Posted August 30, 2010 By Presh Talwalkar. Read about me , or email me .



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"All will be well if you use your mind for your decisions, and mind only your decisions." Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon. .

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I usually bowl a score around 130, but the other day I hit six strikes and wound up with 215.

I took a moment to revel in the high score, but then I got thinking more critically. I was curious about how good the score was in a statistical sense.

I was with some math-minded friends and our discussion brought about many questions.

How many different bowling games are possible? What’s the average bowling score? How many ways can you attain each bowling score (what is the distribution of bowling scores)?

I did a bit of research and was pleased to learn people have already done the math to answer these questions. Here is some of the interesting math.

Writing a bowling game in mathematical terms

The first step in the problem is to translate a bowling game into mathematics. The idea is to develop a short-hand notation to compactly describe the game. I’ll follow the notation developed in this article “Is the Mean Bowling Score Awful.”

Consider the very first frame of a game. What are the possible outcomes when you bowl the ball?

One possibility is you knock down all 10 pins with a strike and the first frame ends.

The other possibility is you don’t knock all 10 pins. Then you get a second chance to hit the remaining pins. If you hit the remaining pins down, it’s a spare.

Mathematically we can model the bowls as a set of two numbers: the number of pins knocked down on the first throw, and the number on the second. We can write this as an ordered set (first throw, second throw) = (x , y).

If you knock down 3 pins and then 4, the set is (3, 4). Or if you knock down 3 and then 7 to make a spare, that would be written (3,7)

In a strike you knock down all 10 pins and you don’t get a second throw. We can write this special case as (10, 0) with the understanding you never actually made a second throw.

With this notation, we can compactly describe the possible outcomes in the first frame. The outcomes can be written as an ordered pair of two numbers, where both numbers are zero or positive, and the two numbers sum to at most 10–since the most you can knock down in a frame is 10 pins.

In set notation, this is written

The first nine frames of the game operate in the same fashion.

The tenth frame of the game is slightly different. If you get a strike in the first throw, or a spare in the second throw, you get to make a bonus third throw in the tenth frame. Therefore the tenth frame has to be represented by three numbers, with special relations depending on whether a strike or a spare is made.

There are four different possibilities:

–Two throws are made, less than 10 are knocked down (no third throw)

–The second throw makes a spare

–The first throw is a strike but the second is not

–Two strikes are made

The notation gets more complicated, but essentially it’s how you would write out these four possibilities in set notation. Here is the formal description:

Therefore, we can write a bowling game as nine pairs of elements from set A and one element from set B.

In other words, a bowling game is a sequence:

And viola, we have a mathematical way to write out a bowling game.

We will address how to account for scoring of spares and strikes in a bit, as this is more complicated.

How many bowling games are possible?

This question is easier to answer since we have a notation system for a bowling game.

We know the first nine frames of a bowling game are elements from set A and the tenth frame is from set B as described above.

It remains to count the number of elements in each of these sets. Then we multiply the number of ways to get the number of total games.

The set A is the number of ways that two positive whole numbers sum to 10 or less. This is a classic combinatorics problem.

There is a clever way to count the number of solutions. I found a derivation via Google Books for finding the number of ways n non-negative integers sum to an integer r:

Link to page 46 of Principles and techniques in combinatorics at Google Books

The formula is C(r + n -1, r)

In our bowling set, we want to find the way two numbers (n = 2) sum to 10 or less (r = 10, 9, 8, …, 0).

We want to calculate the formula for each value of r and then sum them all up. This is less work than it sounds.

For r = 10, we see the formula is C(11, 10) which is 11. For r = 9, the formula becomes C(10, 9) which is 10. The pattern continues for lower values of r, so in the end we want to sum up 11 + 10 + 9 + … 1.

This is readily calculated as 66. Thus there are 66 ways for each of the first nine frames.

We now want to know the number of ways for the tenth frame.

The process is the same as before. I will spare the gory details on this one and just say the answer is 241.

Now we can compute the total number of bowling games by multiplying the numbers together.

The total number of bowling games is (66 x 66 x 66 … x 66) (241) = (66 9) (241), which is approximately 5.7 x 1018

This is no where near the number of possible chess games, but it is still a really large number.

To put it in perspective, it would take the entire world (6.7 x 109), playing a game every day, over 2.3 million years to play that many different games.

What’s the average bowling score?

This part gets even more complicated mathematically.

The trick is to convert the sets into scores, based on the special rules for spares (bonus of next throw) and strikes (bonus of next two throws).

Then the mean can be computed by summing all possible scores by the number of games, which was derived earlier.

The mean bowling score turns out to be about 80 (or to be precise, more like 79.7).

Take that as reassurance that even a modest score like 100 is above average!

The derivation details are explained in the following excerpt:

Link to Mean bowling score at Google Books

The article is dated because it ends with leaving an open question of determining the full distribution of bowling scores. This in fact has been done.

What’s the distribution of bowling scores?

The final and really tough question is finding the bowling distribution.

That is, for each score n, what is the number of ways s(n) to achieve that score.

There are some cases where the answer is obvious. There is only 1 way to get a score of 0, as there is only 1 way to score 300, or 299, or 298, and so on until 291.

The other cases are more complicated to figure out. There are 20 ways you can score 1, and there are 11 ways to score 290.

To figure out the entire distribution requires clever computation. The results are described in this wonderful web page which contains the following nice graphic:

image source: all about bowling scores

Notice that bowling scores are heavily skewed! Scores above 120 are less likely since it requires a player getting a reasonable number of spares and strikes.

Another way to think about this is that even a modest score of 115 is in the 99th percentile.

Remember this the next time you’re out bowling. Given the range of possible bowling scores, your score is probably better than you think!

(The percentiles will change if we base the distribution on actual bowling data. Sadly I was not able to find any stats on this.)