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Stop the four-ball Ehrenfest urn model upon reaching count zero or four. From count $r$, the transitions are to $r-1$ with probability $r/4$ and to $r+1$ with probability $(4-r)/4$. First-step analysis with absorbing boundary values gives the displayed vector $h$ of probabilities of absorption at zero and vector $t$ of expected hitting times of either endpoint. For instance, $t_1=1+3t_2/4$ and $t_2=1+(t_1+t_3)/2$, with $t_1=t_3$, give $t_1=7$. From any interior state, absorption has probability at least $1/4$ within two steps; the resulting geometric tail bound proves both almost-sure absorption and finiteness of the expectations.

For a finite, time-homogeneous Discrete-time Markov chain with row-stochastic transition matrix $P$, a state $a$ is absorbing exactly when $P_{aa}=1$. Nonnegativity and the row sum then force $P_{aj}=0$ for $j\ne a$: once entered, $a$ is never left. Equivalently, $\{a\}$ is a closed communicating class. A larger closed communicating class may keep moving internally; closure means confinement to a set, whereas absorption at a state means remaining at one point.

An absorbing state is useful for recording a terminal outcome: after a toy routine finishes, keep its outcome in the state space so that the same transition matrix still describes subsequent steps. Calling a finite Markov chain absorbing usually also requires that some absorbing state be accessible from every state. Merely containing an absorbing state is insufficient: another closed communicating class could prevent absorption. See Grinstead and Snell, Introduction to Probability, section 11.2 for this convention.

Consider a constructed routine with unfinished modes $u,v$ and terminal outcomes $A,B$. All the following transition probabilities are chosen for illustration, not estimated from observations. In state order $(u,v,A,B)$, takeThe communicating class $\{u,v\}$ is open, while $\{A\}$ and $\{B\}$ are closed communicating classes consisting of absorbing states. Both terminal outcomes are accessible from both unfinished modes, sometimes by an indirect path.

Let $T=\inf \{n\ge 0:X_n\in \{A,B\}\}$ be the terminal hitting time. Write $h_i={\mathbb{P}}_i(X_T=A)$ for the absorption probability and $t_i={\mathbb{E}}_{i}T$ for the expected hitting time, with the subscript denoting the starting state. First-step analysis givesHere $h_A=1$, $h_B=0$, and $t_A=t_B=0$. Solving these two pairs of linear equations yieldsStarting at $u$ therefore ends at $A$ with probability $0.6$ and takes $3.2$ steps on average to reach either terminal state. The terminal hitting time counts the transition into the outcome, but no later self-transitions. Outcome and duration answer different questions: from $v$, outcome $A$ is less likely, yet termination is faster on average.

The matrix power $P^2$ gives the probability distribution after two steps from $u$ asThus ${\mathbb{P}}_u(T>2)=1/2$, and the probability of having already reached $A$ is $3/8$, still below its eventual absorption probability $3/5$. From either unfinished mode, each next step has terminal probability at least $1/4$. Consequently ${\mathbb{P}}_i(T>n)\le (3/4{)}^n$, proving termination almost surely and finite expected hitting time directly. The Fundamental matrix of a finite absorbing Markov chain packages both calculations and counts visits to unfinished modes.

For a biased random walk on the nonnegative integers with upward probability $p>q=1-p>0$, the hitting probability of zero from $i$ is $(q/p{)}^i$. Indeed, first-step analysis gives $h_i=p{h}_{i+1}+q{h}_{i-1}$ for $i\ge 1$, with $h_0=1$. Its characteristic roots are $1$ and $\beta =q/p$, so any solution is $A+(1-A){\beta }^i$. Nonnegativity as $i\to \infty$ forces $A\ge 0$ and hence a value at least ${\beta }^i$. This candidate itself satisfies the recurrence. The theorem on the minimal nonnegative solution for hitting probabilities therefore identifies it with $h_i$. A walk on all of $\mathbb{Z}$ has the same probability until it first reaches zero.

For a nearest-neighbour fortune on $\{0,\dots ,N\}$, with win probability $p$, loss probability $q=1-p$ and $p\ne q$, put $\rho =q/p$. First-step analysis gives $m_i=1+p{m}_{i+1}+q{m}_{i-1}$, with zero endpoint values. Solving the recurrence gives the displayed mean duration. Its symmetric limit is $i(N-i)$.

Take a birth-death chain on $\{0,1,2,3,4\}$. At each interior state, move up one state with probability $p=1/2$ and down one state with probability $q=1/2$. The endpoints are absorbing states: $P_{00}=P_{44}=1$. This is a small gambler's ruin model in which the state can represent capital, and reaching $4$ is success.

Let $T_a=\inf \{n\ge 0:X_n=a\}$ be the hitting time of endpoint $a$, and let $T=\min (T_0,T_4)$ be the first endpoint hitting time. The desired hitting probability is $h_i={\mathbb{P}}_i(T_4<T_0)$. First-step analysis and the Markov property giveThe interior equation says adjacent differences are equal. Four equal increments must bridge the boundary values $0$ and $1$, soFor example, from state $2$ the hitting probability is $1/2$, and its one-step average is $\frac{1}{2}(1/4)+\frac{1}{2}(3/4)=1/2$. Individual sample paths can wander back and forth, but the same averaging rule already accounts for all of them.

Now ask how long it takes to reach either endpoint. The expected hitting time $t_i={\mathbb{E}}_{i}T$ counts steps, including the transition that first reaches an endpoint. First-step analysis givesThis is a discrete Poisson equation, $(I-P)t=1$ in the interior. The extra $1$ counts the step just taken. The quadratic valuessatisfy the equations; from state $2$, $t_2=1+\frac{1}{2}(3)+\frac{1}{2}(3)=4$ steps. The hitting probability is harmonic, whereas the expected hitting time has a nonzero source term. These are special cases of expected duration of symmetric gambler's ruin; see Hao Wu's MIT lecture 16, gambler's ruin, slide 9.

For a biased random walk with $p=0.6$ upward and $q=0.4$ downward, keep the same absorbing states. The interior harmonic equation becomes $h_i=p{h}_{i+1}+q{h}_{i-1}$. Using biased gambler's ruin probability with these boundary values givesThe three interior hitting probabilities are approximately $0.415385$, $0.692308$, and $0.876923$. Thus starting at $2$ now succeeds about $69.23\%$ of the time. A direct check is $0.6(57/65)+0.4(27/65)=9/13$. The curved profile reflects the unequal transition weights: its successive increments shrink by the factor $q/p=2/3$. The formula, with the endpoint denoted $T$ there, is verified in Karger and Lynch's MIT lecture 26, sections 2.3-2.4; the five-state numbers here are direct substitutions.

Why does harmonic averaging produce the correct hitting probability? With the natural filtration ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$, the Markov property gives the conditional expectation identityFor a function harmonic everywhere, this makes $h(X_n)$ a martingale whenever it is integrable. For a function harmonic only in the interior, $M_n=h(X_{n\wedge T})$ is a stopped martingale. In either five-state example, $0\le M_n\le 1$. Also, from any interior state at most three consecutive upward steps reach $4$, with probability at least $p^3>0$. Repeated three-step blocks give ${\mathbb{P}}_i(T>3k)\le (1-p^3{)}^k$, so $T$ is almost surely finite and has finite expected value; compare finite-interval exit times have geometric tails. Applying optional stopping at $n\wedge T$ and then the bounded convergence theorem givesThe boundedness and finite hitting time justify this passage to the limit. See Steven Lalley's notes, harmonic functions and stopped expectations, theorem 1, and stopped martingale in discrete time for this bounded version of optional stopping.

Uniqueness has a short maximum principle explanation. On a finite state space, the difference of two harmonic extensions with the same boundary conditions is zero on the boundary. If its positive maximum occurs inside, the weighted average forces every successor with positive transition probability to have the same maximum. A path to the boundary propagates that maximum to zero, a contradiction. Applying the argument to the negative of the difference proves uniqueness. Reachability of the boundary is essential: an interior closed communicating class can retain an arbitrary constant.

// Target: markov-process.bigb

On $\{0,\dots ,N\}$, stop at zero, move left with probability $q>1/2$ and right with probability $p=1-q$ at interior states, and move deterministically from $N$ to $N-1$. For $m_i={\mathbb{E}}_i{T}_0$, the first-step analysis gives $m_0=0$, $m_i=1+q{m}_{i-1}+p{m}_{i+1}$ and $m_N=1+m_{N-1}$. Setting $D_i=m_i-m_{i-1}$ gives $q{D}_i-p{D}_{i+1}=1$, $D_N=1$, hence $D_i=[1-2p(p/q{)}^{N-i}]/(q-p)$. Summing these differences proves the displayed formula, and more generally $m_i=i/(q-p)-2pq(p/q{)}^{N-i}[1-(p/q{)}^i]/(q-p{)}^2$. A downward block has a uniformly positive probability of reaching zero, which supplies a geometric tail bound and justifies finiteness of these expectations.

For simple random walk on two disjoint complete graphs of size $n\ge 2$ joined through one new vertex $w$, let $h_1$ be the expected hitting time of $w$ from either vertex attached to $w$, and $h_o$ the expected hitting time from any other clique vertex. First-step analysis gives $h_1=1+(n-1)h_o/n$ and $h_o=h_1+n-1$, so $h_1=n^2-n+1$ and $h_o=n^2$. The expected hitting time from $w$ is zero. In discrete time the random walk is an aperiodic Markov chain only for $n\ge 3$.

Outside a target set, first-step analysis for a Markov chain gives $h_i=P_{ii}{h}_i+{\sum }_{j\ne i}{P}_{ij}{h}_j$. Moving the holding term and dividing gives the displayed average over the next distinct state. Thus inserting or removing self-loops preserves eventual hitting probabilities, provided the state is not made absorbing and the relative probabilities of its exits are unchanged. Boundary values are one on the target and zero in closed classes unable to reach it. The waiting time changes, but the sequence of distinct visited states does not.

Consider a random walk on $\{0,1,\dots ,n\}$ with absorbing states $0,n$. At each interior state it stays in place with probability $q$, and moves one position in either direction with probability $(1-q)/2$, where $0\le q<1$. Its mean absorption time from $k$ isProof. A sequence of at most $n$ consecutive leftward moves ensures absorption, with probability at least $((1-q)/2{)}^n>0$. Dividing time into blocks of length $n$ gives a geometric series bound on the survival probabilities, so the expected value is finite. Let $e_k={\mathbb{E}}_k[T]$. The law of total expectation and first-step analysis giveThus $e_{k+1}-2e_k+e_{k-1}=-2/(1-q)$. A particular solution of this linear recurrence relation is $-k^2/(1-q)$; its homogeneous solutions are $A+Bk$. The two boundary conditions give $A=0$, $B=n/(1-q)$, proving both the formula and its uniqueness. Holding slows the usual symmetric random walk by the factor $1/(1-q)$.

Use $T_y=\inf \{n\ge 0:X_n=y\}$ and the positive return time $T_x^{+}=\inf \{n\ge 1:X_n=x\}$. The walk has transitions $p_{uv}=c_{uv}/d(u)$, $d(u)={\sum }_v{c}_{uv}$; its transition matrix defines a reversible Markov chain, although the transition probabilities need not be symmetric.

The voltage $h(u)=P_u(T_x<T_y)$ equals one at $x$, zero at $y$, and is a harmonic function on a graph away from the terminals: $h(u)={\sum }_v{p}_{uv}h(v)$. For this unit electric potential difference the electric current isThe first-step analysis therefore proves the return probability from effective resistance identity:Finite irreducibility ensures eventual hitting of one of the terminals, so no missing escape event appears in this finite-network calculation.

For an infinite locally finite connected graph, exhaust it by finite sets and wire their exteriors to one sink. The event of exiting before the first return decreases to the event of never returning. Thuswith $1/\infty =0$. In particular recurrence is equivalent to infinite effective resistance to infinity.

After deleting an edge, consider any infinite component and choose a root $o$ away from that edge's endpoints. Its vertex degree $d(o)$ is unchanged. On every finite wired exhaustion the Rayleigh monotonicity principle makes the resistance no smaller; the limit of a sequence does too. A recurrent original network consequently leaves a recurrent component. Finite components give recurrent walks as well, with an isolated graph vertex viewed as an absorbing state. This proves that edge deletion preserves recurrence. The phrase “no less recurrent” refers to this recurrence comparison: at a deleted-edge endpoint the vertex degree factor also changes, so monotonicity of its first-return probability alone should not be inferred.

The printed hitting-time formula is false for the stated transition probabilities. For example, take $N=2$ and $q=3/4$, put $p=1-q$, and write $m_i={\mathbb{E}}_{i}T$. A first step givesso $m_2=2/q=8/3$. The printed expression gives $3$, not $8/3$. This discrepancy is present in the original PDF, not just in its converted text. The quantifier mentioning every positive integer must also be restricted to positive starting states $1,\dots ,N$; states larger than $N$ are not in $S$.

Here is the exact solution for the chain as specified. Until $T$, only $\{1,\dots ,N\}$ is visited. From any such state, a sequence of downward steps reaches zero within $N$ steps with probability at least $q^N>0$; the forced step at $N$ only increases that probability. The Markov property therefore givesThus $T$ is finite almost surely and has finite expectation, without needing the supplied transience assumption. Set $m_0=0$. The first-step analysis givesPut $d=q-p=2q-1>0$, $r=p/q$, and $D_i=m_i-m_{i-1}$. The equations becomeBackward iteration, or substitution into this recursion, givesSumming from $1$ to $i$ gives the corrected hitting time of a downward-biased walk reflected at an upper barrier:and in particularThese values also supply the appropriate test function for part (b): $f(i)=m_i$, $f(0)=0$, and $Pf=f-1$ at every state before absorption. The finite-state stopped argument is applied on $\{0,\dots ,N\}$, with zero made absorbing.

For $N\ge 3$, the printed hint cannot produce constant drift by choosing its single constant $c$. With $f(i)=i$ for $i<N$ and $f(N)=N+c$, the expected downward drift is $d$ at ordinary interior states, $d-pc$ at $N-1$, and $1+c$ at $N$. Making the top drift $d$ requires $c=-2p$, but then the drift at $N-1$ is $d+2p^2$, not $d$. For $N=2$ there is no ordinary interior state; a different constant drift can be arranged, but still gives $2/q$ rather than the printed formula. This identifies precisely the missing correction at the neighboring state for the intended drift $d$; the geometric term in the correct test function repairs it throughout the interval.

Let $m_k$ be the expected remaining hitting time of either endpoint, measured in minutes. The first step consumes one minute, after which first-step analysis givesThe expectation is finite: from any interior state, a block of twenty leftward steps guarantees absorption and has probability $2^{-20}$. Thus the survival probability over successive twenty-step blocks has a geometric upper bound. This justifies using finite expectations in the recurrence.

Assume $\mathbb{E}|g(X)|<\infty$, or use nonnegative $g$ with extended expectations. Let $p_{ij}=\mathbb{P}(X=x_i,Y=y_j)$ and $p_j=\mathbb{P}(Y=y_j)$. For $p_j>0$, the discrete conditional expectation is $\mathbb{E}[g(X)\mid Y=y_j]={\sum }_{i}g(x_i)p_{ij}/p_j$. Therefore Fubini's theorem, or Tonelli theorem in the nonnegative case, gives the tower property of conditional expectation:Values of a conditional expectation on zero-probability conditioning events do not affect the sum. An unrestricted signed $g$ whose expectation is undefined is outside this assertion.

Write $M(x)=\mathbb{E}[N(x)]$. The waiting time for two increments larger than $1/2$ dominates $N(x)$ for $0\le x\le 1$ and has mean $4$, so $M$ is finite. Conditioning on the first increment, one step is always consumed; if that increment is below $x$, the remaining independent sequence has the original law. Thus first-step analysis and the tower property of conditional expectation giveThis first makes $M$ continuous and then gives $M^{\prime }=M$, with $M(0)=1$. Hence the uniform-sum crossing time below one hasAn independent explanation uses the tail-sum formula: $\mathbb{P}(N(x)>n)=\mathbb{P}(U_1+\cdots +U_n\le x)=x^n/n!$ for $x\le 1$, since this region is an $n$-dimensional simplex inside the unit cube. Summing from $n=0$ gives $e^x$. In particular the strict-crossing convention gives $N(0)=1$ almost surely and $\mathbb{E}N(1)=e$.

A probability space consists of a sample space $\Omega$, a sigma-algebra $\mathcal{F}$ of events, and a probability measure $\mathbb{P}$ satisfying:

For arbitrary events, Boole's inequality is $\mathbb{P}({\bigcup }_j{A}_j)\le {\sum }_j\mathbb{P}(A_j)$. To prove it, replace each $A_j$ by $D_j=A_j\setminus {\bigcup }_{i<j}{A}_i$. The $D_j$ are disjoint, have the same union, and satisfy $D_j\subseteq A_j$. Countable additivity and monotonicity give $\mathbb{P}({\bigcup }_j{A}_j)={\sum }_j\mathbb{P}(D_j)\le {\sum }_j\mathbb{P}(A_j)$. Finite unions are included by taking subsequent events empty.

Let $H_i$ be the $i$th heads event. For every $m$, the event of infinitely many heads lies inside ${\bigcup }_{i\ge m}{H}_i$. By the union bound, its probability is at most ${\sum }_{i\ge m}{p}_i$, which tends to zero. Thus the probability of infinitely many heads is $0$. This is the First Borel-Cantelli lemma, and requires no independence of the coin tosses.

For the dice experiment there are $4$ ordered outcomes giving a sum of $5$ and $6$ giving $7$, out of $36$. Each trial therefore has $p_5=1/9$, $p_7=1/6$ and irrelevant-outcome probability $13/18$. By independence between trials, summing over the number of initial irrelevant outcomes givesThe geometric series also shows that one of the two relevant sums eventually appears almost surely. Equivalently, first-step analysis gives $p=p_5+(13/18)p$.

Let $N=a+b$ and measure Lionel's fortune in unit stakes. This is gambler's ruin: the fortune moves by $+1$ with probability $p$ and $-1$ with probability $q$, stopping at $0$ or $N$. Let $m_i$ be the expected value of the absorption time starting at $i$. It is finite: from any interior state, a run of $N$ wins or losses has a fixed positive probability of absorption within $N$ steps, giving a geometric bound on survival over successive blocks. First-step analysis therefore givesFor $p\ne q$, the homogeneous solution of the recurrence is $A+B(q/p{)}^i$, and a particular solution is $-i/(p-q)$. Put $\rho =q/p$. Imposing the two endpoint boundary conditions gives the expected duration of biased gambler's ruinConsequently the expected number of games isThe denominator and numerator have matching signs, so this is positive. As a check its continuous limit at $p=q=1/2$ is $ab$, the expected duration of symmetric gambler's ruin.

Let $h$ be the probability of ever hitting $c$ starting at $a$. A first-step analysis gives a direct success with probability $1/3$, a return to the same problem with probability $1/3$, and entry into the closed class $\{b,d\}$ with probability $1/3$, after which $c$ cannot be reached. ThereforeThe possibility of infinitely many self-loops has probability zero and introduces no extra solution of this equation.

Assume the intended initial fortune satisfies $n\ge 2$ and put $X_0=n$. Conditional on current fortune $i>1$, the next gift is uniform on $\{1,\dots ,i-1\}$ and independent of the previous choices. The remaining fortune is therefore uniform on that same set. At fortune $1$ the process stays there. Consequently the future conditional distribution depends only on the current fortune, proving the Markov property, and the transition matrix on $\{1,\dots ,n\}$ isThe uniform decreasing Markov chain has state $1$ as an absorbing state, and until absorption the fortune strictly decreases, so the hitting time is at most $n-1$.

Let $e_i$ be the expected number of additional transitions to hit $1$, taking $e_1=0$. First-step analysis givesIn particular $e_2=1$. For $i\ge 3$, multiplying this recurrence by $i-1$ and the recurrence for $i-1$ by $i-2$, then subtracting, gives $(i-1)e_i=(i-1)e_{i-1}+1$. Hence $e_i-e_{i-1}=1/(i-1)$ for every $i\ge 2$, and telescoping yieldsThe PDF defines $\tau$ using times $i\ge 1$. If $n=1$ were included, that definition would give $\tau =1$ although the displayed empty sum is zero; the initial instruction to choose an integer between $1$ and $n-1$ also requires $n\ge 2$. The result above uses that intended assumption. A hitting time allowing time zero would have expectation zero from state $1$, but that is a different definition.

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

Measure vertices by their Hamming distance from the target and let $h_k$ be the expected hitting time when that distance is $k$. We already know $h_1=2^n-1$. From distance one, one of the $n$ coordinate flips reaches the target and the other $n-1$ flips move to distance two. First-step analysis therefore givesFor $n\ge 2$,

Let $h_k$ be the probability that the simple random walk started at $k$ hits $n$ before zero. The first-step analysis recurrence and boundary conditions areIf $p\ne q$, the characteristic equation of a linear recurrence has roots $1$ and $q/p$, so the boundary conditions giveIf $p=q=1/2$, the repeated root gives the limiting formula $\boxed{h_m=m/n}$. This is the gambler's ruin probability.

For a fixed integer stake $a\le N$, the fact that $a$ divides $N!$ lets us measure Patricia's wealth in units of £$a$. She starts at $M=N!/a$ and succeeds upon reaching $2M$. The formula above factors asFor $p=18/37$, $r=19/18>1$, so this probability increases as $M$ decreases. She should therefore choose the largest stake, $\boxed{a=N}$. For $p=19/37$, $r=18/19<1$, so the probability increases as $M$ increases; she should choose the smallest stake, $\boxed{a=1}$.