Math on Trial came out in the U.S. on March 12. It reached #724 on the Amazon bestseller list, hitting #1 in Criminal Law, #4 in Criminology and #13 in Mathematics!
The book came out in England on March 28, where it's currently at #3 in History of Mathematics and #10 in Probability & Statistics.
Our op-ed was published in the New York Times:
Justice Flunks Math, New York Times, March 27, 2013
We've had a piece published in the Science section of the Huffington Post:
Mathematics in Forensic Science
We've done some podcasts that are online:
Inspired by Math, with Sol Lederman
New Books In Law, with Marshall Poe
And we've had some radio interviews:
Leila was interviewed on the Steve Fast Show (WJBC-AM Bloomington, Indiana) and by Paul Harris on America Weekend (Envision Radio Network).
Coralie was interviewed on BBC Radio 4's Today show, and on the Round Table hosted by Joe Donahue (WAMC/Northeast Public Radio), and she's coming up next week on Radio 4's "More or Less".
It's been an exciting time.
I'm enjoying this! Congratulations on your Lucia chapter. EBook readers: in Netherlands, you can get the eBook from bol.com. There is a Kindle edition (amazon) but in some countries (like mine, NL) unavailable.
ReplyDeleteThanks! We put a lot of effort into all the chapters, but Lucia was one of the most complex to analyse. Thank goodness the story ended as it did.
DeleteLeila and Coralie,
ReplyDeleteWhile it's good to see statistics appreciated in the mass media, please check your articles with an expert before publishing. Firstly - yes, a second test may be informative, but the key assumption is whether the second test is independent of the first. If it is not, then you can continue testing as long as you want, but you will not get any closer to the truth. Secondly - a p-value of 0.08 does not mean that the probability that the coin is biased is 0.08. That's a fairly basic misunderstanding of statistical principles.
Please take a refresher course on the fundamentals of statistical inference before continuing to publish on such weighty questions.
Kind regards,
Stephen
The "second test" requested in court concerned a new DNA test, performed on a new swab taken from the same place as the original swab, by a new set of experts using the latest equipment. The results of such a "second test" would be totally independent from the original test, and if the results are the same as those of the original test, it would have an undoubted confirmational value for that first result.
DeleteAs for the p-value of 0.08, your sentence is confusingly phrased: you need to specify for what event or combination of events the p-value was specified, under what circumstances. But if the probability of a certain combination of circumstances is calculated in the presence of the null hypothesis (i.e. assuming that they occur by pure random chance with no external influence) and the p-value is found to be 0.08, then yes, that does mean that the circumstances in question have a chance of about 8% of occurring naturally and randomly.
As for taking a refresher course, thank you for your kind advice. We are always continuing to educate ourselves on these important and fascinating questions.
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ReplyDeleteI'm a UK secondary school maths teacher who bought your book partly because I didn't understand the example given in the Amanda Knox bbc.co.uk article about the ten coins and two tests. I have been trying to understand the example you gave in the book about calculating the probability of coin being biased having been tossed and landing on heads. My recent knowledge extends to A level statistics (S1 and S2) and I seem to be lacking some knowledge to appreciate the example. Would it be possible for you to post a link to the theory behind the example or maybe provide a more detailed explanation?
ReplyDeleteHi Icosahedron, thank you very much for your message and I hope you enjoy the book!
DeleteFirst of all, the article in the BBC didn't fully explain the premise of the problem (though it has now been altered to include it). What we have is a coin that we know is either fair or has been altered to fall on heads 70% of the time. We also know that there is a 50-50 chance either way.
Now in the first instance, we throw it 10 times and get 9 heads. We want to calculate P(M|R), the probability that the coin has been modified, given our result R (getting 9 heads in 10 throws).
Bayes' theorem tells us that P(M|R)= P(R|M)*P(M)/P(R)
Now - P(R|M) is just the probability of getting 9 heads ot of 10 throws with our modified coin, which is 0.7^9*0.3*10
- P(M) is just equal to 1/2
- P(R) = P(R|M)*P(M) + P(R|not M)*P(not M)
where P(M) and P(not M) are both equal to 1/2 and P(R|M is as above, and P(R|not M) is the probability of 9 heads in 10 tosses with a fair coin, which is 0.5^10 * 10.
If you plug all of these values into the formula given by Bayes, you find a probability of 92% that the coin is biased.
You can then do the same calculation for the result of 8 heads out of 10 tosses (84% chance that the coin is biased), and then for 17 heads out of 20 tosses (98.5% chance that the coin is biased)
I hope that clears it up! Note that the explanation in the book is slightly different and doesn't mention Bayes explictly, but rather uses directly the fact that there is a 50-50 chance of the coin being biased or not.