IBM Research | Ponder This | June 2008 solutions

# Ponder This

## June 2008

<<May June July>>

1. p/(p+(1/n)*(1-p))
2. p/(p+(k/n)*(1-p))
3. .4486

A sketch of the proofs follows.
1) We originally have a fraction p as the actual source and a fraction (1-p) not as the actual source. Testing does not exclude any of the actual source fraction but does exclude all but 1/n of the not actual source fraction. So the new actual source fraction is p/(p+(1/n)*(1-p)).

2) Again we start with a fraction p containing the actual source and (1-p) not containing the actual source. When the database contains the actual source we will get exactly one match (1-1/n)**(k-1) of the time (as there can be no random matches on the remaining people in the database). When the database does not contain the actual source we will get exactly one match k*(1/n)*(1-1/n)**(k-1) of the time (as we must have a random match for one of the k people in the database but not for the others). So the new fraction containing the actual source given exactly one match is p*(1-1/n)**(k-1)/(p*(1-1/n)**(k-1)+(1-p)*k*(1/n)*(1-1/n)**(k-1)) = p/(p+(k/n)*(1-p)). So searching against a database effectively increases the random match probability by a factor k (the size of the database).

3) Plugging n=1100000 and k=338000 into the formula above we obtain .44861337+ or .4486 rounded to four places.

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