Reroll the Ones and the Die Is Worth Four

Roll a fair six-sided die. If it shows a one, roll it again, and keep rerolling until it shows something else. I pay you that face value in dollars. What is the game worth?

Almost everyone answers three and a half, which is the average of a plain die and therefore correct arithmetic about a different object. The answer here is four dollars, and the gap between the two numbers is the whole content of the problem.

The rule deletes an outcome rather than reweighting six

The die is fair and stays fair. What changes is the set of outcomes that can ever pay you. Whatever ends the process is a two, three, four, five or six, because a one always sends you back for another roll.

So the payout is uniform on $\{2,3,4,5,6\}$ and its mean is

The fastest way to get there at a whiteboard is to note that the six faces sum to twenty-one, dropping the one leaves twenty, and twenty over five is four. That is exact rather than an estimate.

The reason three and a half feels right is that people picture the reroll as making the one somehow worth less, which would keep six outcomes and shuffle their weights. It does not. The one is not a low payout, it is not a payout at all.

The slow route agrees, and it is not wrong

You can also count the rejected rolls explicitly. With probability $(1/6)^k$ the first $k$ rolls are all ones, and the next roll contributes its own expectation over the six faces excluding the reroll branch, which is $20/6$. Summing over $k \ge 0$,

Same answer, more work. It is worth saying plainly that this route is valid, because the temptation when comparing two methods is to treat the slower one as an error. Equation (2) is correct; it is simply the long way round, and if you find yourself writing it under time pressure the useful thought is that a geometric series with a single repeated state usually means you should have conditioned instead.

The rerolling is cheap as well as profitable

A natural worry is whether the process ends at all. The chance of needing more than $k$ rolls is $(1/6)^k$, which goes to zero, so the game terminates with probability one and the expectation in (1) is well defined. The number of rolls is geometric with success probability $5/6$, so

Five sixths of the time you roll once. One time in thirty-six you roll three or more. The extra half dollar costs you one fifth of a roll on average, which is about as good a trade as arithmetic offers.

Reject more faces

The argument used nothing about the number one except that it was removed. Reject every face at or below $m$ and the payout is uniform on $\{m+1, \dots, 6\}$, whose mean is the average of its endpoints:

which gives $7/2$ at $m=0$, then $4$, $9/2$, $5$, $11/2$ and finally $6$ at $m=5$, where you are rerolling until you see a six. The expected number of rolls there is $6$, which is the price.

More generally still, reject any subset $S$ of faces and the value is just the mean of the complement. Nothing about the die is being altered, so the answer is always an unweighted average over whatever the rule leaves standing.

That sentence has a name, and knowing it makes the whole problem feel routine. The procedure is rejection sampling: draw from a distribution you have, throw away anything outside the set you want, and what survives is distributed exactly as the original conditioned on that set. Here the original is uniform on six faces and the set is five of them, so the survivor is uniform on five. The reason the argument is so short is that a uniform distribution restricted to a subset is uniform again, which is not true of distributions in general.

The efficiency of a rejection scheme is its acceptance rate, which here is $5/6$, and the expected number of draws is its reciprocal. That is (3) arriving from a completely different direction. If the die had a hundred faces and you rejected ninety-nine of them, the answer would still be a one-line average and you would wait a hundred rolls for it.

A different game: your choice, once

Change one word and the problem changes character. Suppose the reroll is optional and yours to take, once, on any face you like. Now you have a decision, and the right rule is to reroll whenever the face in front of you is worth less than what a fresh roll is worth:

so keep a four, five or six, and reroll a one, two or three. The value is

which beats the forced version, and it should: being allowed to choose is worth more than being made to reroll a specific face. With $k$ optional rerolls the threshold rises each time, computed backwards from (5), and as $k \to \infty$ the value climbs to six, because with unlimited attempts you simply wait for a six.

Confusing the two games is the more sophisticated mistake available here, and it is worth being careful about which one you were asked. The forced game is a question about a conditional distribution. The optional game is a question about a stopping rule, and only the second one has a decision in it.

One note on the framing

Nothing here is a wager. You pay nothing to play and you cannot lose, so the four dollars is the value of a payoff rather than an edge over anybody. If the game did cost something to enter, the interesting number would be four minus the entry price, and the reroll rule would matter only through the half dollar it adds.

Sources and further reading

• The quantity computed in (1) — Expected value
• The distribution of the roll count in (3) — Geometric distribution
• The step in the definition above — Conditional probability distribution
• The framework behind (5) and (6) — Optimal stopping

Every number above was checked before publication three ways: exact conditioning, the geometric series both in closed form and truncated at forty terms, and ten million seeded plays of the literal roll-and-reject procedure, which averaged $4.000$ to three decimals at $1.2$ rolls a play. The generalisation (4) was verified for every $m$ from zero to five by both routes.