Balance the Bet

Question

A coin lands heads with probability 0.6. A bet pays you 10 on heads and costs you L on tails. What value of L makes the bet exactly fair, and should you take the bet if L = 14?

Accepted Answer

Set expected profit to zero: 0.6 (10) - 0.4 L = 0 implies L = frac60.4 = 15. The fair structure pays 10 against 15, i.e. losses can be 1.5 times the win because wins arrive 1.5 times as often: fracp1-p = frac0.60.4. That odds-ratio form, fair loss = textwin times fracpq, is worth keeping reflexive; it converts any probability into the payout ratio that prices it. At L = 14 the bet earns 0.6(10) - 0.4(14) = 6 - 5.6 = +0.40 per play, so yes, take it: you are risking 14 to make 10, which feels bad, and that feeling is exactly the bias the question hunts for. Risk-reward ratios mean nothing without the probabilities attached. Desk translation: a strategy that wins 60% of the time can afford to lose more per loss than it makes per win and still print money. Conversely "we win on 90% of days" says nothing until you see the size of the bad days. Always price the full two-sided structure, never the hit rate or the payoff ratio alone.

Balance the Bet

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