Phishing for Phoolish Mistake

I have been reading Phishing for Phools and found a mathematical error in the book. In footnote 10 of chapter 1, the authors gives out an interesting stat- 2.5% bankruptcy rate every two years. They then go on to say that there is a 62.5% chance that an average American will go bankrupt in his adult lifetime. They go on to say that if those who file one bankruptcy, go on to file for bankruptcy two more times, then the precent of people filing for at least one bankruptcy would be 20% in their life times.

There are two main faults to this statement. The first is that this statement is mathematically incorrect! The life expectancy of an average American is 78 years. An adult is anyone over 18 years. So the adult lifetime would be 60 years. The probability of an event happening in the next y years given p, the probability of it happening in any given year is 1-(1-p)^y. The simple logic can be explained by the similar probability problem for the 100 year flood. The probability of someone not going bankrupt is (1-p). The probability of someone not going bankrupt for y years is (1-p)^y. So the probability of someone going bankrupt in y years is 1-(1-p)^y, which in our case would be 53.2%. Now the extra complication of the linked probability of repeated occurrence. if the average person, who has filed for bankruptcy once, again files for bankruptcy two more times on average, then in any given sample of 3 bankruptcy case, there will be one unique person. Meaning, 17.73% of the population will file for bankruptcy in the next 60 years.

The other fault is a fault of omission. I could not locate the source from where they obtained the stats. These look surprisingly high numbers. I looked online and found abi.org giving an estimate of 0.3% for the bankruptcy rate and debt.org giving a repeat case rate of 8%. Churning these numbers, our probability of a person going bankrupt in his adult lifetime would be 15.2%.

Just to verify my calc, I ran a simple simulation on python. The function np.random.choice generates a sequence of years using 0's(meaning no bankruptcy) and 1's(meaning bankruptcy) with the probability of 1 being 2.5%. I simulated this for a population of 10,000 and added all the positive cases and then halved it for the repeated case. I get an answer of 18.03%(vs17.73%) which is not bad.