Say you plan to roll a die 20 times. Which of these results is more likely: (a) 11111111111111111111, or (b) 66234441536125563152?

I recently enjoyed an interesting comment exchange with a reader named Abs based on my earlier post The Law of Small Numbers. In essence, he challenged me with the question in the title of this post:

Say you plan to roll a die 20 times. Which of these results is more likely: (a) 11111111111111111111, or (b) 66234441536125563152?

Based on probability theory, I answered that they are both equally likely. More specifically, the probability that either sequence of numbers resulted from the rolls is (1/6)^20 (one-sixth to the twentieth power).

However, he pointed out that Marilyn Vos Savant (of the Monty Hall Trap fame) had given the following answer to the same question:

In theory, the results are equally likely. Both specify the number that must appear each time the die is rolled. (For example, the 10th number in the first series must be a 1. The 10th number in the second series must be a 3.) Each number—1 through 6—has the same chance of landing faceup.

But let’s say you tossed a die out of my view and then said that the results were one of the above. Which series is more likely to be the one you threw? Because the roll has already occurred, the answer is (b). It’s far more likely that the roll produced a mixed bunch of numbers than a series of 1’s.

Reference: Ask Marilyn – Sunday’s Column – July 31, 2011

In a follow-up post, a math teacher from Minnesota challenged her answer:

I’m a math instructor and I think you’re wrong about this question: “Say you plan to roll a die 20 times. Which result is more likely: (a) 11111111111111111111; or (b) 66234441536125563152?” You said they’re equally likely because both specify the number for each of the 20 tosses. I agree so far. However, you added, “But let’s say you rolled a die out of my view and then said the results were one of those series. Which is more likely? It’s (b) because the roll has already occurred. It was far more likely to have been that mix than a series of ones.” I disagree. Each of the results is equally likely—or unlikely. This is true even if you are not looking at the result.

Marilyn’s response to him:

My answer was correct. To convince doubting readers, I have, in fact, rolled a die 20 times and noted the result, digit by digit. It was either: (a) 11111111111111111111; or (b) 63335643331622221214.

Do you still believe that the two series are equally likely to be what I rolled? Probably not! One of them is handwritten on a slip of paper in front of me, and I’m sure readers know that (b) was the result.

The same goes for the first scenario: A person rolled a die out of my view and then informed me the result was one of these series: (a) 11111111111111111111; or (b) 66234441536125563152. It was far more likely to be (b), a jumble of numbers.

Reference: Ask Marilyn – Sunday’s Column – October 23, 2011, Parade

As evidenced by the math teacher’s response, Marilyn’s answer seemed to contradict the very well known (and true) laws of probability. But both the math teacher and Marilyn are correct. Marilyn simply could have done a better job in framing her response. Her response was actually not incorrect, but by answering the question from a different perspective, she generated some confusion. (Giving her the benefit of the doubt, perhaps this was her intention?)

First of all, in the first part of her original answer, Marilyn did fully acknowledge that based on probability, both results (a) 11111111111111111111 and (b) 66234441536125563152 are equally likely in the exact sequence that they occur. That is absolutely true and incontrovertible. That is, if you roll the die 20 times in an infinite number of trials, both sequences would occur with the same frequency, either one occurring roughly once every 6^20 trials.

However, in the second part of her original answer, Marilyn continued on to answer the question from a different perspective. She says that after actually rolling a die 20 times, it’s far more likely that the roll produced a mixed bunch of numbers than a series of 1’s, and thus, (b) 66234441536125563152 is more likely than (a) 11111111111111111111.

So, here, Marilyn is actually answering the following question:

Say you plan to roll a die 20 times. Which of these results is more likely: (a) twenty 1’s in a row (of which 11111111111111111111 is the only example) or (b) a mixed bunch of numbers (of which 66234441536125563152 is just one example)?

Note that the exact sequence is no longer in question, but only whether the die yields the same number repeatedly or not.

Here, the answer would clearly be (b), since there’s far more possible combinations of “mixed bunch of numbers” than twenty 1’s in a row. More technically, the entropy is greater for (b) than (a), and thus more likely.

To her credit, Marilyn never actually said anything incorrect in either of her answers. Perhaps this is because the original question was vague enough to be open to interpretation. That is, the original question could be interpreted in either of the following two ways:

(1) Say you plan to roll a die 20 times. Which of these results is more likely: (a) 11111111111111111111, or (b) 66234441536125563152, in the exact sequences as written?

(2) Say you plan to roll a die 20 times. Which of these results is more likely: (a) 11111111111111111111, or (b) 66234441536125563152, when taken as a final result, where the exact sequences are not important?

The answer to (1) is: (a) and (b) are equally likely.
The answer to (2) is: (b) is much more likely than (a).

Marilyn gave both answers. Clever!