Can you solve the counterfeit coin riddle? -
mrearlygold
Posts: 17,858 ✭✭✭
In the video below, we are presented with a version of the 12-coin problem in which we must determine a single counterfeit coin in a dozen candidates. The problem is, we’re only allowed the use of a marker (to make notes on the coins) and three uses of a balance scale. Here are the detailed conditions:
1) All 12 coins look identical.
2) Eleven of the coins weigh exactly the same. The twelfth is very slightly heavier or lighter.
3) The only available weighing method is the balance scale. It can only tell you if both sides are equal, or if one side is heavier than the other.
4) You may use the scale no more than three times.
5) You may write things on the coins with your marker, and this will not change their weight.
6) There’s no bribing the guards or any other trick.
So how do we solve this specific case? Watch the video to find out.
https://www.youtube.com/watch?v=tE2dZLDJSjA&ab_channel=TED-Ed
Coin's for sale/trade.
Tom Pilitowski
US Rare Coin Investments
800-624-1870
Comments
Nope, going to spend the rest of my life in Jail.
bob:(
Video won't play on my computer....
So I spent 20 minutes trying to figure it out.
I decided that if you had claimed they were marbles instead of coins, the post would be off topic.
Therefore, I have chosen to ignore the thread.
.
Insert witicism here. [ xxx ]
Yes, It is a simple binary search, split the coins in two even piles, discard the lighter pile and repeat.
If you were told that the counterfeit was too heavy, that would work. But the problem intentionally makes it unknown if the counterfeit is heavy, or light!
And if you were given as many weighings as you'd like, it also becomes trivial.
The issue is finding it with only 3 weighings, and now knowing the relative weight of the counterfeit.
(I just saw the video...Think I understand the method, though, I had no audio).
Deleted
I saw this a couple of days ago. I had a similar method which worked with 4-5 weighings,
and using the red marker makes it a strange problem.
Like some Sci Fi movies, we might question the "realism" of the problem.
For example, how do you know in advance that there is only one counterfeit coin?
How about checking the die markers? (not the dye markers) :-)
There is a simpler method. Divide the coins into three piles of four. Put one pile on one side of the balance and another pile on the other side. If one pile is heavy pick it up and set the other eight coins aside. If both are equal set those eight coins aside and pick up the other four.
Now balance an two of the heavy pile against the other two. Take the heavy pair, set the other two aside, and balance one against the other.
TD
The way I like to approach a problem like this is to start with a simpler "baby" version of the full problem. In this case, the baby version is to find the counterfeit among 4 coins with only 2 weighings. The relatively simple solution I came up for this version, I think, generalizes to the 12 coins/3 weighings problem.
This wasn't totally clear to me from the description of the problem, but it's important that the scale tells you which side is heavier, if there is one. So when balancing pile A and pile B of coins, it will tell you one of 3 possibilities A > B, B > A or A = B (rather than just A not = B vs. A = B ). Otherwise, 3 weighings wouldn't give enough information to distinguish between the 24 different possibilities for how the weights are arranged.
I think you broke the 3 weighing rule. I was going down that path, and always ended up weighing 4 or more times.
I'm thinking set an equal amount of coins on each side and pull one coin off each side until the balance (or unbalance) is upset or set even. Then I’m thinking once the balance moves you’ve discovered the two coins that cause the “problem”. From there you compare both coins one at a time against one of the coins that didn’t upset the scales. The one coin that doesn’t balance with the verified coin would be the counterfeit.
"A dog breaks your heart only one time and that is when they pass on". Unknown
No, you weigh 4 and 4 the first time, then 2 and 2 the second time, then 1 and 1 the third time.
TD's method would work if we knew in advance if the counterfeit was heavier or lighter than the rest.
The problem happens when the 4 vs. 4 are unequal. You can eliminate the 4 not weighed, but you don't know
immediately which side has the bad one.
Of course, if this was a realistic problem, we would know the counterfeit would be underweight, and there would be no dumb limit on number of comparisons.
For Lincoln cents, a friend of mine once had a similar problem (mid-1990s). She and her dad (a scientist) were trying roll them by weight instead of counting 50. But they noticed that the weight varied even when they had counted 50. They thought the weight difference was due to wear. She told me the exact weights. Then I told her how many pre-1982 and post-1982 coins were in each set of 50!
So this makes it easy to detect the counterfeit Lincoln coins with zero weighings - just look at the date. It's the old die marker method. :-) For half dimes, look at the date and arrows. :-)
I gave up
Collector, occasional seller
How about just send the group to our host and see which one body bags?
My YouTube Channel
I agree! No reason to watch a video... it should only show this method.
Guys: You don't know if the counterfeit coin is HEAVY or LIGHT. This method won't work!
CaptHenway is wrong. That way works if you know whether the coin is heavier or lighter. In this case, you can't split them into pairs in step 2 because you won't know what to do if they don't balance. You have to commingle some from each side. Alternatively, you can weigh three very specially selected sets of 4 against 4.
Give them to two people. When one says he isn't going to waste all that time, hire him.
Oops....Or her.
I stand corrected, I was hasty in thinking we knew that the counterfeit was either under or overweight.
Got it. Scale and even number of coins were very helpful.
This operates under the assumption that the counterfeit is heavier, which is not a priori.
1) fill out PCGS submission form. Bulk submission OK.
2) weigh the package to get the correct postage charge.
3) wait 30 days for the results.
eBay ID-bruceshort978
Successful BST:here and ATS, bumanchu, wdrob, hashtag, KeeNoooo, mikej61, Yonico, Meltdown, BAJJERFAN, Excaliber, lordmarcovan, cucamongacoin, robkool, bradyc, tonedcointrader, mumu, Windycity, astrotrain, tizofthe, overdate, rwyarmch, mkman123, Timbuk3,GBurger717, airplanenut, coinkid855 ,illini420, michaeldixon, Weiss, Morpheus, Deepcoin, Collectorcoins, AUandAG, D.Schwager.
Actually an old problem given in junior high math class.... not using coins... Cheers, RickO