orthogonal subspaces,For ${\ell }_1^{2k}$ the first computation gives the absolute choice $L=256/\pi$. The missing bounded-volume ratio restriction is essential; the literal unrestricted assertion cannot be proved.

For a discrete memoryless channel, its channel capacity in bits per use is $C={\max }_{P_X}I(X;Y)$. The Shannon second coding theorem identifies this with the supremum of rates achievable by arbitrarily long block codes with error probability tending to zero: every rate below $C$ is achievable, and no rate above $C$ can have vanishing error. The channel is used independently at each transmission, and coding rate is measured using base-two logarithms.

Let $q$ be the probability of using the second input symbol. The second output symbol occurs with probability $q/2$, so $H(Y)=h_2(q/2)$. The first input has deterministic output and the second has one bit of conditional output Shannon entropy; hence $H(Y\mid X)=q$ andFor $0<q<1$, its derivative is $\frac{1}{2}{\log }_2[(2-q)/q]-1$, which vanishes at $q=2/5$. Strict concavity of binary entropy makes this the unique global maximum. Substituting givesThus Shannon's theorem supplies operational coding rates arbitrarily close to this value from below, and rules out reliable transmission above it.

The first Borel-Cantelli lemma says that ${\sum }_n\mathrm{Pr}(A_n)<\infty$ implies $\mathrm{Pr}(A_n\text{ infinitely often})=0$, without an independence assumption. Indeed, for every $N$,The decreasing intersection of these tail unions is the infinitely-often event; continuity of probability from above proves the assertion. The second lemma says that for independent $A_n$, divergence of the probability sum implies $\mathrm{Pr}(A_n\text{ infinitely often})=1$.

Convergence in probability supplies a subsequence $X_{n_k}$ converging almost surely to $X$: choose it so $\mathrm{Pr}(|X_{n_k}-X|>2^{-k})<2^{-k}$, then apply the first Borel-Cantelli lemma. For each fixed $N$, the tail of this subsequence lies in $\sigma (X_n:n\ge N)$, and its almost sure limit is $X$. Thus $X$ has a version measurable with respect to the completed tail sigma-field of the independent sequence.

By Kolmogorov's zero–one law, every event in that tail sigma-field has probability zero or one. Consequently $\mathrm{Pr}(X\le r)\in \{0,1\}$ for every rational $r$. Since $X$ is a finite real random variable, its distribution function has limits zero and one at the two infinities; monotonicity then identifies a single threshold $c$ with $\mathrm{Pr}(X=c)=1$. Hence $X$ is almost surely constant. Completion only adds null events and does not alter the zero-one conclusion.

Under complete convergence of random variables, for each integer $k\ge 1$ the first Borel-Cantelli lemma gives $\mathrm{Pr}(|X_n-X|>1/k\text{ infinitely often})=0$. Intersecting these probability-one conclusions over the countably many $k$ shows that every tolerance is eventually satisfied. Therefore complete convergence implies almost sure convergence, without independence.

Almost sure convergence implies convergence in probability, so part (b) identifies the limit as a constant $c$ almost surely. For each $ϵ>0$, the events $A_n=\{|X_n-c|>ϵ\}$ are independent. Almost sure convergence gives $\mathrm{Pr}(A_n\text{ infinitely often})=0$. If their probability sum diverged, the second Borel-Cantelli lemma would instead give probability one. Hence the sum is finite for every $ϵ$, proving almost sure convergence implies complete convergence for independent sequences.

When $R=0$ all rates vanish. Every state is absorbing and the communicating classes are the $s$ singletons, all closed. There are no open communicating classes. Every probability distribution is stationary in this degenerate case.

At a state with positive departure rate ${\lambda }_i=R-r_i$, the next-jump probabilities areIf at least two rates are positive, every state has positive departure rate and the normalizing constant $Z={\sum }_i{r}_i(R-r_i)=R^2-{\sum }_i{r}_i^2$ is positive. The jump-chain equilibrium law isIndeed ${\hat{\pi }}_i{\hat{p}}_{ij}=r_i{r}_j/Z$ is symmetric, proving both stationarity and reversibility. Jump observations weight the continuous-time equilibrium by the departure rate, so $\hat{\pi }$ is generally different from $\pi$.

If exactly one rate, $r_k$, is positive, state $k$ is absorbing and the denominator $R-r_k$ vanishes. There is no next jump from that state: the genuine embedded jump sequence terminates there. Under the usual extended-chain convention set ${\hat{p}}_{kk}=1$; every other state jumps to $k$ with probability one. Its equilibrium law is ${\delta }_k$ and it is trivially reversible there. The normalized $r_i(R-r_i)$ formula cannot be used in that absorbing case.

The unrestricted categorical MLE is ${\tilde{\pi }}_j=N_j/n$, so the multinomial combinatorial factor cancels from the likelihood ratio andAssume the true cell probabilities are positive and the null has identifiable dimension $d$ with full-rank information. With probability tending to one, all $N_j$ are positive. Put ${\delta }_j=n{\hat{\pi }}_j-N_j$. Regular estimation gives ${\delta }_j=O_p(\sqrt{n})$ and $N_j$ of order $n$. Expanding $-2N_j\log (1+{\delta }_j/N_j)$ yieldsThe linear terms cancel exactly because ${\sum }_j{\delta }_j=0$. ThereforeThe full probability simplex has dimension $K-1$, explaining the degrees of freedom. Existence and uniqueness of an MLE alone do not supply all Wilks regularity assumptions; positivity, smoothness and the local rank condition are also needed. The displayed approximation is not a finite-sample prescription for division by a zero observed count.

An arbitrage is a portfolio with nonpositive initial cost, nonnegative terminal payoff almost surely, and a strict benefit either from a negative initial cost or a positive payoff with positive probability. An equivalent martingale measure is a probability law equivalent to the physical law under which discounted traded prices are martingales. The bond has constant price one, so no discount factor is needed here.

Order the four states as $UU,UD,DU,DD$. Equivalence requires positive probability in every state, since the physical product law has full support. Each stock has risk-neutral up probability $1/2$, obtained by solving $1=(1+ϵ)p+(1-ϵ)(1-p)$. Thus all equivalent martingale laws areThe two risky returns need not be independent under a pricing measure.

A replicating portfolio pays $b+\theta S_2+\phi S_3$, so its four payoffs satisfyConversely this condition suffices: take $\theta =(Y_{UU}-Y_{DU})/(2ϵ)$, $\phi =(Y_{UU}-Y_{UD})/(2ϵ)$, and choose the bond holding to match $Y_{DD}$. The relation then matches the fourth state as well. Its unique price is $(Y_{UU}+Y_{DD})/2=(Y_{UD}+Y_{DU})/2$.

For the same-direction claim, the payoff vector is $(1,0,0,1)$ and its price under $q$ is $2a$. Thus the possible strictly arbitrage-free prices fill $(0,1)$ and the lower arbitrage bound is the infimum zero. It is not achieved by an equivalent measure, and pricing the nonnegative nonzero claim at zero would itself permit arbitrage by buying it. Its payoff is not replicable, since the displayed payoff relation fails. A buyer may pay a positive amount for its insurance against joint downward moves, for a view about dependence, or according to risk preferences. Under the stated physical independent law its expected payoff is $13/25$, illustrating that a zero lower bound does not make the payoff economically worthless.

First-season recruits in year $n+1$ contribute $\alpha \sigma \gamma X_n$. Seeds produced in year $n-1$ that survive two winters, fail first-season germination and germinate in their second season contribute $\beta {\sigma }^2(1-\alpha )\gamma X_{n-1}$. HenceThis treats recruitment from older seeds as absent; otherwise the two-season recurrence would need further terms. The characteristic polynomial is $r^2-ar-b$. For distinct roots,with constants determined by two initial values. For $b=0,a>0$, the recurrence is first order after the initial step; for $a=b=0$ it gives $X_{n+1}=0$. These describe the zero-root degeneracies without interpreting $0^0$ in the formula.

Since $a,b\ge 0$, $|r_{-}|\le r_{+}$. The positive root is less than one exactly when $a+b<1$: the polynomial is nonpositive at zero, has only one positive root, and its value at one is $1-a-b$. ThereforeThis is the decay condition for the two-season seed-bank recurrence. The deterministic model describes expected or averaged abundance; literal eventual integer extinction requires that interpretation. A corresponding branching model with geometrically decaying expected abundance becomes extinct almost surely, since the probability of any surviving plants is at most the expected number.

This is a gambler's ruin hitting problem. For $0<p<1$, let $h_j$ be the probability of reaching $+3$ before $-3$ when the current difference is $j$. Conditioning on the next independent game givesStopping occurs almost surely: conditional on being inside the barriers, three consecutive wins by either player force absorption within the next three games. This event has probability $p^3+q^3>0$, so the chance of surviving $m$ such blocks is at most $(1-p^3-q^3{)}^m$, which tends to zero.

If $p\ne q$, the characteristic roots of the interior recurrence are $1$ and $r=q/p$, so $h_j=A+B{r}^j$. Imposing the two boundary values yieldsIn particular,For $p=q=1/2$ the recurrence instead has affine solutions, and the boundary values give $h_j=(j+3)/6$, hence $h_0=1/2$. The deterministic cases $p=0,1$ give zero and one respectively. The formula covers all of them continuously:The stopping argument and the boundary recurrence show why the answer concerns the first barrier reached rather than the win difference after a predetermined number of games.

Put $q=1-p$ and track the difference $S_n$ between the numbers of wins. It is a biased random walk with increments $+1$ and $-1$, stopped on first reaching $3$ or $-3$.

To finish at $+3$ on game five, there must be four wins for $A$ and one for $B$, with the final game won by $A$. The unique loss for $A$ can be in position one, two or three. If it were in position four or five, the first three games would already have stopped the contest. Each of the three allowed paths remains strictly inside the barriers until the last step and has probability $p^{4}q$. By interchanging players, there are also three paths ending at $-3$, each with probability $p{q}^4$. ThereforeFor a fair game this is $3/16$. Counting all five-game sequences with final win difference three would overcount paths on which the contest ended earlier.