36.8% in Probability

I made a post a little while ago On the Occurrence of Improbable Events. Recently, I saw from WordPress’s stat-tracking system that someone was able to get to that post by a search for 36.8 probability (as a few references to 36.8% were made in the post). Interested, I went to Google and typed 36.8 probability myself (the percent sign doesn’t affect the results) and, sure enough, I was surprised to find my own post as the first result.

Perhaps this is just demonstrating the Law of Large Numbers, that given enough posts I made, the chance that one of them would be on the front page of some Google search increases to reasonable levels.

Curious as to what would come up if I simply typed 36.8 into Google, I got the following results:

36.8 Search

The results are also unrelated if 36.8 is replaced with .368 or 0.368; I don’t want to throw in that many Google search screenshots in here, so you can take my word for it, or try searching them.

What Is 36.8%?

Noticing this discrepancy of 36.8, I decided to write a post on 36.8% in probability. Here it is.

First of all, from where did this number appear? In the last post, I was discussing the probability that improbable event would (not) occur.

Suppose you were to flip a fair coin. You have a 50% chance of heads (success) and a 50% chance of tails (failure). Since there is a 1/2 chance you’ll get heads, let’s flip the coin twice, to increase the chance of heads. Here, the chance that both trials will fail, i.e. that both coins will land tails, is 1/4 or 25%. Note that we may think of this as the chance that the first coin will land tails (1/2) times the chance that the second coin will land tails (1/2). So, 1/2 times itself is 1/4.

But for our purposes, a coin has too high of a chance of success. Let us modify the experiment so that we use a fair six-sided die. Suppose rolling a one is success; all other outcomes are failures. The chance of success is thus 1/6, and the chance of failure is 5/6. Now we throw six dice, which is the expected number for one success. Expectedly, the dice should read 1-2-3-4-5-6, but that’s probably not going to happen. Instead, we could end up with 1-1-1-1-1-1, 6-6-6-6-6-6, or anything in between. Now let us calculate the chance that all six throws will fail. The chance that the first will fail is 5/6, the second 5/6, the third 5/6, etc. So, we multiply 5/6 by itself several times—mathematically, we raise 5/6 to the 6th power. The chance that all trials fail is 33.5%.

In the last post I supposed the probability of snow in Austin on any given day was 0.1%, or 1/1000. The chance of failure is 99.9%, or 999/1000. The chance that 1000 days will all fail is 999/1000 to the 1000th power, or 36.8%.

36.8%! Let’s compare the numbers we tabulated so far. When there was a 1/2 chance of success, the probability that all trials fail is 25%. For 1/6, it is 33.5%, and for 1/1000, it is 36.8%.  But in mathematics we love to find patterns. And a simple pattern so far is: as the chance of success decreases, the chance that all trials fail in the expected number of trials increases, and seems to approach a number near 36.8%.

Let us say this with symbols. Let p be the probability that an event will occur. For the sake of simplicity, let us assume that p is “one over a positive integer,” or {\frac{1}{n}}. The expected number of trials is n, the denominator of the probability. The chance that the event will not occur is one minus the chance it will occur, or 1 - \frac{1}{n}. And the goal is to find the probability that all trials fail, that is, the event never occurs, P(0). This is not so difficult. It is simply the chance that a single event does not occur, raised to the power equal to the expected number of trials. Actually, since we want to differentiate between the results for different values of n, such as n = 2, n = 6, and n = 1000 for the examples above, let us use a subscript. We define P_{n}(0) as the probability that there are no successes in n trials when the probability is {\frac{1}{n}}.

\displaystyle P_{n}(0) = \left( 1 - \frac{1}{n} \right) ^ n

And here is what we know so far:

\displaystyle P_{2}(0) = \left( 1 - \frac{1}{2} \right) ^ 2 = .25

\displaystyle P_{6}(0) = \left( 1 - \frac{1}{6} \right) ^ 6 \approx .335

\displaystyle P_{1000}(0) = \left( 1 - \frac{1}{1000} \right) ^ {1000} \approx .368

What happens when n is really, really large, like 649740? In this case, the probability for success in one trial is 1/649740. A poker fan will recognize this as the chance of obtaining a royal flush. We plug n = 649740 into the formula and obtain

\displaystyle P_{649740}(0) = \left( 1 - \frac{1}{649740} \right) ^ {649740} \approx .368

.368 appears again. But what if n is infinitely large? We’ll call that probability {x}. We can write this as a limit:

\displaystyle x = \lim_{n \rightarrow \infty} \left( 1 - \frac{1}{n} \right) ^ {n}

Of course, if we just plug in n = \infty into the equation, we would end up with 1^\infty, which is mathematically meaningless (it doesn’t necessarily equal one). Thus we use a limit, which allows us essentially to treat the infinite n as a finite number. But this doesn’t help us with the value very much. What is {x} exactly? For this task, we shall use the binomial expansion formula.

For instance, it is possible to expand (a + b)^4 as a^4 + 4a^3b + 6 a^2b^2 + 4ab^3 + b^4. To state it generally, i.e. when n is any positive integer,

\displaystyle (a + b)^n = \sum_{k = 0}^n \binom{n}{k} a^{n-k}b^k

The \binom{n}{k} notation means the number of ways to choose {k} elements from a set of n elements, and is equal to \frac{n!}{k!(n-k)!}. Now we simply substitute a = 1 and b = -\frac{1}{n}:

\displaystyle \left( 1 - \frac{1}{n} \right) ^ {n} = \sum_{k = 0}^n \binom{n}{k} \left( -\frac{1}{n} \right)^k =\sum_{k = 0}^n \frac{{(-1)^k} n!}{k!(n-k)!n^k}

When n is extremely large, n, n-1, n-2, etc. are virtually the same. That is, n! \approx (n-k)!n^k. Thus we can say

\displaystyle x =\sum_{k = 0}^n \frac{(-1)^k}{k!}

Then we’ll expand the first few terms:

\displaystyle x = \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots

In other words, {x} is the alternating sum of the reciprocals of factorials. The value turns out to be \frac{1}{e} (where e is Euler’s number), or 0.367879441...

There, we’ve established the origins of 0.368. It’s a rounding of the reciprocal of e, a very famous number. And it’s also the probability that an event will never occur in n trials when the chance to occur for each trial is {\frac{1}{n}}, when n is large.

36.8% is also the solution to the marriage problem; this problem is expressed and named in a number of ways. Wikipedia, for example, calls it the secretary problem. Roughly, the idea is to obtain the best choice among a set of choices, but to do so in an optimal manner. The answer is to reject the first 36.8% of the choices and then pick the next choice that is better than all the previously rejected choices. The probability of picking the best choice using this method is 1/e. Because we like to think in terms of integers, the solution is also known as the 37% rule.

But let us return to the improbable event. There is actually a distribution that handles such events (including events with higher chances of occuring)—the Poisson distribution. What we figured out above was that given a sequence of events with a total expected value of one occurrence, the chance that it will occur zero times is 36.8%. So the chance, as we know, of drawing no royal flushes in 649740 random hands is 36.8%. What about the chance of it happening once? That is, in 649740 hands, what is the chance that there is exactly one royal flush? The answer, surprisingly (or unsurprisingly), is also 36.8%. And unfortunately for gamblers, this trend does not continue: the chances of two royal flushes are 18.4% (1/2 of 36.8%), three is 6.1% (1/3 of 18.4%) , four is 1.5% (1/4 of 6.1%), and five 0.3% (1/5 of 1.5%). But it is amazing nonetheless, that the probability of something like this not happening is exactly equal to the probability that it happens once.

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