Excuse the rather basic question, but I was reading this article on thyroid cancer in Fukushima and it was reported that 3 cases in 38,000 children were found in the previous fiscal year. Another web site tells me that for 15-19 year olds in the UK (I cannot find Japanese figures) the rate is about 1.6 per 100,000 per year, which would be 0.6 per 38,000.

The rational person in me thinks that three cases rather than zero or one in a year is just a statistical hiccup, but the Think Of The Children person in me says it's a five times higher incidence.

So, can someone statistically assure me/worry me about these figures?

(PS: I realise that increased testing specifically looking for thyroid problems may skew the stats for last year by finding tumours earlier, but can we ignore that here, please?)


1 Answer 1


Let's say each kid flips a biased coin to determine whether or not they have cancer. If we assume that the probability of heads (cancer) is 1.6/100,000, we can find the distribution of cancer counts we'd expect using a binomial distribution.

In R code, we can find the distribution with the dbinom command:

dbinom(x = 0:5, size = 38000, prob = 1.6/100000)

Here, x is the number of cases (0:5 means we're looking at the probability of 0 cases, 1 case, etc. up to 5). Size is the number of kids, and prob is the baseline probability you cited.

After cleaning up the output slightly, we get a table like this:

    number_of_cases probability
               0     0.54444
               1     0.33102
               2     0.10063
               3     0.02039
               4     0.00310
               5     0.00038

So you'd expect to find 3 cases out of 38,000 children only 2% of the time under this model--and you'd almost never find more than that.

In short, (assuming the figures are comparable), it does seem on the high side, and might be worth investigating further. But you wouldn't necessarily need to invoke any special factors beyond random chance to explain the difference.

Edited to add: Per EpiGrad's comment, I added an image that shows how these probabilities could change if we were uncertain about the baseline probability of 1.6 cases per 100k. The red points are the values I listed above, and the cloud of points represent what we'd expect if the

For this example, I sampled baselines from a beta distribution using rbeta(1000, 1.6, 100000 - 1.6), which has a mean of 1.6 cases per 100k and some spread on either side but doesn't drop below 0. The amount of spread may or may not be reasonable, depending on what assumptions you'd like to make. My gut feeling is that I included more variation than I should have, but who knows.

As you can see from the plot, if the British figures substantially underestimated the "real" pre-Fukushima rate of cancer incidence in Japan, we might expect to see 3 cases per 38k as often as 20% of the time. Whether you think that's likely depends on other information outside the scope of this problem, including whether I included an appropriate amount of uncertainty from the British estimate.

  • 1
    $\begingroup$ Thank you - that is a nice and simple to understand answer! (BTW, the doctor leading the survey said that it was highly likely that the tumours were existing pre-meltdown conditions, even though he also gave the cancer incident rate as "one in a million".) $\endgroup$
    – Ken Y-N
    Commented Feb 15, 2013 at 2:48
  • 3
    $\begingroup$ Keep in mind however that the "1.6/100,000 number" also has some uncertainty behind it. And I really wouldn't discount the impact of intensified surveillance for cancer in detecting new cases. $\endgroup$
    – Fomite
    Commented Feb 15, 2013 at 5:16
  • $\begingroup$ @EpiGrad two very good points! (+1) $\endgroup$ Commented Feb 15, 2013 at 23:02
  • $\begingroup$ @Epigrad I updated my answer to include explore what might happen if we allow for uncertainty in the 1.6/100k number $\endgroup$ Commented Feb 17, 2013 at 7:30
  • 1
    $\begingroup$ @DavidJ.Harris I'd upvote you again for the sensitivity analysis, but the software doesn't let me do that :) $\endgroup$
    – Fomite
    Commented Feb 17, 2013 at 7:31

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