# The maths of parkrun Bingo

Dec 30, 2018 · 7 min read

parkrun Bingo is perhaps the silliest of all the parkrun challenges. In this blog, we look at the maths behind the challenge and provide you with some tools to compare your Bingo performance against the average parkrunner, some celebrity parkrunners and every other person in the world.

For the uninitiated, Bingo is one of a series of unofficial challenges that has recently become very popular among obsessive parkrunners.

The idea is simple, every time you finish a parkrun, you discard the “minutes” bit of your finishing time and use the “seconds” to complete a virtual bingo card, that contains sixty numbers from 00 to 59. For example, my (rather unimpressive) finish time at a recent parkrun was “36:49”, so I get to tick 49 seconds off my parkrun Bingo card. Or at least I would have if I hadn’t run 29:49 at Oak Hill back in 2013.

It is a silly challenge. Unlike many of the other parkrun challenges (such as the Alphabet Game) for which people travel thousands of miles to complete, the “seconds” part of your time is essentially random and (unless you try very hard) you don’t have much influence over it. Which is what makes parkrun Bingo interesting from a mathematical probability point of view.

In this blog, I’ll try to answer the following important questions:

1. How many parkruns should it take me to get a Full House?
2. How does my performance at parkrun Bingo compare with other parkrunners and some parkrun celebrities?
3. If every single person on the planet tried to complete parkrun Bingo, what is the fewest (and the greatest) number of parkruns in which people would complete a full house?

A bit like real Bingo, parkrun Bingo seems deceptively easy to “win”. I’ve done 174 parkruns, and I have 56 of the 60 numbers checked off. Just FOUR more numbers to go! I’m nearly there!

But I’m not nearly there. I’m not even close. Let me explain why…

At the start of your parkrun career, you will probably be ticking off a number every week. As you get closer and closer to a Full House, unticked numbers become progressively more difficult to find. On average it will take me about 15 parkruns to get my next number (my 57th), another 20 parkruns to get the 58th number, 30 more parkruns to get my 59th number. To get the final number will take about 60 parkruns.

So all things considered, I have about another 125 parkruns to go before I complete parkrun BINGO on about 299 runs. This is not very surprising, because on average it will take a brand new parkrunner 280.79 runs to complete parkrun Bingo:

Or in graphical form (if you click on the link you can access an interactive version of this chart)

Now, of course, I could influence this in one of two ways:

1. I could sandbag as I approach the line to try to target one of my missing times. This is not easy because of the difficulty in matching your desired finishing time to the one recorded the stopwatch. However, if I only required one number to complete my Bingo card, it would take me much less than the (on average) 60 attempts required by relying on probability alone.
2. I could go into WebFMS (the parkrun results system) and manually edit the seconds part of a few of my runs.

I think that we can all agree that (1) and (2) are both highly immoral, and against the rules of a stupid and arbitrary game with no prizes. Plus to do so would be literally no fun.

So assuming that the number of seconds in your finishing time is random. parkrun Bingo is essentially the same as drawing numbered balls from a bag.

The bag contains 60 balls, numbered from 00 to 59. In some ways, it’s very similar to the way they to do the FA Cup Draw but with one essential difference: every time you pull a ball out of the bag, then you have to replace it (because of course, you will record many duplicate numbers on the way to a full house).

I’ll bore those of you that are interested with the maths of this at the end. However, I solved this problem in two ways:

1. I wrote a computer program to simulate 7.44 billion parkrunners trying to complete parkrun Bingo (yes that’s everyone in the whole world).
2. I solved the problem analytically using probability theory.

Both approaches give the same answer (although (2) is more elegant than the pure brute force approach of (1). Although they both give the same result, they each offer some subtly different insights though.

1. By the time they join the 50 Club: no-one will have completed Bingo (obviously)
2. By the time they join the 100 Club: 1 in 12 million parkrunners will have completed Bingo
3. By the time they join the 250 Club: 39.3% of parkrunners will have completed Bingo
4. By the time they join the 500 Club: 98.7% of parkrunners will have completed Bingo
5. By the time they join the (as yet non-existent) 1000 Club: 99.9997% of parkrunners will have completed Bingo

Since 60% of us will complete parkrun Bingo somewhere between 250 and 500 runs, it could be viewed as a welcome distraction while you wait at least 5 years to earn your dark blue t-shirt.

Of course, there are no guarantees that you will ever complete a Full House. In my simulations, one unlucky parkrunner took 1817 parkruns to get a Full House (but that did involve 7.44 billion parkrunners). You have to be pretty unlucky not to have done in by the time you reach 500.

I started this by saying that parkrun Bingo feels deceptively easy. On my total of 174 parkruns, only 2.3% (or 1 in 43) parkrunners will have completed a full house. Like I said earlier, I’m really not close at all.

For a little bit of validation we can look at some famous parkrunners:

PSH’s performance is quite impressive because only 12% of people will manage to complete a full house by 204 runs.

Nicola’s performance is rather less impressive as 75% of people manage a full house before 321 runs.

Finally (before the proper maths bit), what are the chances of someone completing the Bingo challenge in exactly 60 parkruns? The answer is 1 in 5.873 septillion: 1 in 587354000000000000000000

5.873 septillion is about 6000 times the stars in the universe. It isn’t going to happen anytime soon.

In my simulation of every person in the world trying to complete parkrun Bingo, the smallest number of runs in which anyone completed a full-house was 89. Of the entire population of the world, only 266 virtual parkrunners managed to complete a full house before 100 runs. If anyone tells you that they completed a full house in fewer than 100 runs, they are either cheating or lying.

This is the maths bit. Unless you are some sort of geeky weirdo (like me) you can probably ignore this section.

If you imagine each of the sixty balls has a different color, then the analytical solution is described on this answer by Arnaud Mégret On Stack Exchange:

For our case, the number of colors (or numbers) is k= 60 and we are interested in the case where m is also 60 (ie a full house). So we are looking for P(n,60) across n parkruns. It took me about 10 minutes to do this in a spreadsheet. You have to start with P(1,1), and then P(1,2) and work your way down to P(1000,60) say. My spreadsheet is here if you fancy taking a look.

This generates the following cumulative probability distribution, which shows the probability of completing a full house before n parkruns:

We can also work out the probability of completing a full house on any given parkrun.

Although the mean number of runs required to complete a full house is 280.79, the median value is 269.5 (ie 50% of parkrunners will have completed their Bingo card between 269 and 270 parkruns). The modal value is 245 parkruns (0.641% of parkrunners will complete their card on this number).

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