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There are 4 players in the game. First player 1 plays with player 2, with loser exiting the game. Next winner of previous game plays with player 3, with loser exiting. Further, winner of previous game plays with player 4, and the winner of this game is tournament victor. Given that a game between players $i$ and $j$ is won by player $i$ with probability ${i \over {i + j}}$.

Find expected number of games played by player 1, [and player 3]. [EDIT]

Approach: Player 1 if exits in first round. He plays 1 match with probability $ {1\over {1+2}}$ = $1 \over 3$. Similarly, if exits in round two, then plays 2 matches with probability, (wins first round and then second round)${ 1 \over 12} $. Now for playing 3 matches, probability = ${1 \over 60}$.
So, weighted mean or expected value is,
$E(player 1)$ = $1 \times {1 \over 3} + 2 \times {1 \over {12}} + 3 \times {1 \over {60}} = {{11} \over {20}}$.
But solution is incorrect.

1 Answers1

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In order to play exactly one game he needs to loose the first one which happens with probability $\frac{2}{3}$. To play exactly $2$ games he needs to win the first one and loose the second one, so $\frac{1}{3}\cdot\frac{3}{4}$. Now compute the chance for playing exactly $3$ (where it doesn't matter whether he wins or looses the third game) in a similar way and you should get the correct answer.

For player 3 you can consider the tournament as a whole. There are always 3 games played in total. First game is always between 1 and 2, so player 3 will never play it. Second game is the winner from game 1 against player 3 so he will always play it. To play in game 3 he needs to win game 2. This happens either if player 1 wins game 1 and looses in game 2 or if player 2 wins game 1 and looses in game 2. Compute the probability for these two cases and add 1 for playing game 2 will give you the expected number of games played by player 3.

quarague
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  • What about player 3. I got answer for player 1, but expected number of games played by 3 seems tricky. For this player three plays 1 game either with 1 or 2, with incoming probability of 1 and 2, and then loose match with either of them. For playing 2 games,win either with 1 or 2 and then loose or win with player 4. I used this logic and answer came incorrect. Help me with player 3 expectation value. –  Jul 17 '19 at 06:58
  • For player 3, the answer is coming out to be 23/10 but the answer which is given is 99/60. –  Jul 17 '19 at 08:08
  • Check your algebra, I get $33/20$. – quarague Jul 17 '19 at 13:05