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.


1 Answer 1


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: $$\frac18+4\times\frac1{32}=\frac14$$

  • $\begingroup$ Does 1/8 come from $1/2^3$? And could you explain what does 1/32 mean? $\endgroup$
    – Oscar Wan
    Oct 2, 2022 at 15:49
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 2, 2022 at 15:59
  • $\begingroup$ 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 $\frac1{32}$. $\endgroup$
    – drhab
    Oct 3, 2022 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.