I just came across the following die probability problem at the xkcd blog:
Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6, the game is over. Sue rolls first. If she doesn’t roll a 6, Bob rolls the die; if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.
Bob rolls a 6 before Sue.
What is the probability Bob rolled the 6 on his second turn?
No solution is given in the post, so I figured I’d knock one out. As usual, I’ll over-explain.
The author emphasizes that the answer is not 5/36, presumably because it’s a common intuitive response. Though I don’t see why it would be. That’s the unconditional probability that Bob wins on his first turn. That is, the probability that Sue rolls something other than 6 (with probability 5/6), then Bob rolls a 6 (with probability 1/6), which gives: (5/6)(1/6) = 5/36.
It seems to me a more attractive intuitive answer (and one that a few commenters do indeed give) would be that Sue misses, then Bob misses, then Sue misses, then Bob hits on his second turn: (5/6)(5/6)(5/6)(1/6) = (53/64). But this is also wrong. It fails to account for the knowledge that Bob wins. In other words, this is the unconditional probability that Bob wins on his second turn, when what we need is the probability that Bob won on his second turn given that—i.e., conditional on the knowledge that—he won. Put yet another way, we need to find the proportion of time that Bob wins—that is, the probability that he wins, period—then find the proportion of that time that he wins on his second turn.
Finding the probability that Bob wins on his second turn is easy. We’ve already done it: (53/64). Finding the probability that Bob wins, period, requires some extra tools, which I’ll get to shortly.
And sometime soon, I’ll take the opportunity in a follow-up post to expand on those tools to solve two common probability questions involving a die: How many rolls of a fair, 6-sided die are expected to see a 6? How many to see all six sides?
Back to Sue and Bob. The first thing to notice, aside from “where’s Alice?,” is that this is a conditional probability problem. An easier example of which is: What’s the probability Bob rolled a 2 given that he rolled an even number?
We can reason this out intuitively. We have the following sample space for rolling a fair die, with equal probability for each outcome: {1,2,3,4,5,6}. Now condition on the knowledge that Bob’s roll gave an even number. In other words, update the sample space to: {2,3,6}. The 2 is one of three remaining, equiprobable options. So there’s a 1/3 probability Bob rolled a 2 given that he rolled an even number.
There are other ways to effectively restrict the sample space. A useful one is to just calculate the fraction of time (to put it one way) you find yourself in a “rolled an even number” world when rolling a die. That’s just the probability of rolling an even number. There are six possible outcomes in the standard sample space, three of which are even. So there’s a 3/6 = 1/2 probability of rolling an even number. This’ll be our denominator. For the numerator, pop in the fraction of time you’ll find yourself in a “rolled a 2” world when rolling a die. This ratio will yield the probability that you rolled a 2, given that you’ve found yourself in a “rolled an even number” world:
We can use either of these methods to answer the Sue vs Bob question. I definitely prefer the latter.1 I hope it makes sense why it works. If you rolled the die 60 times, you’d roll an even number 30 times and you’d roll a 2 ten times. Since 2 is even, this means that the proportion of times you rolled an even and got a 2 is 10/30 = 1/3. (This also supports the obvious conclusion that the probability you rolled an even given that you rolled a 2 is 1.) We arrive at 1/3 just as easily by using simplified fractions to begin with. But we can start with 30 and 20 to get there as well:
We can do essentially the same for Bob winning on his second turn given that he won. We just need to find the probability that he wins (regardless of what turn he wins on). Then we can put that into the denominator with the probability that he wins on his second turn in the numerator, like so:
The denominator is the tricky part. We need to work out the probabilities for all the outcomes in which he wins, and add those together. Conveniently, if we start listing the possibilities, a simple pattern quickly emerges. We need something like this: Sue misses, then Bob misses, then Sue misses, then Bob hits and wins. Every miss has a probability of 5/6 and the final, winning hit has a probability of 1/6. The idea is to track, over every possible scenario, how many instances there are of misses leading up to the winning roll, then multiply their corresponding probabilities. Like so:
Sue goes first, then Bob wins on his….
first turn: (5/6)(1/6)
second turn: (5/6)(5/6) × (5/6)(1/6)
third turn: (5/6)(5/6) × (5/6)(5/6) × (5/6)(1/6)
fourth turn: (5/6)(5/6) × (5/6)(5/6) × (5/6)(5/6) × (5/6)(1/6)
And so on into infinity. I’ve separated each set of turns by multiplication signs. Notice that the part in bold remains constant, while each successive term is multiplied by a new instance of (5/6)2. This is known as the “common ratio,” and from here on I’ll express it as 25/36 (similarly, I could express the constant as 5/36, but will keep it as (5/6)(1/6)). Notice also that the exponent on 25/36 can be easily determined by subtracting 1 from the turn number. For example, the probability Bob wins on his fourth turn is (25/36)3 × (5/6)(1/6). Winning on his 277th turn has probability (25/36)276 × (5/6)(1/6).
The only thing that changes from turn to turn is the power to which we raise 25/36. It will be raised to 0 then 1 then 2 then 3… onto infinity.2
We can generalize this pattern into a formula for finding the probability that Bob wins on the nth turn:
Now just add the probability of all those infinitely many outcomes together to find the probability that Bob wins. Maybe this is obvious, but in case it’s helpful, I’ll first apply this method to the simpler die case. The probability of rolling an even number is the probability of rolling a 2 or a 4 or a 6. Each of those mutually exclusive outcomes has a probability of 1/6, so we get: 1/6 + 1/6 + 1/6 = 3/6 = 1/2.3
We obviously can’t manually list out and sum up infinitely many probabilities. One fix for this would be to add up the first several instances and see where that takes us. Let’s do Bob’s first four turns:
So there’s about a 30% chance Bob wins on his first, second, third, or fourth turn; in other words, on or before his fourth turn. This is tedious work, so I’ll just pop the formula into Desmos‘ sigma function4:
Technically, the upper limit should be ∞ but Desmos doesn’t have that feature and 74 is plenty for getting an estimate to several decimal places. Actually, Desmos outputs .454545454545. So I assume the precise answer to be 45/99 = 5/11.
While understanding how to use this brute summation method is intuitively useful, there’s luckily a simpler way to sum the series. Just use this formula: a/(1–r). Where a is the constant and r is the common ratio. Recall that, above, we saw a constant of (5/6)(1/6), and a common ratio—i.e., the number by which we newly multiplied each successive term—of 25/36. Plug those in:
Again we get a probability of 5/11 that Bob wins (which means Sue wins with a probability of 1–(5/11) = 6/11). This result is the sum of an infinite geometric series. If you’d like to learn how to derive this powerful little formula and when it’s valid, here’s a fun Khan Academy video. (And see the Post Script below, where I offer an elegant alternative solution for finding the probability that Bob wins, from which this formulation naturally emerges.)
All that’s left now is to input the numbers into the ratio we stated at the offset:
So the final answer is around 21%. Some commenters on the xkcd blog post got it right, providing varying degrees of justification for their answers, but you’ll mostly find a small set of repeatedly cited wrong answers there.
Ok, that’s that. As I said, I’ll soon follow this up with a post covering two problems that expand on the tools explored here: How many rolls of a fair, 6-sided die are expected in order to see a 6? How many to see all six sides? I’ll link it once posted.
Post Script:
Here’s an alternative solution for finding the denominator—that is, the probability that Bob wins, period. Let “
The key observation here is that if neither Sue nor Bob wins on their first turn, the game resets. So we can get everything we need by looking at those two turns. Here are the relevant scenarios:
Either…
…Sue wins game on her first turn, in which case Bob then goes on to win the game with probability 0. We can disregard this scenario, as it adds 0 our equation.
…or Sue’s first turn misses, then Bob wins on his first turn. This scenario has probability (5/6)(1/6).
…or Sue’s first turn misses, then Bob’s first turn misses, then everything resets and Bob’s probability of winning the game resets to P(B). In other words, it’s as if they are starting a new game. The probability of this scenario is (5/6)(5/6)
Now build an equation by setting
This is exactly where we ended up above when plugging the starting term a and common ratio r into a/(1–r). We’ve already shown that this simplifies to 5/11.
Finally, we could have also used this streamlined approach to find the probability that Sue wins, and then subtract that from 1 to find the probability that Bob wins. It’s good practice, so I’ll go ahead and do that. Let 
Either…
…Sue wins game on her first turn with probability 1/6.
…or Sue’s first turn misses, then Bob wins on his first turn, then Sue goes on to win the game with probability 0. Disregard this one.
…or Sue misses, then Bob misses, then everything resets and Sue goes on to win the game with probability (5/6)(5/6)
Now build an equation that will look very similar to the one for finding
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Footnotes:
- Though I will give a snapshot of the former in the promised follow-up post.
- Recall that any non-zero number raised to the power of 0 equals 1.
- This only works when events are mutually exclusive. You can’t roll a 2 and a 4 at the same time, so you can get the probability of landing a 2 or a 4 by adding up their respective probabilities. But notice that you can simultaneously roll a 2 and an even, so you can’t find the probability of landing a 2 or an even by adding 1/6 and 1/2. What to do in that case is a discussion for another day.
- If you need a refresher on sigma notation, here’s a quick and clear one.