IN ONE OF THE COLUMBO EPISODES ON T.V. COLUMBO WAS SET A PUZZLE.I CAN'T REMEMBER WHAT THE EPISODE TITLE WAS,BUT I CAN REMEMBER PART OF THE PUZZLE,I BELIEVE IT WAS SET BY A MAGICIAN/ILLUSIONIST? FOR COLUMBO TO SOLVE.
BASICALLY,A MAN HAS 5 BAGS OF GOLD COINS (THE NUMBER I BELIEVE WAS IMMATERIAL),ONE OF THE BAGS OF COINS ARE NOT REAL,BUT THEY ARE HEAVIER THAN THE REAL ONES.
HE HAS A SET OF SCALES,BUT I THINK? HE COULD ONLY USE THEM ONCE? HOW DOES HE FIND OUT THE COUNTERFEIT BAG OF COINS?
PART OF THE ANSWER AS I RECALL,WAS TO TAKE 1 COIN FROM THE FIRST BAG,2 FROM THE SECOND BAG,3 FROM THE THIRD BAG,AND SO ON----
CAN ANYONE HELP ME OUT WITH THE COMPLETE PUZZLE AND OBVIOUSLY THE SOLUTION.
REGARDS MICKC
WEST MIDLANDS
ENGLAND.

Welcome to the site, Mick, and to the Forum.

This is actually in our FAQ, linked from the home page. You remembered the puzzle very well, although it is from a different episode than the one with the magician (which was "Now You See Him," with Jack Cassidy).

The "phony gold" puzzle comes from an episode where the killer and victim belonged to a sort of "genius club," similar to MENSA.

Quoting from the FAQ:

14) What is the "phony gold" puzzle, from "The Bye Bye Sky High IQ Murder Case"?

The puzzle:

In a room are several sacks of gold pieces, as many sacks as you like. Each sack contains several of these gold pieces -- again, as many as you like. One sack, however, is full of artificial gold pieces, and they weigh differently. The solid gold pieces weigh, let's say, a pound each. And the artificial pieces weigh, let's say, a pound and an ounce.

Now you have a penny scale. You put the penny in, and you get a card, and that tells you how much the weight of the gold is. But, you only have one penny. You have one reading on the weight.

Which sack has the artificial gold pieces?

The answer (as deduced by Mrs Columbo):

Well let's just say there are three sacks. And each real gold piece weighs one pound. But one of these sacks has the phony gold. And the phony gold weighs one pound and one ounce.

We're going to the scale. Now, we take one piece from bag one, we take two pieces from bag two, we take three pieces from bag three. That's six pieces. We take all six pieces, we put em on the scale.

Now...if all six pieces were real the gold, it would weigh six pounds. But if the first bag had the phony pieces, it would weigh six pounds and one ounce, because I only took one piece from the first bag. And if the second bag has phony gold pieces, it would weigh six pounds and two ounces, because I took two pieces from the second bag. And if the third bag were the phony bag, it would weigh six pounds and three ounces. And so on, and so on, and so on.

Sir...it's a terrific puzzle.

According to the Columbo Newsletter, isn't there another solution to the puzzle?

Thanks TED for such a prompt reply.
My daughter works at the local Assay office,so it's an ideal question for them to solve.
regards Mickc

Thank you Ted for stating the puzzle verbatim as I even had thoughts of buying episode 39 in order to get this puzzle exactly as it was said. I cannot recall the original question of the puzzle word for word by Columbo as I felt that some information was missed. Many other viewers have expressed the same opinion. Nevertheless in Columbo's answer to the question all the relevant facts to the puzzle are given. I know this because I now understand the puzzle and solution from a Mathematical formula dealing with probability. Unfortunately however in Columbo's answer to puzzle, a mistake is made in semantics thus giving the impression that the bags would weigh 6 pounds and 1 ounce, or 6 pounds and 2 ounces etc etc. Well actually it is not the bags that would have that weight; we actually do not know what the bags weigh and neither do we need to know. What the weight of the 6 coins reveal is which bag contains the fake gold coins. That's it.

Actually the answer was that you take one piece from bag 1, 2 from bag 2 ect.. You leave one bag untouched, and based on how many ounce over one pound the weight is, is equal to the number of pieces you took from the fake bag ie 1lb 3oz, bag 3 is fake. If it's even, then the untouched bag is fake.
However if this were the case, there are at least 2 issues. What if the number of bags is higher than the coins in each bag? Or if there were 17 bags, and you end up at exactly 16 lbs. You only have a 50% chance of being right. It would either be the bag you took 16 coins from, or the bag untouched.
It is an imperfect riddle.

Nm, I just watched the episode. No bag is left untouched. So that rules out the second flaw, but the first is still valid.

On the May 20, 1977 episode of "The Tonight Show Starring Johnny Carson," Peter Falk poses the puzzle to Johnny Carson and Ed McMahon. Johnny Carson thinks he knows the answer but Peter Falk begs him to not reveal the answer.

The facts are stated correctly. I believe some of the confusion about this puzzle come from the fact that this is a conversation between the killer and Lt. Columbo. Columbo interjects the weights, and what a penny scale is. The rest of the puzzle is stated by the "genius". The episode is S6 Ep3, entitled "The Bye Bye Sky High IQ Murder Case".

1 pound = 16 ounces. If there are more than 16 sacks it would be difficult (or impossible) to determine which sack has the phony coins. The scale reading would add whole pound to the result if the "different sack" is not one of the first 16 to be picked from.
btw-I am a huge Columbo fan;

No. If each piece of real gold weighs a pound then 16 bags should weigh 136 pounds. If bag 16 had the fakes, then you would have 137 pounds on the scale. You see you just subtract the expected weight from the actual weight, in ounces, and it reveals the bag number, regardless of working in pounds or not. It would work equally well if you added up in ounces so that one would expect 2176 ounces.

You would be in trouble if you had more bags than you had coins in each bag. Eg 50 bags each with 20 coins and thus only 20 fakes.

So what if you decided there were 8 sacks, but each sack only had 4 pieces of gold,