# Find probability of winning of Player A and Player B

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

• This site isn’t a HW mill. Show your work. Commented Nov 8, 2019 at 14:25
• @Max0815 I tried considering cases like WLWL...WW etc,but this doesn't seem to work
– atin
Commented Nov 8, 2019 at 14:43
• @AtinLM10 Trying WLWL combinations won't work as there are infinite possibilities Commented Nov 8, 2019 at 14:51

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.

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

• @Damien is the answer and approach correct
– atin
Commented Nov 10, 2019 at 13:21
• You simplified my approach! Simple solution at the end Commented Nov 11, 2019 at 8:12
• @Atin: If a Markov chain is used, the answer that emerges is different. Commented Mar 19, 2021 at 17:22
• @trueblueanil I'm not familiar with Markov chain, can you send the solution or maybe any hints? I'll also look up markov chain.
– atin
Commented Mar 19, 2021 at 17:56
• @trueblueanil Also I don't have any official solution/answer for this, it's just a random question my friend asked me :/
– atin
Commented Mar 19, 2021 at 18:02

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.