Find probability of winning of Player A and Player B

Question

Two player $A$ and $B$ are playing a tournament.

For Player $A$ (for a single game)

\begin{align}P(\text{Win})=\frac12\\ P(\text{Draw})=\frac16\\ P(\text{Lose})=\frac13\end{align}

The player who will win two consecutive games win the tournament.

Question

Find the probability of

(i) $A$ winning the tournament

(ii) $B$ winning the tournament

Answer 1

For player $A$, we have the following probability, for each game:

$$ P(Win) = p_W = \frac{1}{2}$$ $$ P(Draw) = p_D = \frac{1}{6}$$ $$ P(Loose) = p_L = \frac{1}{3}$$

There is a way to address this problem by considering states.

For example, at a given time, if the last game was a draw (D), then the two players are in the same situation, independently of the previous games, assuming no one as gained before. This corresponds to a neutral state. This neutral state is also attained at the start of the tournament.

We will call :

• $P_0$: the probability of the neutral state
• $P[W]$ the probability of the state attained when $A$ has gained the last game, but not the previous one
• $P[WW]$ the probability of the state corresponding to a final victory for $A$, i.e. he gained the last two games
• $P[L]$ the probability of the state attained when $A$ has lost the last game, but not the previous one
• $P[LL]$ the probability of the state corresponding to a final loss for $A$, i.e. he lost the last two games

Then, it is straightforward to get the following relations:

$$P_0(t=0) = 1$$ $$P[W](t=0) = P[L](t=0) = P[WW](t=0) = P[LL](t=0) =0$$ $$P_0(t+1) = 1 + p_d(P_0 + P[W] + P[L])(t) $$ $$P[L](t+1) = p_L (P_0(t) + P[W])(t) $$ $$P[W](t+1) = p_W (P_0 + P[L])(t) $$ $$P[WW](t+1) = p_W P[W](t)$$ $$P[LL](t+1) = p_L P[L](t)$$

Finally, the global probability $P[WW]$ is obtained by summing all $P[WW](t)$ over (t).

At this point, much work is still needed to get $P[WW]$, but the problem has been simplified.

Note: a figure representing the states and their different relations help understanding the method and the relations above. This figure is relatively easy to draw by hand, I encourage you to do it.

There is sometimes the possibility to address this kind of problem by defining polynomials representing the status of a state at all time, for example:

$$P_0(X) = 1 + P_0(1) X + P_0(2) X^2 + P_0(3) X^3 + \dots$$

and to translate the previous relations in terms of polynomials.

For example: $$P_0(X) = 1 + p_D\,X(P_0(X) + P[W](X) + P[L](X)) $$

In this case, at the end, $P[Final Win] = P[WW](X=1)$

I will let you explore it if you want.

Answer 2

Taking help from @Damien's approach
First I noticed that

P($A_w$)+P(B$_w$)=1

Because the probability that the tournament will continue forever i.e. Nobody wins is zero.
The summation of probabilities of all possible combination till the last draw is same for both A and B
$$X=\sum_{}{} P(......D)$$
Cases after last Draw for A's win={LWW,LWLWW,....} U {WW,WLWW,....}
$$P(A_w)=X*(1/4)*\sum_{n=0}^{\infty}(1/6)^n+X*(1/4)*(1/3)*\sum_{n=0}^{\infty}(1/6)^n$$ $$P(A_w)=X*(1/4)*(6/5)*(1+1/3)$$ $$P(A_w)=2X/5$$ Similarly $$P(B_w)=X/5$$
Now putting in above equation $$P(A_w)+P(B_w)=1$$ $$X=5/3$$ $$P(A_w)=2/3$$ $$P(B_w)=1/3$$

Sorry for the bad formatting I'm new at this site.
Any edit would be appreciated

Answer 3

For (i), you can try to consider all possible cases and find a pattern. For the question, all possible cases would mean letting the tournament "run to infinity".

For (ii), you can make use of the idea of complement.