The Risk from Consecutive Independent Events is not Additive

In a discussion about risks from medical procedures, I heard someone make the following statement,

“If there is a 1% chance of an adverse reaction from the procedure, that means if the procedure is done 100 times, an adverse reaction is nearly certain—assuming each application of the procedure is an independent event.”

While I immediately recognized that this was not true—it incorrectly assumes that the risk from each event is additive—it did seem to be an approximation that most people would use; I have certainly heard it said before. What I did not know was how wrong the approximation is.

The true probability of having an adverse reaction is one minus the probability of applying the procedure 100 consecutive times without an adverse reaction. In this case, we get

1 - (\frac{99}{100})^{100} \approx 0.634

So experiencing an adverse reaction would be more likely than not, but it is far from a certain event. One obvious question is how well does this generalize across a range of probabilities. If an event is said to occur once out of every N times, how often will it occur if we repeat it N times. For simplicity, we will refer to this as risk. This is a generalization of the above equation to

\mbox{risk}(n) = 1 - (\frac{n-1}{n})^n

If we plug in different values for N, we can see how steady this value is for any meaningful range. For values of N from 100 to over 1 trillion, the risk is about 63%. By going down to 10, the risk goes up to 65%, but those numbers are still close enough for informal conversational purposes.

Now I know if I have a one in a million chance in winning the lottery, I only need to play one million times in order to have a 63% chance of winning.

Advertisements
This entry was posted in Math Envy and tagged , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s