This problem is a combination of a traditional weights and scales puzzle and a probability puzzle.
You are given the traditional scale balance and a set of 10 coins. You know that in this set of coins there are four fakes - two that are heavier than the others, and two that are lighter. Furthermore, you know that a light coin plus a heavy coin will perfectly balance two genuine coins. You are permitted only two weighings with the balance and asked to pick one of the 4 fake coins. What strategy should you use to maximize your success rate, and what would your success rate be?
What if you were given one genuine coin to start with (though still leaving you with 10 unknown coins)?
(In reply to
Better than 50% by Brian Smith)
With a genuine coin for reference:
Choose 2 coins from the set of 10. If either coin does not match the genuine coin, that coin is a fake. Otherwise both are genuine so choose one of the 8 remaining coins. The probability of not finding a fake is ((6*5)/(10*9))*(4/8) = 1/6. The success rate of finding a fake is then 5/6.
Without a genuine coin for reference:
Choose 3 coins at random. Call the first coin A, the second coin B and the third coin C. Weigh A vs B and A vs C.
If both weighings are equal (prob 120/720) then A, B and C are all genuine. Take one of the other 7 coins (4 of 7 are fake).
If one weighing is equal and the other is not equal (prob 304/720) take the coin which appears only in the unequal weighing (prob fake 64/76). Example: if A=B and A>C take C.
If the weighing results are {A>B,A>C} or {B>A,C>A} (prob 248/720) then take A (prob fake 112/124).
If the weighing results are {B>A,A>C} or {C>A,A>B} (prob 48/720) then B and C are both fake with one heavier and one lighter. Take either B or C.
Total success rate = (120/720)*(4/7) + (304/720)*(64/76) + (248/720)*(112/124) + 48/720 = 29/35.
re: Better than 50% - details
