# Are you guilty? Yes/No?

I used simultaneous equations for what I think was the first time in my career last week, having learnt them as a teenager. It was to figure out a base cost c and the variable cost m where I knew three equations that should rightfully have satisfied the equation y=mx+c, x and y known in each of the scenarios.  Although rusty, my knowledge didn’t fail me.

And I resorted this week to a statistical technique that I remember learning back in 1992–3 while at university in Newcastle.  The technique aims to estimate percentages of people who share an attribute where the people sampled aren’t necessarily up for answering the question truthfully.

The question to which I wanted to know the answer was “Of those people accused of terrorism who walk free, what percentage were actually guilty?”

If you ask the freed suspects outside the Old Bailey “Did you do it?”, you’re unlikely to get a considered response, particularly from those that did.  Why not instead give them a die and ask them to roll it in secret.  If they get a six, they should answer the question dishonestly; if they get anything other than a six, they should be honest in their response.

If 70% (y) of respondents said they were guilty, then the estimated percentage of people that were guilty would actually be 80% (x).  Here’s the math(s).

0.8*(5/6) + 0.2*(1/6) = 0.7.  Or (5/6)x + (1/6)(1-x) = y

And here’s the English.  The estimated proportion of people saying they were guilty is:

The proportion of guilty people answering truthfully * the odds of them answering truthfully

plus

The proportion of innocent people answering dishonestly * the odds of them answering dishonestly

But having conducted the experiment with a sufficiently large set of people, you know the y, so re-arranging and simplifying, the estimated proportion of guilty people is:

(y-(1/6))*1.5

It’s quite a nice little solution.  If, of course, you can trust people to act honestly based on their rolling of the die.