Porisma

Worked solution

Sum mutually exclusive winning rounds as a geometric series

Two players A and B each have three cards showing elephant, cat and mouse. They flash their cards simultaneously. Elephant defeats cat, cat defeats mouse, and mouse defeats elephant. Matching animals draw. In each round, A chooses elephant, cat and mouse with probabilities 0.3,0.3,0.4; B’s corresponding probabilities are 0.2,0.5,0.3. Play continues until there is a winner.

Find the probability that

(i) a round ends in a draw;

(ii) A wins the first round;

(iii) B wins the first round given that it is not a draw;

(iv) B wins the contest.

[2008/JJC/II/7]

CJC · Discussion · Question 7

Answer

(i) 0.33; (ii) 0.32; (iii) 35/6735/67; (iv) 35/6735/67, under the source’s independence/stationarity assumptions.

State the independence model used in the source products

The printed solution treats A’s and B’s choices as independent within each round. For the repeated-round geometric series it also assumes independent rounds with the same probabilities after each draw. Simultaneous flashing alone does not prove these assumptions; they are the model underlying the printed calculations.

How did this part go?

(i) Add the three matching-animal outcomes

P(draw)=P(E,E)+P(C,C)+P(M,M)=(0.3)(0.2)+(0.3)(0.5)+(0.4)(0.3)=0.33P(\text{draw})=P(E,E)+P(C,C)+P(M,M)=(0.3)(0.2)+(0.3)(0.5)+(0.4)(0.3)=0.33.

How did this part go?

(ii) Add A’s three winning outcomes

A wins with elephant against cat, cat against mouse, or mouse against elephant. Thus P(A wins round1)=(0.3)(0.5)+(0.3)(0.3)+(0.4)(0.2)=0.32P(A\text{ wins round1})=(0.3)(0.5)+(0.3)(0.3)+(0.4)(0.2)=0.32.

How did this part go?

(iii) Reuse(i) and (ii), as the source annotation suggests

The three disjoint outcomes are A win, B win and draw. Hence P(B wins round1)=1−0.32−0.33=0.35P(B\text{ wins round1})=1-0.32-0.33=0.35. A B win automatically is a non-draw, so

P(B wins round1∣not draw)=P(B wins round1∩not draw)P(not draw)=0.351−0.33=3567.\begin{aligned}&P(B\text{ wins round1}\mid\text{not draw})\\ & =\frac{P(B\text{ wins round1}\cap\text{not draw})}{P(\text{not draw})}\\ & =\frac{0.35}{1-0.33}\\ & =\frac{35}{67}.\end{aligned}

How did this part go?

(iv) Retain the source’s complete winning-round table

The disjoint possibilities are:

Winning roundDescriptionProbability
1B wins immediately0.350.35
2One draw, then B wins0.33×0.350.33\times0.35
3Two draws, then B wins0.332×0.350.33^2\times0.35
⋯\cdots⋯\cdots⋯\cdots
nnn−1n-1 draws, then B wins0.33n−1×0.350.33^{n-1}\times0.35

Therefore P(B wins contest)=0.35+0.33(0.35)+0.332(0.35)+⋯P(B\text{ wins contest})=0.35+0.33(0.35)+0.33^2(0.35)+\cdots. This infinite GP has first term 0.35 and ratio 0.33, giving 0.35/(1−0.33)=35/670.35/(1-0.33)=35/67. Since 0.33n→00.33^n\to0, an endless run of draws has probability 0 in this model.

How did this part go?

Common mistakes

A draw does not decide the contest. The eventual winner requires summing over every possible number of preceding draws.

Solution notes

All asks, exact animal match-ups/probabilities, coloured reuse-parts(i)/(ii) annotation, entire source table including the generaln row and infinite-GP interpretation retained. Independent choices/rounds and unchanged post-draw probabilities are explicit modelling qualifications. No marks printed.

Read the hint