**If you find any of this useful, please consider donating via PayPal to help keep this site going.**

**Email news@statisticool.com to sign up to receive news and updates**

# Relative Frequency Exploration

**6/28/18**

Consider flipping a coin for i = 1 to N flips. For trial i, if the coin lands on Heads, write down X_{i} = 1. If the coin lands on Tails, write down X_{i} = 0. Do this for all N flips. Next, at each flip i, calculate the relative frequency of Heads as f_{i} = sum(X_{j}, j = 1 to i)/i. For example, if the observed results are 0, 1, 1, 1, 0, 1, then f_{i}s are 0, 1/2, 2/3, 3/4, 3/5, and 4/6, respectively.

We intuitively recognize that this relative frequency will "settle down". It might not settle down to a theoretical p = 50%, however, and that is OK. For example, the coin could be biased or the object might not be a coin (it could be a tack for example).

Call the last relative frequency f_{N}. One question is, how close to p does f_{N} have to be before you conclude f_{N} = p? This is both an absurd question and an important question. It is an absurd question, because we know f_{N} is *not* p (although it could be by chance). It *is* an important question because we understand that .5, for example, is for all intents and purposes equal to .499999 or .500001 (and if those don't do it for you, just add more 9s and 0s), much like we use 3.141595 to estimate pi for many important engineering tasks, and this type of frequentist interpretation of probability as a long-term relative frequency is the most dominant and objective.

Consider a tolerance or epsilon, e, and a distance, d = |f_{N}-p|. We might conclude f_{N} = p if |f_{N}-p| < e for some small e>0. For example, if e = .001, and after N trials we find that f_{N} is in the interval [.499, .501], we conclude f_{N} = p.

I set to test this out for a coin flip for different e, as well as for a varying number of flips N. I used a "digital coin", the random function in Excel. You can, however, do this exact same thing for a real, physical coin. For each (e, N) combination, I ran 10,000 repetitions. However, I only did 100 repetitions for the (e = .000001, N = 100,000) pair, since it was taking too long on any computer I used due to the computations involved with the N coin flips being refreshed. I kept track of the percent of time f_{N} landed in the interval. Let's call this percentage P%.

Here are the results:

We can see that 95.4% of the 10,000 simulations of 10,000 flips had relative frequency of Heads in [.49, .51]. Also, it is clear that the P% decreased as e decreases. Also, for a fixed e, as N increases the P% gets larger. These results matched my intuition pretty well, as well as what I know about the Strong Law of Large Numbers. However, I was somewhat surprised that the P% got small very fast when I decreased e by an order of magnitude or decreased N by an order of magnitude.

As I mentioned, I did these simulations in Excel. You can download the spreadsheet here. It has a Visual Basic macro you can run by doing Ctrl-J, after you enter the information in the yellow highlighted boxes you want (p, e, and number of simulations).

You will also need to fill down the data accordingly for how many flips, N, you want to use.

Thanks for reading.

### Please anonymously VOTE on the content you have just read:

Like:Dislike:

If you enjoyed *any* of my content, please consider supporting it in a variety of ways:

**PLEASE**take a moment to check out two GoFundMe fundraisers I set up. The idea is to make it possible for me to pursue my passions. My goal is to be able to create free randomized educational worksheets and create poetry on a__full-time basis__.**THANK YOU**for your support!- Email news@statisticool.com to sign up to receive news and updates
- Donate any amount via PayPal
- Take my Five Poem Challenge
- Subscribe to my YouTube channel
- Visit my Amazon author page
- Buy what you need on Amazon using my affiliate link
- Follow me on Twitter here
- Buy ad space on Statisticool.com

AFFILIATE LINK DISCLOSURE: Some links included on this page may be affiliate links. If you purchase a product or service with the affiliate link provided I may receive a small commission (at no additional charge to you). Thank you for the support!