Price the Carnival Dice Game

Question

A carnival game: pick a number from 1 to 6 and bet 1; three fair dice are rolled. If your number shows on k ge 1 dice you win k (and keep your stake); if it shows on none, you lose the stake. What is your expected profit per 1 bet?

Accepted Answer

Matches K sim textBinomialleft(3, frac16right): P(K=0) = frac125216,quad P(K=1) = frac75216,quad P(K=2) = frac15216,quad P(K=3) = frac1216. Profits are -1, +1, +2, +3 respectively: mathbbE = frac-125 + 75 + 30 + 3216 = -frac17216 approx -0.079, a house edge of about 7.9%, worse than double-zero roulette. Why the game feels fair: "each die shows my number with probability frac16; with three dice that's an even-money shot, plus bonus payouts for doubles and triples." The expected number of matches is indeed 3 cdot frac16 = frac12, but matches cluster on the same outcomes, so P(textwin anything) = frac91216 approx 0.42, well short of frac12, and the enhanced payouts are far too rare (frac16216 combined) to close the gap. The "add up the chances" argument silently assumes disjoint events. Trader takeaway: price the payoff distribution, never the narrative around it. The fastest way to expose a bad game, or a bad structured product, is the full outcome table dotted with the payout vector, which is the entire three-line computation above. House edges live in exactly the places where a story double-counts overlapping good states.

Price the Carnival Dice Game

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