Probability & Statistics

Part (i).

Within each game there are three points. Xavier wins point 1 with probability $p$; thereafter, whoever won the previous point wins the next with probability $p$, and loses it with probability $1-p$. The match ends as soon as one player takes two consecutive points.

Tracing all game outcomes. Condition on who wins point 1.

*Xavier wins point 1* (probability $p$):
- Xavier wins point 2 (prob $p$): sequence XX — Xavier wins the match. Overall probability $p^2$.
- Younis wins point 2 (prob $1-p$): now Younis has the advantage.
- Younis wins point 3 (prob $p$): sequence XYY — Younis wins. Probability $p(1-p)\cdot p=p^2(1-p)$.
- Xavier wins point 3 (prob $1-p$): sequence XYX — neither player has two consecutive; the game is replayed. Probability $p(1-p{)}^2$.

*Younis wins point 1* (probability $1-p$):
- Younis wins point 2 (prob $p$): sequence YY — Younis wins. Probability $(1-p)p$.
- Xavier wins point 2 (prob $1-p$): Xavier now has the advantage.
- Xavier wins point 3 (prob $p$): sequence YXX — Xavier wins. Probability $(1-p{)}^{2}p$.
- Younis wins point 3 (prob $1-p$): sequence YXY — replayed. Probability $(1-p{)}^3$.

Collecting probabilities:$$P(\text{Younis wins one game})=p^2(1-p)+(1-p)p=p(1-p)(1+p),$$$$P(\text{rematch})=p(1-p{)}^2+(1-p{)}^3=(1-p{)}^2(p+(1-p))=(1-p{)}^2.$$Recurrence. Since every game is identically distributed, $w=P(\text{Younis wins match})$ satisfies$$w=p(1-p)(1+p)+(1-p{)}^2\cdot w.$$For $p\ne 0$, rearrange using $1-(1-p{)}^2=p(2-p)$:$$w=\frac{p(1-p)(1+p)}{p(2-p)}=\frac{(1-p)(1+p)}{2-p}=\frac{1-p^2}{2-p}.$$Comparing $w$ with $\frac{1}{2}$: compute$$w-\frac{1}{2}=\frac{2(1-p^2)-(2-p)}{2(2-p)}=\frac{p-2p^2}{2(2-p)}=\frac{p(1-2p)}{2(2-p)}.$$For $0<p<1$, the factor $p>0$ and $2-p>0$, so the sign is determined entirely by $(1-2p)$. Hence $w>\frac{1}{2}$ when $p<\frac{1}{2}$, and $w<\frac{1}{2}$ when $p>\frac{1}{2}$.

Monotonicity of $w$ in $p$: differentiate:$$\frac{dw}{dp}=\frac{-2p(2-p)+(1-p^2)}{(2-p{)}^2}=\frac{p^2-4p+1}{(2-p{)}^2}.$$The numerator vanishes at $p=2-\sqrt{3}\approx 0.27$. For $0<p<2-\sqrt{3}$, the numerator is positive, so $\frac{dw}{dp}>0$ — meaning $w$ is increasing in $p$ on this interval, i.e. decreasing $p$ actually decreases $w$. Therefore no, $w$ does not always increase when $p$ decreases.

Part (ii).

With $p=\frac{2}{3}$:$$w=\frac{1-\frac{4}{9}}{2-\frac{2}{3}}=\frac{\frac{5}{9}}{\frac{4}{3}}=\frac{5}{12}.$$For a fair game, Younis's expected gain is zero. He wins $\pounds k$ with probability $\frac{5}{12}$ and pays $\pounds 1$ with probability $\frac{7}{12}$:$$0=\frac{5}{12}k-\frac{7}{12}⟹k=\frac{7}{5}=\mathrm{1.4.}$$Part (iii).

When $p=0$, the player who won the previous point wins the next with probability $0$ — so the outcome of each point strictly alternates. Within every game, no player ever secures two consecutive points, so every game ends in a rematch. The rematch probability equals $(1-p{)}^2=1$, confirming that the match never ends.

Part (i).

An alternating run of length 1 means the first two tosses show the same face (so the alternation breaks immediately after the first toss). The probability is$$\mathrm{P}(A=1)=p^2+q^2,$$since both tosses are heads (probability $p^2$) or both tails (probability $q^2$).

A straight run of length 1 means the first two tosses differ. The probability is$$\mathrm{P}(S=1)=pq+qp=2pq.$$Now $p^2+q^2-2pq=(p-q{)}^2>0$ since $p\ne q$, so $\mathrm{P}(S=1)<\mathrm{P}(A=1)$.

Part (ii).

A straight run of length exactly 2 requires two identical tosses followed by a different one:$$\mathrm{P}(S=2)=p^{2}q+q^{2}p=pq(p+q).$$An alternating run of length exactly 2 requires two alternating tosses followed by a repeat of the second:$$\mathrm{P}(A=2)=pqq+qpp=p{q}^2+q{p}^2=pq(p+q).$$Hence $\mathrm{P}(S=2)=\mathrm{P}(A=2)$.

For length 3: a straight run of length exactly 3 is three identical then one different, giving $p^{3}q+q^{3}p$. An alternating run of length exactly 3 is three alternating then a repeat of the third toss. The two patterns (starting with heads or tails) give $p\cdot q\cdot p\cdot p+q\cdot p\cdot q\cdot q=p^{3}q+q^{3}p$. Hence $\mathrm{P}(S=3)=\mathrm{P}(A=3)$.

Part (iii).

For a straight run of length $2n$: exactly $2n$ identical tosses then one different:$$\mathrm{P}(S=2n)=p^{2n}q+q^{2n}p=pq(p^{2n-1}+q^{2n-1}).$$For an alternating run of length $2n$: the $2n$ tosses alternate, then the $(2n+1)$th matches the $2n$th. A sequence of $2n$ alternating tosses starting with heads consists of $n$ heads and $n$ tails (since $2n$ is even, it ends on tails), and has probability $p^n{q}^n$; the next toss must be tails, contributing an extra factor $q$. By symmetry:$$\mathrm{P}(A=2n)=p^n{q}^n\cdot q+p^n{q}^n\cdot p=p^n{q}^n(p+q).$$Now compute the difference:$$\mathrm{P}(S=2n)-\mathrm{P}(A=2n)=pq(p^{2n-1}+q^{2n-1})-p^n{q}^n(p+q).$$Factoring out $pq$:$$=pq[p^{2n-1}+q^{2n-1}-p^{n-1}{q}^{n-1}(p+q)]=pq[p^{n-1}(p^n-q^n)-q^{n-1}(p^n-q^n)],$$where the last step uses $p^{2n-1}-p^{n-1}{q}^n=p^{n-1}(p^n-q^n)$ and similarly for the $q$ terms. Thus:$$\mathrm{P}(S=2n)-\mathrm{P}(A=2n)=pq(p^{n-1}-q^{n-1})(p^n-q^n).$$Since $p\ne q$, both factors $(p^{n-1}-q^{n-1})$ and $(p^n-q^n)$ are nonzero and have the same sign (both positive if $p>q$, both negative if $p<q$). Hence their product is strictly positive, and$$\mathrm{P}(S=2n)>\mathrm{P}(A=2n).$$For odd length $2n+1$ (with $n>1$): a straight run gives $\mathrm{P}(S=2n+1)=p^{2n+1}q+p{q}^{2n+1}$, whilst an alternating run of length $2n+1$ has $n+1$ of the starting face and $n$ of the other, ending on the starting face; the next toss repeats it. Computing both starting cases:$$\mathrm{P}(A=2n+1)=p^{n+2}{q}^n+p^n{q}^{n+2}=p^n{q}^n(p^2+q^2).$$The difference is:$$\mathrm{P}(S=2n+1)-\mathrm{P}(A=2n+1)=pq(p^{2n}+q^{2n})-p^n{q}^n(p^2+q^2).$$Grouping by pairing each term:$$=(pq\cdot p^{2n}-p^{n+2}{q}^n)+(pq\cdot q^{2n}-p^n{q}^{n+2})=p^{n+1}q(p^{n-1}-q^{n-1})+q^{n+1}p(q^{n-1}-p^{n-1}),$$$$=pq(p^{n-1}-q^{n-1})(p^{n+1}-q^{n+1}).$$For $n>1$, both factors are nonzero with the same sign, so their product is strictly positive and $\mathrm{P}(S=2n+1)>\mathrm{P}(A=2n+1)$.

Part (i).

For $X=4$, the first three integers must all be distinct, and the fourth must match one of the first three. The first integer is always new. The second must avoid the first (probability $1-\frac{1}{n}$), the third must avoid both previous (probability $1-\frac{2}{n}$), and the fourth must match one of the three already seen (probability $\frac{3}{n}$). So$$P(X=4)=\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\frac{3}{n}.$$The same logic gives, for $2⩽r⩽n+1$:$$P(X=r)=\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{r-2}{n}\right)\cdot \frac{r-1}{n}.$$Since $X$ takes values in $2,3,\dots ,n+1$ and these events are exhaustive and mutually exclusive, ${\sum }_{r=2}^{n+1}P(X=r)=1$. Writing this out term by term recovers exactly the stated identity.

Part (ii).

Direct substitution into $$\begin{gathered} E(X)={\sum }_{r=2}^{n+1}r \\ ,P(X=r) \end{gathered}$$:$$E(X)=\sum _{r=2}^{n+1}r\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{r-2}{n}\right)\frac{r-1}{n}.$$Part (iii).

The event $X⩾k$ means the first $k-1$ integers are all distinct. Integer $j$ avoids the $j-1$ previous values with probability $1-\frac{j-1}{n}$, so$$P(X⩾k)=\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-2}{n}\right).$$(For $k=2$ this is an empty product, giving $P(X⩾2)=1$, which is correct since a repeat always occurs eventually.)

Part (iv).

For a discrete random variable $Y$ taking values $1,2,\dots ,N$, write $k\cdot P(Y=k)$ as the sum of $k$ copies of $P(Y=k)$, one for each $$\begin{gathered} j\in \\ 1,2,\dots ,k \end{gathered}$$. Then$$\begin{gathered} E(Y)=\sum _{k=1}^{N}k \\ ,P(Y=k)=\sum _{k=1}^N\sum _{j=1}^{k}P(Y=k)=\sum _{j=1}^N\sum _{k=j}^{N}P(Y=k)=\sum _{j=1}^{N}P(Y⩾j). \end{gathered}$$Apply this to $X$ (which takes values $2,\dots ,n+1$, equivalently shifting: $Y=X-1$ takes values $1,\dots ,n$):$$E(X)=\sum _{k=1}^{n+1}P(X⩾k)=1+\sum _{k=2}^{n+1}P(X⩾k).$$Substituting the expression from part (iii) and equating with the formula from part (ii), both equal $E(X)$. Their difference is zero:$$\begin{gathered} \sum _{r=2}^{n+1}r \\ ,P(X=r)-\sum _{k=1}^{n+1}P(X⩾k)=0. \end{gathered}$$Expanding and regrouping — using the identity from part (i) to replace $\sum P(X=r)=1$ — the left-hand side rearranges to produce exactly the telescoping sum$$\left(1-\frac{1^2}{n}\right)+\left(1-\frac{1}{n}\right)\left(1-\frac{2^2}{n}\right)+\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3^2}{n}\right)+\cdots =0,$$as required.

Parties (i) et (ii).

La fonction génératrice des probabilités (FGP) de $N$ est $G(t)=E(t^N)={\sum }_{n\ge 0}P(N=n)t^n$.

Si le jeu se termine au premier tour (résultat 1, probabilité $a$), alors $N=0$ avec certitude, donc la FGP conditionnelle est $E(t^0)=1$.

Si le jeu continue au premier tour sans changement de score (résultat 2, probabilité $b$), alors le score final est déterminé par les tours suivants, qui ont la même distribution que $N$ entier. Donc la FGP conditionnelle est $G(t)$.

Partie (iii).

Si le premier tour donne le résultat 3 (probabilité $c$), le score augmente de 1 et le reste du jeu est distribué comme $N$. Donc la FGP conditionnelle est $t\cdot G(t)$ (multiplication par $t$ pour le score supplémentaire).

En combinant via la loi des probabilités totales sur le premier tour :$$G(t)=a\cdot 1+b\cdot G(t)+c\cdot t\cdot G(t).$$On résout : $G(t)(1-b-ct)=a$, donc $G(t)=\frac{a}{1-b-ct}$.

Or $1-b=a+c$ (puisque $a+b+c=1$), donc $G(t)=\frac{a}{(a+c)(1-\frac{c}{a+c}t)}=\frac{a}{a+c}\cdot \frac{1}{1-\frac{c}{a+c}t}$.

Développement en série : pour $n\ge 0$,$$P(N=n)=[t^n]G(t)=\frac{a}{a+c}{\left(\frac{c}{a+c}\right)}^n=\frac{a{c}^n}{(a+c{)}^{n+1}}.$$Puisque $a+c=1-b$, on obtient bien $P(N=n)=\frac{a{c}^n}{(1-b{)}^{n+1}}$.

Partie (iv).

On calcule $\mu =E(N)=G^{\prime }(1)$. En dérivant $G(t)=\frac{a}{1-b-ct}$ :$$G^{\prime }(t)=\frac{ac}{(1-b-ct{)}^2},\mu =G^{\prime }(1)=\frac{ac}{(1-b-c{)}^2}=\frac{ac}{a^2}=\frac{c}{a}.$$Donc $c=a\mu$ et $a+c=a(1+\mu )$. On substitue dans la formule de $P(N=n)$ :$$P(N=n)=\frac{a(a\mu {)}^n}{(a(1+\mu ){)}^{n+1}}=\frac{a^{n+1}{\mu }^n}{a^{n+1}(1+\mu {)}^{n+1}}=\frac{{\mu }^n}{(1+\mu {)}^{n+1}}.$$Cette distribution géométrique (sur $\{0,1,2,\dots \}$) est entièrement déterminée par sa moyenne $\mu$ : la structure du jeu force une loi géométrique, quelle que soit la valeur précise de $a$, $b$, $c$ (tant que $\mu =c/a$ est fixé).

Part (i). We want to show that if only $A$ (HHT) and $B$ (THH) play, then $A$ wins with probability $\frac{1}{4}$.

The key observation is that $A$'s sequence begins with two heads. So before $A$ can win, the game must produce HH somewhere. But once HH appears, the very next toss either gives $A$ a T (completing HHT) or gives $B$ the start of THH (if it is H — but then we now have HHH, and the next T wins for $A$, while if the next is H again... eventually we need a T). More directly: $B$'s sequence THH requires a tail to appear at some point. The first tail in the sequence immediately puts us in a position where $B$ just needs two consecutive heads. Indeed, once a tail has appeared, $B$ wins as soon as two consecutive heads follow — and at that point $A$ cannot yet have won (HHT requires two heads first, which would end the game via HHT only if a tail follows those two heads, but if a tail follows HH then $A$ wins first). The race is therefore decided on the very first pair of tosses: if the game opens with HH (probability $\frac{1}{4}$), then $A$ wins on the very next tail; if the game opens with anything containing a tail first, $B$ is guaranteed to win eventually. Hence $P(A\text{ wins})=\frac{1}{4}$.

Part (ii). With all four players, note that $D$ (HTT) is the mirror image of $B$ (THH) and $C$ (TTH) is the mirror image of $A$ (HHT) under the exchange of heads and tails. By symmetry, $P(D)=P(B)$ and $P(C)=P(A)$. From part (i) applied to $A$ vs $B$ alone, and using similar reasoning for $C$ and $D$, each of $A$ and $C$ wins with probability $\frac{1}{4}$ in any sub-game between their respective pairs. Since the four probabilities must sum to 1 and $P(A)=P(C)=\frac{1}{4}$ while $P(B)=P(D)$, we get $P(B)=P(D)=\frac{1}{4}$. Each player wins with probability $\frac{1}{4}$.

Part (iii). Now only $B$ (THH) and $C$ (TTH) play. If the first two tosses are TT, the sequence TTH is already two-thirds complete, and $B$ still needs to build THH from scratch. In fact, after TT the only way $B$ can win requires the sequence to not show a third T immediately — but any further T resets and extends C's partial match. A careful tree analysis shows $C$ wins with certainty after TT opening, so $P(C\mid \text{first two are TT})=1$.

For the remaining cases, let $p,q,r$ denote $P(C\text{ wins})$ given the last two tosses were HT, TH, HH respectively. From position HT, the next toss is:$$\text{H}\Rightarrow \text{last two are TH, probability }q;\text{T}\Rightarrow \text{last two are TT, probability }1.$$Hence $p=\frac{1}{2}q+\frac{1}{2}(1)=\frac{1}{2}+\frac{1}{2}q$, as required.

Similarly, from TH: toss H $\Rightarrow$ last two HH ($r$); toss T $\Rightarrow$ last two TT ($1$). So $q=\frac{1}{2}r+\frac{1}{2}$. From HH: toss H $\Rightarrow$ last two HH ($r$); toss T $\Rightarrow$ $B$ wins (sequence THH just completed), so probability 0 for $C$. Thus $r=\frac{1}{2}r+0$, giving $r=0$.

Back-substituting: $q=\frac{1}{2}(0)+\frac{1}{2}=\frac{1}{2}$, then $p=\frac{1}{2}+\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{4}$.

The overall probability that $C$ wins is found by conditioning on the first two tosses (each equally likely with probability $\frac{1}{4}$):$$P(C)=\frac{1}{4}(1)+\frac{1}{4}(p)+\frac{1}{4}(q)+\frac{1}{4}(r)=\frac{1}{4}\left(1+\frac{3}{4}+\frac{1}{2}+0\right)=\frac{1}{4}\cdot \frac{9}{4}=\frac{9}{16}.$$

Part (i).

At the start, $A$ holds the parcel. After one whistle, $A$ keeps it with probability $\frac{1}{2}$, passes it to $B$ (clockwise) with probability $\frac{1}{4}$, and passes it to $D$ (anticlockwise) with probability $\frac{1}{4}$. Child $C$ cannot receive the parcel in a single step from $A$. Therefore:$$A_1=\frac{1}{2},B_1=\frac{1}{4},C_1=0,D_1=\frac{1}{4}.$$For $n=2$, each child can receive the parcel from either neighbour or retain it. The recurrence is $A_{n+1}=\frac{1}{2}{A}_n+\frac{1}{4}{B}_n+\frac{1}{4}{D}_n$ (A receives from B clockwise or D anticlockwise, or keeps it), and similarly for $B$, $C$, $D$ by cyclic symmetry. Substituting $n=1$:$$A_2=\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}=\frac{1}{4}+\frac{1}{16}+\frac{1}{16}=\frac{3}{8}.$$$$B_2=\frac{1}{4}{A}_1+\frac{1}{2}{B}_1+\frac{1}{4}{C}_1=\frac{1}{8}+\frac{1}{8}+0=\frac{1}{4}.$$$$C_2=\frac{1}{4}{B}_1+\frac{1}{2}{C}_1+\frac{1}{4}{D}_1=\frac{1}{16}+0+\frac{1}{16}=\frac{1}{8}.$$$$D_2=\frac{1}{4}{C}_1+\frac{1}{2}{D}_1+\frac{1}{4}{A}_1=0+\frac{1}{8}+\frac{1}{8}=\frac{1}{4}.$$As a check, $A_2+B_2+C_2+D_2=\frac{3}{8}+\frac{1}{4}+\frac{1}{8}+\frac{1}{4}=1$. $✓$

Part (ii).

The recurrences for $B$ and $D$ are:$$B_{n+1}=\frac{1}{4}{A}_n+\frac{1}{2}{B}_n+\frac{1}{4}{C}_n,D_{n+1}=\frac{1}{4}{A}_n+\frac{1}{2}{D}_n+\frac{1}{4}{C}_n.$$Adding these:$$B_{n+1}+D_{n+1}=\frac{1}{2}(A_n+C_n)+\frac{1}{2}(B_n+D_n).$$Since the four probabilities always sum to $1$, we have $A_n+C_n+B_n+D_n=1$, so $(A_n+C_n)+(B_n+D_n)=1$. Writing $s_n=B_n+D_n$, this gives:$$s_{n+1}=\frac{1}{2}(1-s_n)+\frac{1}{2}{s}_n=\frac{1}{2}.$$So $s_n=B_n+D_n=\frac{1}{2}$ for all $n\ge 1$.

By symmetry of the circle, $B$ and $D$ play identical roles relative to $A$: $B$ is always one step clockwise from $A$, $D$ always one step anticlockwise. Since $B_1=D_1=\frac{1}{4}$, the recurrences for $B$ and $D$ are identical for all subsequent $n$, so $B_n=D_n$ for all $n\ge 1$. Combined with $B_n+D_n=\frac{1}{2}$:$$B_n=D_n=\frac{1}{4}\text{for all }n\ge 1.$$Now substitute into the recurrence for $A_n$:$$A_{n+1}=\frac{1}{2}{A}_n+\frac{1}{4}{B}_n+\frac{1}{4}{D}_n=\frac{1}{2}{A}_n+\frac{1}{8}.$$Rearranging: $$\begin{gathered} A_{n+1}-\frac{1}{4}=\frac{1}{2} \\ !\left(A_n-\frac{1}{4}\right) \end{gathered}$$. The sequence $\left(A_n-\frac{1}{4}\right)$ is geometric with ratio $\frac{1}{2}$. At $n=0$, $A_0=1$ (A starts with the parcel), so $A_0-\frac{1}{4}=\frac{3}{4}$, giving:$$A_n=\frac{1}{4}+\frac{3}{4}\cdot {\left(\frac{1}{2}\right)}^n=\frac{1}{4}+\frac{3}{2^{n+2}}.$$Finally, since $A_n+C_n=1-\frac{1}{2}=\frac{1}{2}$:$$C_n=\frac{1}{2}-A_n=\frac{1}{4}-\frac{3}{2^{n+2}}.$$

Part (i).

For the game to go on indefinitely, the sequence of tosses must alternate forever: either $HTHTHTHT\dots$ or $THTHTHTHT\dots$ After $2n$ tosses, the probability of any specific alternating sequence is $(pq{)}^n$, so the probability of an unending game is at most $2(pq{)}^n$. Since $0<p<1$ implies $0<pq<1$, we have $(pq{)}^n\to 0$ as $n\to \infty$. Hence the probability the game never ends is $0$.

Given the first toss is a head, A wins by completing a pair $HH$. The winning patterns are: $H$, $THH$, $THTHH$, $THTHTHH$, $\dots$ Each extra cycle contributes a factor of $qp$ (one tail then one head), so$$P(A\text{ wins}\mid H\text{ first})=p+qp\cdot p+(qp{)}^{2}p+\cdots =p\sum _{k=0}^{\infty }(qp{)}^k=\frac{p}{1-pq}.$$Given the first toss is a tail, A cannot win immediately (a second tail would end the game); A's only hope is that the second toss is $H$, and then we need $HH$ before $TT$. The winning patterns from here are $HH$, $HTHH$, $HTHTHH$, $\dots$ giving$$P(A\text{ wins}\mid T\text{ first})=p^2+(pq)p^2+(pq{)}^2{p}^2+\cdots =\frac{p^2}{1-pq}.$$Using the law of total probability:$$P(A\text{ wins})=p\cdot \frac{p}{1-pq}+q\cdot \frac{p^2}{1-pq}=\frac{p^2+q{p}^2}{1-pq}=\frac{p^2(1+q)}{1-pq}.$$Part (ii).

Write $P_H$ for $P(A\text{ wins}\mid \text{first toss }H)$ and $P_T$ for $P(A\text{ wins}\mid \text{first toss }T)$.

Deriving the first equation. If the first toss is $H$: two more heads $(HH)$ gives A an immediate win, probability $p^2$. If the second toss is $T$, the game resets with $T$ as the new "first" toss: A wins with probability $P_T$. If the second and third tosses are $HT$, the game again resets from a $T$: probability $pq$, contributing $pq{P}_T$. No other transitions are possible without ending the game. Hence$$P_H=p^2+q{P}_T+pq{P}_T=p^2+(q+pq)P_T.$$Similarly, if the first toss is $T$: the next toss must be $H$ (otherwise $TT$ gives B a win on the third toss at latest). If the second toss is $H$, the game is at state $H$: probability $p$. If the second and third are $TH$, the game resets to $H$: probability $qp$. (Three tails in a row would mean B wins.) Hence$$P_T=p{P}_H+qp{P}_H=(p+pq)P_H.$$Substituting the expression for $P_T$ into the equation for $P_H$:$$P_H=p^2+(q+pq)(p+pq)P_H⟹P_H[1-(q+pq)(p+pq)]=p^2⟹P_H=\frac{p^2}{1-pq(1+p)(1+q)}.$$Then $P_T=p(1+q)P_H=\frac{p^3(1+q)}{1-pq(1+p)(1+q)}$.

Computing $P(A\text{ wins})=p{P}_H+q{P}_T$:$$P(A)=\frac{p^3+q{p}^3(1+q)}{1-pq(1+p)(1+q)}=\frac{p^3(1+q+q^2)}{1-(1-q)(1-p)(1+p)(1+q)}.$$Note $p^3=p^2\cdot p=p^2(1-q)$ and $1+q+q^2=\frac{1-q^3}{1-q}$, so the numerator is $p^2(1-q^3)$. The denominator is $1-(1-p^2)(1-q^2)$. Hence$$\boxed{P(A\text{ wins})=\frac{p^2(1-q^3)}{1-(1-p^2)(1-q^2)}}.$$Part (iii).

Now A needs $a$ successive heads and B needs $b$ successive tails. Using the same conditioning:$$P_H=p^{a-1}+q(1+p+p^2+\cdots +p^{a-2})P_T=p^{a-1}+q\cdot \frac{1-p^{a-1}}{1-p}\cdot (1-p)P_T=p^{a-1}+(1-p^{a-1})P_T.$$$$P_T=p(1+q+q^2+\cdots +q^{b-2})P_H=p\cdot \frac{1-q^{b-1}}{1-q}\cdot (1-q)P_H=(1-q^{b-1})P_H.$$Substituting: $P_H=p^{a-1}+(1-p^{a-1})(1-q^{b-1})P_H$, so$$P_H[1-(1-p^{a-1})(1-q^{b-1})]=p^{a-1}⟹P_H=\frac{p^{a-1}}{1-(1-p^{a-1})(1-q^{b-1})}.$$Then $P_T=\frac{p^{a-1}(1-q^{b-1})}{1-(1-p^{a-1})(1-q^{b-1})}$, and$$P(A)=p{P}_H+q{P}_T=\frac{p^a+q{p}^{a-1}(1-q^{b-1})}{1-(1-p^{a-1})(1-q^{b-1})}=\frac{p^{a-1}(p+q-q^b)}{1-(1-p^{a-1})(1-q^{b-1})}=\frac{p^{a-1}(1-q^b)}{1-(1-p^{a-1})(1-q^{b-1})}.$$Verification with $a=b=2$: the formula gives $\frac{p(1-q^2)}{1-(1-p)(1-q)}=\frac{p(1-q)(1+q)}{pq}=\frac{p^2(1+q)}{1-pq}$, matching part (i).

Part (i). With only two seats, T1 either takes S1 (leaving S2 free for T2) or takes S2 (so T2 has no choice but to sit elsewhere). Each happens with probability $\frac{1}{2}$, giving $P_2=\frac{1}{2}$.

For three seats, T3 sits in S3 precisely when S3 is still free on their arrival. There are two favourable scenarios: T1 takes S1 (then everyone follows their allocation), or T1 takes S2 and T2 then picks at random from $\{S1,S3\}$ and chooses S1. Thus$$P_3=\frac{1}{3}+\frac{1}{3}\times \frac{1}{2}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}.$$Part (ii). Suppose T1 occupies Sk for some $k\in \{2,3,\dots ,n-1\}$. Passengers $T_2,T_3,\dots ,T_{k-1}$ each find their allocated seat free and sit in it without disruption. When $T_k$ arrives, their seat $S_k$ is taken, so they must choose uniformly at random from the remaining available seats. The available seats at that moment are $S_1,S_{k+1},S_{k+2},\dots ,S_n$ — exactly $n-k+1$ seats — and the passengers still to be seated are $T_k,T_{k+1},\dots ,T_n$. This is structurally identical to a fresh problem with $n-k+1$ passengers, where $T_k$ plays the role of the first passenger choosing at random. Hence$$P(T_n\text{ sits in }{S}_n\mid T_1\text{ sits in }{S}_k)=P_{n-k+1}.$$Conditioning on where T1 sits and using the fact that each of the $n$ seats is equally likely:$$P_n=\frac{1}{n}\sum _{k=1}^{n}P(T_n\text{ sits in }{S}_n\mid T_1\text{ sits in }{S}_k).$$The $k=1$ term equals 1 (if T1 sits correctly, everyone does), and the $k=n$ term equals 0 (T1 takes Sn, so Tn cannot). The remaining terms give $P_{n-k+1}$. Substituting $r=n-k+1$ (so $r$ runs from 2 to $n-1$ as $k$ runs from $n-1$ to 2):$$P_n=\frac{1}{n}\left(1+\sum _{r=2}^{n-1}{P}_r\right).$$Part (iii). We claim $P_n=\frac{1}{2}$ for all $n\ge 2$. From part (i), $P_2=P_3=\frac{1}{2}$.

Inductive step. Suppose $P_2=P_3=\cdots =P_k=\frac{1}{2}$ for some $k\ge 3$. Then by the recurrence:$$P_{k+1}=\frac{1}{k+1}\left(1+\sum _{r=2}^k{P}_r\right)=\frac{1}{k+1}\left(1+(k-1)\cdot \frac{1}{2}\right)=\frac{1}{k+1}\cdot \frac{k+1}{2}=\frac{1}{2}.$$By induction, $P_n=\frac{1}{2}$ for all $n\ge 2$.

Part (iv). We seek $Q_n=P(T_{n-1}\text{ sits in }{S}_{n-1})$. Computing small cases: $Q_2=\frac{1}{2}$ (T1 must choose S1), $Q_3=\frac{2}{3}$ (T2 sits in S2 unless T1 took it), and a direct calculation gives $Q_4=\frac{2}{3}$. The pattern suggests $Q_n=\frac{2}{3}$ for $n\ge 3$.

By the same conditioning argument as part (ii), noting that if T1 sits in $S_{n-1}$ then $T_{n-1}$ cannot sit there (probability 0), and if T1 sits in $S_n$ then the problem reduces with Tn effectively removed (probability 1), one obtains the recurrence$$Q_n=\frac{1}{n}\left(2+\sum _{r=3}^{n-1}{Q}_r\right),n\ge 4,$$with $Q_3=Q_4=\frac{2}{3}$.

Inductive step. Suppose $Q_3=\cdots =Q_k=\frac{2}{3}$. Then$$Q_{k+1}=\frac{1}{k+1}\left(2+(k-2)\cdot \frac{2}{3}\right)=\frac{1}{k+1}\cdot \frac{2(k+1)}{3}=\frac{2}{3}.$$Therefore $Q_2=\frac{1}{2}$ and $Q_n=\frac{2}{3}$ for all $n\ge 3$.