OK, so DiMaggio had one chance in 125,000 to hit 56?

No, that's not what I said. What I said is that in any given string of 56 games, there is one chance in 125,000 that every one will be successful. A player gets more than one opportunity per season.

In any given season of length "x", there are x chances to hit in one consecutive game. There are therefore x-1 chances for two game streaks, and x-2 chances for three game streaks. The number of opportunities to achieve a streak of y in a season of length x is  (x-(y-1)) or, more simply x+1-y.

Is the math giving you a headache? It's simple. How many chances do you get for a 56 game streak in a 139 game season. The formula says 84 chances. Forget the formula and do it in your head. You can achieve such a streak in games 1-56, games 2-57, and so forth until games 84-139. That's it. That's all. 84 different starting dates. You can't start a 56 game streak in game 85 because streaks do not carry over to the following year, so the streak can never reach 56. Playing in 139 games, DiMaggio had exactly 84 chances to hit in 56 consecutive games in 1941.

Let's look at that over a million seasons, since if we use a smaller number we will be dealing in tiny decimals. DiMaggio had about one chance in 125,772 to achieve a "56 for 56" string, and he would achieve 84 million such opportunities in a million seasons. Therefore, that million years would produce 668 such strings.

So he has about one chance every 1500 years or so? No.

Those 668 strings include streaks of 56 games, 57 games, 58 games, and more. Each 57 game streak produces TWO strings of "56 for 56". Each 58 fame string produces THREE strings of "56 for 56".

I think you can best understand this point by answering this trivia question. "How many 45 game hitting streaks have there been in the history of baseball?" If you are a baseball wizard, you answered either "zero" or "one", depending on whether you assumed that I meant "exactly 45" or "at least 45". You are wrong. You are not even close. If you are a mathematical wizard, you know that the correct answer is twelve. All twelve of them were amassed by Joe DiMaggio in the summer of 1941. He had such a streak in games 1-45 of his famous hit streak. He had another one in games 2-46, and so forth until the one he had in games 12-56. If you search through all the forty five game series in baseball history, you will find 12 strings in which all 45 were successful, yet there has never been a hit streak of exactly 45 games, and there has only been one of 45 games or longer.

Therefore, we are nowhere near the end of our calculation. Knowing that we have 668 strings of "56 for 56" is no more useful than knowing we have twelve strings of "45 for 45". We have to find a way to convert that to the number of streaks of exactly 56 games, no more. Once we know that, we can repeat the entire process for 57 game streaks, 58 game streaks, and so forth, which will enable us to deduce the number of 56 or larger.

Let's not get ahead of ourselves. For now, let's calculate the number which are exactly 56 in length.

We have 668 strings of "56 for 56". What percentage of them will be from larger streaks.

The math sounds complicated here. It's actually very simple math, but it sounds complicated.

In a 139 game season, there are 84 possible ways to accumulate a 56 game streak, as noted earlier.

• Of those 84, two of them are "one-tailed": the first one and the last one. If you find a perfect string of 56 in the first 56 games, that streak is exactly 56 games long if there is no hit in game 57. The chance that DiMaggio will not hit in game 57 is 18.9%. Therefore, there is an 18.9% chance that this streak is a stand-alone, and an 81.1% chance that it is part of a longer streak. The same logic applies to a string of "56 for 56" found in the last 57 games of the year.

• The remaining 82 series are "two-tailed". If you find a perfect string of 56 in games 2-57, that streak is exactly 56 games long if there is no hit in game 1 AND no hit in game 58. The chance that DiMaggio will not hit in either game is (18.9% to the second power), or about 3.6%. Therefore, there is an 3.6% chance that this streak is a stand-alone, and an 96.4% chance that it is part of a longer streak.