Parkrun Bingo Probability

[ALD55]

I run Parkrun almost every week - this a free timed weekly 5k run in many countries. I downloaded an app called 5k and this includes a number of challenges including Parkrun bingo. This means that you have to get a finish time ending in 0, 1, 2 seconds all the way to 59.

I have completed 235 runs and was surprised to have a finish time for every second.

I am not sure how to calculate the probability of this happening in 235 results as other people I know are lacking one or two results and find it hard to pick them up.

So the question is: what is the probability of achieving all 60 secondary results in 235 runs?

Thanks

 

[ksdhart2]

To truly calculate such a probability, a lot more information is needed. At minimum, we'd need to know the probabilities of running a 5k with a time ending in \(0, 1, 2, \cdots\) seconds. However, we can generate a very very rough, ballpark estimate by assuming uniformity - assume that the probability of a time ending in 0 is the same as the probability of a time ending in 1 is the same as...

If we make this assumption, then it becomes a variant of the Coupon Collector's Problem. That states that the expected number of races needed to get one of each time is:

\(\displaystyle E(T) = 60 \cdot H_{60} = 60 \left( \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{60} \right) \approx 281\)

And the variance is:

\(\displaystyle V(T) = \sum\limits_{k = 1}^{60} \left( \frac{60(k-1)}{(60 - (k - 1))^2} \right) = 10 \cdot (60 \pi^2 - 360 \psi^{(1)}(60) - 6 \psi^{(0)}(60) - 6 \gamma \approx 5581.4676\)

Therefore the standard deviation is:

\(\displaystyle S(T) = \sqrt{V(T)} \approx 74.7092\)

With all of this information, we can perform a One Sample t-test:

\(\displaystyle t = \frac{281 - 235}{s_{\overline{x}}}\) where \(\displaystyle s_{\overline{x}} = \frac{S(T)}{\sqrt{60}}\)

\(\displaystyle t \approx 4.7694\)

The p-value associated with this t-score is:

\(\displaystyle p \approx 0.0206\)

So, with these incredibly unjustified assumptions made, the model suggests roughly a 2.06% chance of such an occurrence happening.

 

[ALD55]

Thanks for your reply. I think we have to assume equal provability of each of the seconds as I cannot think why there should be any bias.

 

[ALD55]

May I ask a further question? How did you calculate the p value from the t-score as I am getting different numbers? Excel functions seem to give a different answer, but I think I have the wrong inputs.

Thanks

 

[ksdhart2]

May I ask a further question? How did you calculate the p value from the t-score as I am getting different numbers? Excel functions seem to give a different answer, but I think I have the wrong inputs.

Without seeing exactly what you did, it's hard for me to say if you did something wrong or not, but I have a theory. It might be the difference between performing a one-tailed t-test vs. a two-tailed t-test. These will result in different p-values, even though the test statistic remains the same. I chose to do a lower tail only t-test because I knew your "score" of 235 was less than the mean "score" of 281. If you performed a two-tailed t-test instead, the associated p-value would be twice as much or \(p \approx .04126\).

 

[ALD55]

Euler-Mascheroni constant				0.577215664901532
$L$10: 0.577215664901532
Expected runs to get one of each time E(T) Maximum = 236			280.302
$L$12: ' =Maximum*(LN(Maximum)+Euler_Mascheroni)+1/(2*Maximum)
Variance V(T)					5,641.452
$L$14: ' =(PI()^2)/6*Maximum^2-Maximum*(LN(Maximum)+Euler_Mascheroni)-1/Maximum
Standard Deviation S(T)					75.110
$L$16: ' =SQRT(L14)
One Sample t-test (t)					4.569
$L$18: ' =(L12-Max_Number)/(L16/SQRT(Maximum))
Probability of Result (p)					2.24%​
$L$20: ' =TDIST(t,2,1)