Optimal Chip Allocation in a Round-Robin Tournament

Game Theory · Hard · Free problem

You enter a round-robin poker tournament with

01$ players: you and
00$ poker pros. Every player plays every other player exactly once.

Each pro starts each of their matches with

00$ chips. You receive
0{,}000$ chips total before the tournament begins, and you may distribute them however you like across your
00$ matches. For example, you could put
{,}000$ chips on each of
0$ matches and $0$ on the rest, or
00$ on each match, etc.

In each match, the probability of winning is proportional to chips: if you bring $a$ chips and your opponent has $b$ chips, your win probability is

$P(\text{win}) = \frac{a}{a + b}$

Chips do not transfer between matches. A win awards

$ point; a loss awards $0$ points. After all matches, prizes are awarded by rank:

00{,}000$
  • 6th--10th place: $\
    00{,}000$ each
  • Everyone else gets nothing

    • Ties in points are broken by a uniformly random permutation of the tied ranks.

    1. What is the optimal allocation of your 0{,}000$ chips across the
      00$ matches?
    2. Under this optimal allocation, what is your expected number of wins?
    3. What is the expected number of wins for each pro?

    Hints

    1. Your win probability $x/(x+100)$ is a concave function of $x$. What does concavity imply about how to allocate a fixed budget?
    2. Think about Jensen's inequality: for a concave $f$ with a fixed sum constraint, $\sum f(x_i)$ is maximized when all $x_i$ are equal.
    3. Set $x_i = 200$ for all $i$. Compute your expected wins as
      00 \times 200/300$. Then find each pro's expected wins: they play you once (winning with probability
      /3$) and $99$ fair matches against other pros.

    Worked Solution

    How to Think About It: You have twice as many total chips as each pro has across their matches ( 0{,}000$ vs.

    00 \times 100 = 10{,}000$). The question is how to spread them. Should you dump everything into a few matches and guarantee those wins, or spread them evenly? Your win probability in match $i$ is $x_i / (x_i + 100)$, which is a concave function of $x_i$. Concavity means diminishing returns: the 101st chip in a match helps less than the 1st chip. Whenever you see concavity plus a fixed budget, the optimizer is screaming "equalize."

    Quick Estimate: If you spread evenly, you put

    00$ chips on each match. Your win probability per match is 00 / (200 + 100) = 2/3$. Expected wins:
    00 \times 2/3 \approx 66.7$. Now compare: if you put $400$ on $50$ matches and $0$ on the other $50$, you get $50 \times (400/500) + 50 \times 0 = 50 \times 0.8 = 40$ expected wins. That is much worse. What about
    0{,}000$ on
    $ matches? You get
    \times (10{,}000/10{,}100) + 0 \approx 1.98$ wins. Catastrophic. The more you concentrate, the fewer expected wins you get. The uniform allocation dominates.

    Approach: We prove uniform allocation is optimal using concavity and Jensen's inequality, then compute expected wins for you and for each pro.

    Formal Solution:

    Let $x_i$ be the chips allocated to match $i$, with $\sum_{i=1}^{100} x_i = 20{,}000$ and $x_i \geq 0$. Your total expected wins are

    $W = \sum_{i=1}^{100} \frac{x_i}{x_i + 100}$

    Define $f(x) = x / (x + 100)$. We check concavity:

    $f'(x) = \frac{100}{(x + 100)^2} > 0$

    $f''(x) = \frac{-200}{(x + 100)^3} < 0$

    Since $f$ is strictly concave on $[0, \infty)$, by Jensen's inequality:

    $\frac{1}{100} \sum_{i=1}^{100} f(x_i) \leq f\!\left(\frac{1}{100} \sum_{i=1}^{100} x_i\right) = f(200) = \frac{200}{300} = \frac{2}{3}$

    Equality holds if and only if all $x_i$ are equal, i.e., $x_i = 200$ for all $i$. Therefore $W \leq 100 \times 2/3 = 200/3$, with equality at the uniform allocation.

    Alternatively, think of it via reallocation: if $x_j > x_k$, then transferring a small amount $\epsilon$ from match $j$ to match $k$ increases total expected wins, because the marginal gain at the lower-chip match exceeds the marginal loss at the higher-chip match (this is exactly what concavity says). The only allocation where no such improving transfer exists is $x_j = x_k$ for all $j, k$.

    Expected wins for each pro:

    Under optimal play, each pro loses to you with probability

    /3$ (and wins with probability
    /3$). Among the $99$ other pros, every match is
    00$ vs.
    00$ chips, so each is a fair coin flip. A pro's expected wins are:

    $E[\text{pro wins}] = \frac{1}{3} + 99 \times \frac{1}{2} = \frac{1}{3} + 49.5 = 49.83$

    Note this is strictly less than $50$, which means you are expected to finish with more wins ($66.7$) than every pro ($49.8$). You are a heavy favorite for first place.

    Answer:

    1. The optimal strategy is to allocate chips uniformly: $x_i = 200$ for each of the
      00$ matches.
    2. Your expected number of wins is
    00/3 \approx 66.67$.
  • Each pro's expected number of wins is
    /3 + 99/2 = 299/6 \approx 49.83$.

    • Intuition

    This problem is a clean application of the principle that concavity plus a budget constraint implies equalization. The function $x/(x+100)$ has diminishing returns: doubling your chips in a match does not double your win probability. So every chip you pile onto an already-strong match is wasted relative to spreading it to a weaker one. This is the same logic behind diversification in portfolio theory -- if returns are concave in allocation (as they are with risk), you spread your bets. Jensen's inequality makes this rigorous in one line.

    The deeper lesson for quant interviews is to recognize the shape of the objective before optimizing. Once you see that the per-match payoff is concave, the answer is immediate -- no Lagrange multipliers needed, no calculus of variations. The reallocation argument ("move a chip from a heavy match to a light one") is the kind of clean economic reasoning interviewers love, because it shows you understand why the math works, not just that it works.

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