Tom plays Jerry in a chess tournament, which is won by the first player who wins two consecutive games.?

U

Answering the question as directed...

For the tournament to continue,
Tom can not win two consecutive games or loose two consecutive games.
Tom must alternately win and and loose games.
Tom may win or loose the first game but must alternate from then onward.
Thus, for the tournament to continue beyond any named game we must sum the two ways to reach that game.

P[W>1] = 0.6 + 0.4

P[W>2] = (0.6)(0.4) + (0.4)(0.6)
P[W>2] = 2(0.6)(0.4)

P[W>3] = (0.6)(0.4)(0.6) + (0.4)(0.6)(0.4)
P[W>3] = (0.6^2) (0.4) + (0.4^2) (0.6)

P[W>4] = (0.6)(0.4)(0.6)(0.4) + (0.4)(0.6)(0.4)(0.6)
P[W>4] = 2(0.6)(0.4)(0.6)(0.4)
P[W>4] = 2(0.6^2) (0.4^2)

P[W>5] = (0.6)(0.4)(0.6)(0.4)(0.6) + (0.4)(0.6)(0.4)(0.6)(0.4)
P[W>5] = (0.6^3) (0.4^2) + (0.4^3) (0.6^2)

By inspection
P[W>2m] = (0.6^m) (0.4^m)
P[W>2m+1] = (0.6^[m + 1]) (0.4^m) + (0.4^[m + 1]) (0.6^m)


With this sort of question you don't know where it is going, you just have to do what is asked and see what happens.

That's not how I'd solve this. You can avoid infinite series with a bit of "recursive thinking".

The tournament has three "states", depending on who won the previous game.

State 0: The starting state, where nobody has won a game.

State 1: Tom won the last game (but not the tournament).

State 2: Jerry won the last game (but not the tournament).

There are two additional states for the end of the tournament, but they are not relevant for probability calculuation.

Let p1 be the probability that Tom wins once State 1 as been reached and p2 be the probability that Tom wins from State 2. Then:

p1 = p + (1-p)p2

That is, Tom has a probability p of winning the tournament immediately, and a (1-p)*p2 probability of losing this game, but still winning the tournament from State 2.

p2 = p*p1

That is, Tom wins from state 2 only by first winning this game (probability p) and then winning from state 1.

That makes two equations in two unknowns. Solve for p1 by subsitution:

p1 = p + (1-p)p2 = p + (1-p)(p*p1)
p1 = p + (p - p^2)p1
(1 - p + p^2)p1 = p
p1 = p / (1 - p + p^2)

And then p2 = p*p1 = p^2 / (1 - p + p^2).

Now, from state 0, Tom has probability p0 = p*p1 + (1-p)p2 of either winning this game and then winning the tournament, or losing this game but still winning the tournament.

p0 = p*[p / (1 - p + p^2)] + (1-p)*[p^2 / (1 - p + p^2)]
= p^2 / (1 - p + p^2) + (p^2 - p^3) / (1 - p + p^2)
= (2p^2 - p^3) / (1 - p + p^2)

When p=0.6, that's about 0.66316.

The expected length can be computed the same way. Let L1 be the expected number of games left from state 1, and L2 be the expected number of games left from state 2.

L1 = p*1 + (1-p)*(1 + L2)
L2 = q*1 + p*(1 + L1)

Solve for L1,L2 and then the expected duration of the tournament is:

p*(1 + L1) + (1-p)*(1 + L2).