Sometimes insight from probability allows us to solve problems or prove theorems from branches of mathematics outside of probability. Sklansky gave an example of this with a problem that asked "If 1 billion Chinese couples continued to have children until they had a boy, and then stopped, how many children would be born on average, assuming boys and girls are equally likely?". You could solve this problem if you knew how to sum the series 1*1/2 + 2*1/4 + 3*1/8 + .... But if you don't know how to sum this series, you could simply notice that each family will have exactly 1 boy, and since half of the children will be boys, there must be a billion boys and a billion girls for a total of 2 billion children born. Therefore we can conclude that the series equals 2, which is correct.
Here is one I just discovered. I claim that for all 0 < x < 1, and all integers n > 1,
(1 - x/n)^n > 1 - x
This could probably be proved for all values of n by induction, but then it would have to be proved for all x, perhaps by considering derivatives and logarithms. Here is a simple probabilistic argument:
Proof: Suppose we have a set of n possible events, each with an equal probability of occurring on a given trial. Take x to be the probability that at least one of the events will not occur in some number of independent trials N > n. For example, the n events could be the 38 numbers on a roulette wheel, and x would be the probability that at least one of them does not occur in N > 38 spins. Then the probability that any particular event fails to occur in N trials must be >= x/n, with equality being taken when the events are mutually exclusive. This is because if this probability were x/n, simply summing n probabilities x/n to yield x will always overestimate the probability that some event fails to occur by over counting the cases where more than one event fails to occur, unless the events are mutually exclusive. Therefore since we know that the probability that some event fails to occur in N trials is indeed x, the probability that a particular event does not occur in N trials must be >= x/n.
Now consider the probability that all of the n events occur in N trials, 1 - x. Note that the probability that a particular event occurs in N trials must be > (1 - x)^1/n. This is because if this probability were (1 - x)^1/n, simply raising this to the nth power to yield 1 - x would produce the probability that all of the events occur only if the events occur in N trials independently of one another. This is not the case, however, since the occurrence of an event will allow fewer chances for another event to occur, so raising the probability that a single event occurs to the nth power must overestimate the probability that all events occur. Therefore, since we know that the probability that all events occur is indeed 1 - x, the probability that a particular event occurs must be > (1 - x)^1/n, so the probability that a particular event does not occur in N trials must be < 1 - (1 - x)^1/n. Since we already proved that this probability is >= x/n, x/n < 1 - (1 - x)^1/n. By rearrangement, QED.
The interesting thing is that this inequality doesn't depend on probability theory; we just have some functions of numbers. Yet we are able to say something about how these functions behave because of our knowledge of independence and mutually exclusive events from probability theory.
This has been verified up for n = 2 to 1,000,000,000 and x = 1/1,000,000,000, at which point Excel fails to recognize the truth of this theorem; however, the windows calculator, with it's 32 digits of accuracy, still maintains the small difference between the two terms, at that point on the order of 10^-13 (smallest for n=2). Then it verifies it for x = 1/1,000,000,000,000 for which the difference is on the order of 10^-25 at n=2. This inequality is a good way to check the precision of your calculator.
This was used in the solution to
this problem.