Thursday, September 24, 2009

Why is your team more likely to get seven in a row than you are?

In this post, I explore a question I posed in my last post. Unless you really like statistics and probability, you might want to skip this one, but obviously, this is the kind of stuff I love. In the last post, I noted how unusual it was that as a team we have strung together five or more strikes five times. Individually, it has only happened once. As a team this week, we got seven in a row, something no individual bowler has accomplished. This was contrary to my expectations because certain individuals are on average better bowlers (more likely to get a strike) than the team as a whole. So, the chance of getting seven in a row should be highest for the best bowler on the team and lower for the team as a whole.

Well, that is true, and it isn't true. It would be true if individuals had the same number of opportunities to get seven in a row as the team as a whole, but they do not. If we look at a game of team bowling as the sum of four individuals, both the team and individuals have the same number of opportunities to get a strike, but they don't have the same number of opportunities to get seven strikes in a row.

For an individual, in any given game, you only have six chances to get seven strikes in a row. They are in frames: 1-7, 2-8, 3-9, 4-10, 5-10, and 6-10. In comparison, the team has many more opportunities, beginning with the first bowler in the first frame and ending with the 2nd bowler in the 10th frame. In fact, for a four man team in a single game, there are as many as 42 chances to get seven in a row.

So, even though the skill of the team will be lower than that of the best bowler on the team, in many cases the team will have a better chance of stringing together a large number of strikes in a single game than the best individual bowler.

Here's a demonstration of that. The best bowlers on our team have a strike percentage of about 30%. The probability of getting 7 in a row for that bowler is about 0.022%. If you have six chances in a single game of getting seven in a row, this equates to a probability of 0.153%, meaning it should happen roughly once in every 654 individual games (If you really want to know how I calculated that, leave a comment. I won't be holding my breath.)

Our team as a whole rolls strikes in approximately 25% of frames. The probability of the team rolling 7 in a row then is 0.006%. If the team has on average 40 chances to roll seven in a row in a single game, this equates to a probability of 0.244% meaning it should happen on average once every 410 team games.

That's why your team will almost always be able to string more together than individual bowlers. If you don't really follow the math, think of it this way. Who is more likely to get 14 strikes in a row in a single game, one person, or the team? It is impossible for an individual to do it, right? It has to be the team.


  1. I wouldn't estimate the team's probability of seven consecutive strikes by raising the average team probability of .25 to the 7th power.

    Rather, I would multiply together each bowler's personal strike probability. So, if we refer to Bowler A's probability as "a," Bowler B's as "b," etc., and each bowler rolls two frames, then the equation would take the form:


  2. Hey Alan,

    You always give me more work to do. I made the mistake of assuming that (abcd/4)^4 was an equivalent expression. Obviously, it's not.

    This makes the problem a bit more complex since to calculate the precise probability would require taking into account the 10th frame in which 7 in a row could be bowled by only 3 bowlers.

    If we ignore that case, and calculate the probabilities assuming that any of the bowlers could start the streak, p's range between .0002 and .0009. Then, if you consider that bowlers 1 and 2 have at least 1 more opportunity to start a 7 in a row than bowlers 3 and 4 (say 10 vs. 9 chances) and sum the probabilities, it comes out to approximately p=.0325 per game. So, it should happen once every 308 games or so, a slightly better chance than I previously estimated.


  3. Sorry. Make that ((a+b+c+d)/4)^4.


Note: Only a member of this blog may post a comment.