Ten to One on Four Coin Calls Costs 31 Cents on the Dollar

Four coin flips. A ticket costing one unit returns ten if you call all four correctly, and nothing otherwise. Is it worth buying?

No, and the gap is not close. One chance in sixteen needs fifteen to one to break even, so at ten to one the buyer hands over 31.25 cents of every dollar committed. Ten to one only sounds generous because ten is a big number next to four.

Four halvings and one survivor

Each call is independent and each is exactly even, so the probability that all four come right is the product:

Counting says the same thing without multiplying anything. Four flips generate $2^4 = 16$ sequences, all equally likely, and exactly one of them is the sequence you called. Every flip halves whatever is still alive: sixteen down to eight down to four down to two down to one.

The payout that would make it fair

Set the expected profit of a binary payoff to zero. With probability $p$ you gain $R$, otherwise you lose 1:

At $p = 1/16$ that is $(15/16)/(1/16) = 15$. Fifteen to one, and the arithmetic is visible: the fair net payout is the number of losing outcomes divided by the number of winning ones, so it counts the fifteen sequences that fail against the one that does not.

There is a third way to see the same gap, and it is the one that scales to comparing several offers. Invert the quoted payout to recover the probability it implies:

The offer is priced as though the four calls came right nine times in a hundred. They come right six and a quarter times in a hundred. You are paying for a chance 45 percent larger than the one you are actually buying, and reading a price back into an implied probability is the general move: it puts every quote on one scale, whatever units it was quoted in.

Read gross instead of net and nothing changes except the words. A ticket returning eleven in total for one is worth $11/16 = 0.6875$, so a dollar buys 68.75 cents of value. The 31.25 cents in (3) and the 68.75 cents here add to a dollar, which is the check that the two conventions have been kept apart.

Every extra leg doubles the fair payout

For $k$ independent even calls, substituting $p = 2^{-k}$ into (2) gives:

One leg needs 1 to 1, two legs need 3 to 1, three legs need 7 to 1, four legs need 15 to 1. The fair payout doubles and adds one with every leg added, so it runs away from any payout a seller is likely to quote. That is the structural reason multi-leg tickets are attractive to the side writing them: the honest price grows exponentially in the number of legs while the buyer's sense of what sounds generous grows roughly linearly.

Six legs already need 63 to 1. Ten legs need 1023 to 1.

How long the loss stays invisible

A negative expectation of 31.25 cents is not something a buyer notices quickly, and it is worth knowing why. The payoff per unit committed is $+10$ with probability $1/16$ and $-1$ otherwise, so:

The edge is 0.3125 a ticket and the spread of a single ticket is 2.663, more than eight times as large. The cumulative loss reaches one standard deviation of the running total only when $0.3125\,n = 2.663\sqrt{n}$, that is at $n \approx 73$ tickets. Before that the arithmetic is buried in noise, and a buyer with a run of luck has genuine evidence of nothing at all.

Where the verdict flips

Equation (1) used two assumptions and both are worth attacking. The first is that each call is exactly even. Suppose instead you have a genuine edge and call each leg right 55 percent of the time:

The break-even payout has dropped below ten, so at ten to one the same ticket now returns +0.00657 per unit. A five-point edge per leg turns a 31-cent loss into a positive expectation, which tells you how much of this problem is really about the assumption rather than about the arithmetic. Four coin flips is the no-information case, and it is the case the question poses.

The second assumption is independence, and dropping it moves the answer much further. If the four calls were perfectly correlated, so that they all resolve the same way, the chance of winning is a half and ten to one is enormously profitable. Correlation between legs is the single largest source of mispricing in tickets of this shape, in both directions, because multiplying probabilities is only legitimate when the events do not talk to each other.

One thing that does not flip the verdict is preference. A buyer may prefer a small chance of a large payout to its expected value, and there is nothing incoherent about that. Equation (3) is a statement about the price, not about anyone's appetite for variance, and the two arguments should not be allowed to change places mid-discussion.

Sources and further reading

• The quantity equation (3) computes — Expected value
• Why the four probabilities may be multiplied — Independence (probability theory)
• The net and gross conventions, side by side — Odds
• A single call, as a distribution — Bernoulli distribution

The sample space has sixteen points, so the probability, the break-even payout and the expected value were all settled by exact enumeration in rational arithmetic. A simulation of two million tickets measured a win rate of 0.062189 against 0.0625 and a loss of 0.31593 per ticket against 0.3125. The same draws priced at the fair fifteen to one returned 0.00498, inside one standard error of nothing, which is the control that keeps the loss attributable to the price. The doubling rule in (5) was enumerated for one through twelve legs.