There's a game with four players, numbered by their increasing difficulty: 1,2,3 and 4.

The probability of any player (x) beating another player (y) in a 1v1 game is x / (x + y). For example if player 3 plays player 4 the chance of player 3 winning is 3/7. There are no draws - every game ends in one side winning.

There's a tournament where in the first round the four players are organised in to two 1v1 groups (group A and group B). The winner of each group play each other in the final. The winner of the final wins the tournament.

Player 1 gets to choose who they want to play in the first round. For example they might choose to play the easiest opponent (2), and hope that they don't meet the hardest opponent (4) in the final.

The possible groupings of the first round are:

A: 1v2
B: 3v4


A: 1v3
B: 2v4


A: 1v4
B: 2v3

And in the final the winner of group A will play the winner or group B:

Final: AvB

Can player 1 increase their chance of winning the tournament through their choice of opponent in the first round? (And how can I calculate their chances in each scenario?)


Informally, team $1$ really wants to avoid $3,4$ so it's better if those two play each other. At least that way, one of the two tough teams is eliminated.

But let's work it out.

I. $1$ plays $2$ first.

Then, in order to win the whole match, $1$ needs first to beat $2$, probability $\frac 13$. Then $1$ needs to beat the winner of $3$ vs $4$, which happens with probability $$\frac 14 \times \frac 37+\frac 15\times \frac 47=\frac {31}{140}$$

Thus the probability that $1$ wins along this path is $$\frac 13\times \frac {31}{140}=\frac {31}{420}\approx \boxed{.0738}$$

II. $1$ plays $3$ first

A similar computation gives $$\frac 14\times \left( \frac 13\times \frac 26+\frac 15\times \frac 46\right)=\frac {11}{180}=\boxed {.06111\cdots}$$

III. $1$ plays $4$ first.

We get $$\frac 15\times \left( \frac 13\times \frac 25+\frac 14\times \frac 35\right)=\frac {17}{300}=\boxed {.05666\cdots}$$

Warning: While straightforward, the above calculations are somewhat error prone. I advise checking them all carefully.

  • $\begingroup$ FWIW, I independently set up a small spreadsheet that confirms these numbers. $\endgroup$ – David K Jun 28 '18 at 17:34
  • $\begingroup$ @DavidK Thanks! $\endgroup$ – lulu Jun 28 '18 at 17:41

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