# Probability of winning the series given not playing the first game

Suppose three pool players A, B, C are equally good, meaning that they each have a 0.5 probability of winning against each other. They are playing a game as follows: the winner of a game plays whoever didn't play the last round. For example, if A plays B, and B loses, then A plays C. Whoever wins 3 games first wins the series.

The question is, what is the probability of A winning the series, given that A does not play the first game?

It is clear that A is in disadvantage. Consider a smaller series: whoever wins 1 game first, then A has no chance of winning.

I have run some simulations in python, and found that the probability of A winning a 3-game series is about 0.25, and the other two players equally share the remaining 0.75. I'm wondering how to do this calculation mathematically.

def play():
t = rr(0,1) # randomly picks 0 or 1
p = {1 : 0, 2 : 0, 3 : 0} # tracks each player's number of wins
all = {1,2,3} # all players
match = {2,3} # players in match currently

# player 2 and 3 play first
p[2] += t
p[3] += 1 - t
winner = 2 if t == 1 else 3

while max(p.values()) < 3:
t = rr(0,1)
remaining = (all - match).pop() # whoever does not play in the game
match = {winner, remaining} # winner of the match plays the remaning player
p[winner] += t
p[remaining] += 1 - t
winner = winner if t == 1 else remaining

return max(p, key = p.get)enter code here

wins = np.array([0,0,0])
for ii in range(int(1e6)):
wins[play() - 1] += 1
print(wins / sum(wins))
#[0.250308 0.375203 0.374489]

P.S. This question arose when I lost the same game, and I did not play the first game. When I lost a game and the score became 2-2-2, I realized that I had no chance of winning.

WLOG let it be that $$B$$ wins the first match.

Then by a victory of $$A$$ only the folowing sequences of winners are possible:

• $$BAAA$$
• $$BAABCA$$
• $$BACBAA$$
• $$BACCAA$$
• $$BBCAAA$$

So the probability on a victory for $$A$$ is: $$\frac18+4\times\frac1{32}=\frac14$$

• Does 1/8 come from $1/2^3$? And could you explain what does 1/32 mean? Commented Oct 2, 2022 at 15:49
• @OscarWan my take: first game doesnt matter who wins sin we'd just rename C to B. After that everything has to go exactly right. In case 1 three games going your way is 1/2^3 = 1/8. Cases 2-5, five games need to go your way, since if C loses to B at the wrong time, B wins early, so 1/2^5 = 1/32. Commented Oct 2, 2022 at 15:59
• As @SpencerFleming explains. If $E$ denotes the event that $B$ wins the first game then $P(BAAA|E)=\frac12\frac12\frac12=\frac18$. Similar story for the $4$ other cases but then with outcome $\frac1{32}$. Commented Oct 3, 2022 at 5:45