Playing a Coin Flip With a Slight Edge

Probability · Easy · Free problem

I offer you a game: we flip a coin. If you win you gain $\

$; if you lose you pay $\
$. The coin is very slightly biased in your favor, so your win probability is a little above $50\%$.

Would you play this game? Does your answer change if you can play it many times? What if a single loss could be catastrophic relative to your wealth?

Hints

  1. Don't answer yes or no immediately -- start by writing the expected value of a single play in terms of the win probability.
  2. With $p = 0.5 + \epsilon$ the per-play edge is
\epsilon$; now think about what happens to total profit versus its standard deviation as you repeat the game.
  • The edge grows linearly in $n$ while the noise grows like $\sqrt{n}$, so signal-to-noise scales as $\epsilon\sqrt{n}$ -- but reconsider if a single loss is huge relative to your bankroll.

    • Worked Solution

    How to Think About It: The interviewer is not after a yes/no -- they want to see you reason about edge, repetition, and bet sizing. With a genuine edge the expected value of one play is positive, so a pure EV maximizer plays. But the realistic answer depends on how the stakes compare to your bankroll and how many times you can play.

    Quick Estimate: Let your win probability be $p = 0.5 + \epsilon$ for small $\epsilon > 0$. The expected profit per play is

    $E[\text{profit}] = (+1)\,p + (-1)(1-p) = 2p - 1 = 2\epsilon > 0.$

    So each play has a small positive edge of \epsilon$ dollars. Over $n$ independent plays the expected total is \epsilon n$ with standard deviation about $\sqrt{n}$ (each play has variance close to

    $). The signal-to-noise ratio grows like
    \epsilon n / \sqrt{n} = 2\epsilon \sqrt{n}$, so with enough repetitions the edge overwhelms the noise and you are very likely to finish ahead.

    Approach: Decide based on (a) sign of the edge, (b) number of plays, (c) stake relative to bankroll.

    Formal Solution: One-shot: positive EV, so a risk-neutral player accepts. A risk-averse player weighs the \epsilon$ edge against the chance of an unpleasant loss; for a tiny edge and a meaningful stake they might decline. Repeated play: by the law of large numbers the average outcome converges to $+2\epsilon$ per play, so playing many times is clearly attractive -- this is the whole logic of a trading desk grinding a small edge across thousands of trades. Catastrophic-loss caveat: if a single $\

    $ loss is large relative to your wealth (or if losses can compound), maximizing expected wealth is the wrong objective; you should size by Kelly or refuse, because risk of ruin destroys the ability to keep playing the edge.

    Answer: Yes, play -- the edge is positive (

    \epsilon$ per flip) and repetition makes winning nearly certain by the law of large numbers. The only reason to decline is if the stake is large enough relative to your bankroll that variance or ruin risk matters, in which case you size down (Kelly) rather than refuse outright.

    Intuition

    This is the foundational mindset of quantitative trading: a small, reliable edge played many times is a money machine, because variance averages out while the edge accumulates. The signal-to-noise ratio improving as $\sqrt{n}$ is exactly why high-frequency desks want enormous trade counts on tiny per-trade edges. The subtlety the best candidates raise unprompted is bankroll and ruin: a positive expectation is worthless if a bad run knocks you out before the law of large numbers kicks in, which is why sizing (Kelly, risk limits) matters as much as having the edge in the first place.

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