Question

A large jug contains 2018 nickels and 2019 dimes. Next to the jug is a large pile of dimes (treat it as an unlimited number of dimes). The following procedure is to be performed repeatedly until a single coin remains in the jug.

Two coins are chosen at random from the jug:

i. If both coins are dimes, one is put back and the other is discarded.

ii. If both coins are nickels, both are discarded and a dime from the pile is added to the jug.

iii. If one coin is a dime and the other is a nickel, the nickel is put back and the dime is discarded.       

Will the last coin remaining in the jug be a nickel or a dime?

Answer 1

In step 1, we are effectively discarding one dime.

In step 2, we are effectively reducing 2 nickels and adding one dime

In step 3, we are effectively discarding one dime.

Therefore as we are given that initially we have even number of nickles, there is a finite probability that we keep drawing nickles 2 at a time and discard all of them, and so we would be left with dimes.

But there is also a possibility that we keep taking out a nickel and a dime and keep discarding the dime and so at the end only left with nickel but there is no possibility that we would be left with 1 nickel as we are never effectively taking out 1 nickel, we are either taking out 2 nickels or not taking out nickel

Therefore dime is the only possibility for the last coin to remain.