Best Answer k-man, 17 January 2013 - 05:02 PM

Let's number the 4 suspect coins 1, 2, 3 and 4 and let's use the notation H-heavy, L-light and N-normal. Among the 4 suspect coins we have 12 possible distributions: HLNN, LHNN, NLHN, NHLN, NNLH, NNHL, HNNL, LNNH, HNLN, LNHN, NLNH, NHNL.

We also have 2 known normal coins (NN). Let's use one of the known normal coins to pay for the first use of the scale and weigh 1,2 vs. 3,N. We will use the other known normal coin in the first weighing. The possible distribution of outcomes from the first weighing is

Balanced: HLNN, LHNN, NLHN, NHLN

Left is heavy: HNLN, HNNL, NNLH, NHNL

Right is heavy: LNHN, LNNH, NNHL, NLNH

If balanced, we know that the coin 4 is normal and the cheat coins are either 1,2 or 2,3. We can now use the second known normal coin for the second and final weighing. We will weigh 1,2 vs.3,4. If balanced then 1,2 is the cheat pair. If not, then 2,3 is the cheat pair. **Done in 2 weighings.**

If left or right is heavy we will need 1 or 2 more weighings to identify cheat coins, but we need to identify one normal coin in the second weighing to use for the third. Let's look at the case when the left side was heavy. It's symmetrical, so the same logic can be applied for the right side too.

Let's make the second weighing 1,4 vs 2,3.

Balanced: HNNL. 1 and 4 are the cheat coins. **Done in 2 weighings.**

Left: HNLN or NNLH. Need the third weighing, but we now know that 2 is a normal coin. Use it to weigh 1 vs 4 and identify the cheat pair. **Done in 3 weighings.**

Right: NHNL. 2 and 4 are the cheat coins. **Done in 2 weighings.**