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A Hard Lesson Learned

Have you ever had a problem stuck in your head, and you couldn’t find the answer? I was recently reminded of a problem I first came up with while doing a statistics workbook the summer of my 3rd grade year (yes, my math-teacher mother gave us workbooks to do during summer break… hey, it got results). The book dealt with dice and the probabilities of a 1 showing on a 6-sider, or the sum of 2 rolled dice being 7, etc. but my question had a twist I couldn’t quite solve. Now it’s easy to see that the probability of rolling a 1 on a 6-sided die is 1 in 6, but what probability exists, in rolling 2 dice, of seeing at least one 1 (on either die, or both)?

So there I was, 7 years old and stuck with a math problem I couldn’t solve. Well, what would you do? That’s right, I asked people. Over the next 9 years, I asked math teachers and other adults I thought might be able to help, but no one seemed able to explain it to me. Of course, I could brute force the answer for 2 or even 3 dice quite easily as well (and did), but by that time I wasn’t interested in just the single answer to my problem, but a more general solution for N dice. The difficulty of the problem comes from the “at least one” phrasing. Using discrete math, you could end up having to compute the probability for each of the numbers of dice smaller than your requested count and use alternating subtraction and addition to account for subset solutions and overlap.

Of course, I did come up with (or was given) attempts at a quicker solution. They usually fell into 2 types. The first type was the simplest; since we were rolling two dice, and the probability with 1 die was 1 in 6, the new probability must be 2 in 6. Unfortunately, this easily extrapolates to say that when rolling 6 dice, you absolutely must get a 1, 2, 3, 4, 5, and 6. Anyone who plays Yahtzee can certainly see the flaw there. The second type of false solution came from, at some point, someone recalling some remnant of a college statistics course (the part where previous rolls shouldn’t affect subsequent rolls) and concluding that the answer must be still 1 in 6. This answer also fails to satisfy quickly when you consider more than just 2 dice being rolled. How could the probability of getting at least one 1 when rolling 10 or even 100 dice still be just 1 in 6?

It wasn’t until I reached college and took an actual course in statistics that I found the answer. Fortunately, there’s an easy method. If you ever find a statistics problem that uses the “at least one” phrase, the best bet is to turn it around. What are the odds of not getting any 1s on those dice? As it happens, that’s merely 5 in 6 for each die rolled, multiplied together. In my problem, with only 2 dice, the probability of not getting a 1 is 5/6 * 5/6 = or 25 in 36. Now, since all the other possible results must contain a 1, we’re left with a solution of (36-25) = 11 in 36! This method works for N dice as well, with the probability of getting at least one 1 out of N dice rolled being:

P(N) = 1 – (5/6)^N.

Imagine that, all this time I’d been asking relatively smart people for the answer to a question I’d found a long time ago and the trick wasn’t really in finding the solution, but in re-wording the question so the solution became obvious. This is a lesson every math student will learn along the way. I wish I could go back and tell my 7-year-old self how easy it was to solve, without him having to see how hard it was to find the answer.

The Count

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