[Guest]

The local casino has started featuring an experimental high stakes game called 'Fools Gold'. The game works as follows: You pay the house 25 dollars, and a 3000$ deposit, and are given 12 identical looking coins. You are told that, in fact, 11 of those coins are identical. The odd coin out appears the same as the others but in some games it will be lighter than the others and in some games it will be heavier than the others (though you aren't told which it is!). If you can guess which coin is diferent you will win 3000$, in addition to getting your deposit back. You are only permitted one guess per game and if you guess wrong, you win nothing and the house keeps your deposit.

To help you discern between the different coins, the house has graciously offered to loan you it's 2 pan balance scale... at 1000$ per weighing! Of course the rules forbid you from using any other device to determine the weight or relative weight of the chips. You cannot leave the table but you can back out of the game any time before you make a guess and get your deposit back.

Anyways, not wanting to miss an opportunity, you sit down to play. After 170 games or so, and about 10000$ in winnings, you are politely but firmly asked to leave. How did you beat the house?

-Tuck

[Guest]

Three weighings are necessary, if I remember correctly. One numbers the coins in the ternary scale, and weighs coins in three groups of four against each other. It takes a bit of explaining.

[Guest]

If you use the scale 3 times in a game and then correctly guess the odd coin out, you've lost 25$. But the 'you' from the question has evidently found a way of doing things that has allowed him to win 10000$ in 170 games (or so). That's enough consistency that it probably wasn't an accident.

Edited September 12, 2009 by Tuckleton

[Guest]

If:

you split the 12 into two groups of 6, weigh 3 against 3 of one group, If they weigh the same, then get a group of 3, and weigh two. Same? Then do it again! Now a 50/50 chance! Maybe this isn't right, but this is the best I can come up with.

Edited September 12, 2009 by new guest

[Guest]

If:

you split the 12 into two groups of 6, weigh 3 against 3 of one group, If they weigh the same, then get a group of 3, and weigh two. Same? Then do it again! Now a 50/50 chance! Maybe this isn't right, but this is the best I can come up with.

In fact your solution works very well if you know before you start if the coin is lighter or heavier. If that was the case then, by your method, you'd be making 475$ a game on average!! Unfortunately we don't know if the odd coin is lighter or heavier at first, we just know it's different. But you seem to have the right frame of mind. Try again!

Edited September 13, 2009 by Tuckleton

[Guest]

I suspect that you are complicating an original problem, namely "With how many weighings, using a balance only, is it necessary to detect a false coin out of 11 other true coins, including whether the false coin is heavier or lighter than the true ones." The answer is 3.

[Guest]

While it's true that my riddle is built around an already existing one, the answer is not the well know solution to the original problem. If a player uses that method (which finds the odd coin out in 3 weighings), that player will lose 25$ every game he plays. The solution will actually have to be more efficient than that.

Your solution doesn't have to find it in the same number of weighings each time. Maybe sometimes it finds it in 2 weighings, or just one. And other times it takes 4 or even 5 times. As long as, on average, it's low enough to make you money.

You don't neccessarily need to know the relative weight of the odd coin. Just which one it is.

-Tuck

Edited September 13, 2009 by Tuckleton

[Guest]

Hi Tuck!

I liked this riddle and worked on it.

My ultimate solution allows me to gain a few pennies only if I pay 25/3 $ per game instead of 25$.

Maybe I can improve it a bit, but I don't think to have any profit after any number of games while pay per game is 25$.

Still I plan to work on this, though being hopeless :-)

But -for clarification- I want to ask that wheather in a single game, after one guess, having paid 25$, if I back out of the game, do I get my all deposit of 3000$ or do I get (3000 - 1000)=2000$ ?

Since my english is limited and poor, I could have not enlightened this myself in OP.

My guess is 2000$, but a solution seems to be impossible, then.

Bye.

[Guest]

In fact your solution works very well if you know before you start if the coin is lighter or heavier. If that was the case then, by your method, you'd be making 475$ a game on average!! Unfortunately we don't know if the odd coin is lighter or heavier at first, we just know it's different. But you seem to have the right frame of mind. Try again!

I've been following this thread and must be confused on the rules because I see no way of even coming close and the above method by 'new guest' (even knowing if it is lighter or heavier) not working either, but even if it does work I don't see how you could make any money using it. Here is it with the costs added in the <<>>: "if you you split the 12 into two groups of 6, weigh 3 against 3 of one group <<-$1,000>> , If they weigh the same, then get a group of 3, and weigh two <<-$1000>>. Same? Then do it again! <<-$1000>> Now a 50/50 chance <<so 1/2 with $3,000-$25 and 1/2 -$3,025>>! Maybe this isn't right, but this is the best I can come up with."

So based on the above you spend $3,000 per game and on average lose $25 per game for a grand (average) loss of $3,025 per game.

So what am I missing?

[Guest]

I've been following this thread and must be confused on the rules because I see no way of even coming close and the above method by 'new guest' (even knowing if it is lighter or heavier) not working either, but even if it does work I don't see how you could make any money using it. Here is it with the costs added in the <<>>: "if you you split the 12 into two groups of 6, weigh 3 against 3 of one group <<-$1,000>> , If they weigh the same, then get a group of 3, and weigh two <<-$1000>>. Same? Then do it again! <<-$1000>> Now a 50/50 chance <<so 1/2 with $3,000-$25 and 1/2 -$3,025>>! Maybe this isn't right, but this is the best I can come up with."

So based on the above you spend $3,000 per game and on average lose $25 per game for a grand (average) loss of $3,025 per game.

So what am I missing?

I apologize for any confusion. First it's important to note that the 3000$ deposit you give at the start of the game is refunded unless you guess incorrectly. And I'll also go over the method described by New Guest as I understood it. Perhaps it can help shed some insight on this problem.

...that the coin was heavy before you started. So you weigh 3 against 3. If they don't balance, then you can narrow it down to 3 coins (Whichever scale went down.) If that is the case it only take one more weighing to find the fake. So in 6 out of 12 cases, it only takes 2 weighings. If, instead, they balance, take 2 of the remaining 6 against each other. If they don't balance you've found your fake. So 2 more of 12 are found in 2 weighings. But if it still balances, take 2 of the remaining 4. If they don't balance, then you've found the fake again. So 2 out of 12 cases will take 3 weighings. Finally, if it still balances, it will take one more weighing to find it in the remaining 2. So 2 out of 12 cases takes 4 weighings. So:

8/12 takes 2, 2/12 takes 3 and 2/12 takes 4. For an average of 2.5 weighings.

So you pay 25$ to get in, a 3000$ deposit which you always get back since you never guess wrong, an average of 2500$ to use the scale 2.5 times and you win 3000$. That's 475$ in winnings, on average, per game.

So basically, in 8 out of 12 games, you will win 975$, in 2 out of 12 games you will lose 25$ and in 2 out of 12 games you will lose 1025$.

You see, it's not about winning money in every single game, it's about having a higher chance of winning money than losing it and playing lots of games.

Anyways, hope that helps clear things up a little. If I had my time back I probably would have gotten rid of the deposit and just said a game costs 3025$ to enter and you win 6000$ on a correct guess. Oh well. I'm thinking maybe I should just stick to trying to solve riddles . But it's good to know people are still thinking about this one.

[plainglazed]

EDIT: Think I might be onto something but have to rethink some. Have been convinced this is not doable but now have a glimmer of hope. Not sure of what I just posted. Hopefully be back soon.

Edited September 16, 2009 by plainglazed

[plainglazed]

OK, got a little cold feet for a moment. Think this works.

Label the chips 1-12 and chips determined to be normal label N:

1) Compare 1,2,3,4,5 vs 6,7,8,9,10

if 1,2,3,4,5 = 6,7,8,9,10

2) Compare 11 vs N

if > 11 is a heavy coin (one chip determined heavy in 2 weighings)

if < 11 is a light coin (one chip determined light in 2 weighings)

if = 11 is normal

3) Compare 12 vs N

if > 12 is a heavy coin (one chip determined in 3 weighings)

if < 12 is a light coin (one chip determined in 3 weighings)

if 1,2,3,4,5 > 6,7,8,9,10

Assume 1 is your lucky coin and it is absolutely normal. Afterall you are a gambler.

2) Compare 2,3,4 vs 6,7,8

if > then one of 2,3 or 4 are heavy

3) Compare 2 vs 3; if 2>3 then 2H, if 2<3 then 3H, if 2=3 then 3H (three chips determined in 3 weighings)

if < then one of 6,7, or 8 are light

3) Compare 6 vs 7; if 6>7 then 7L, if 6<7 then 6L, if 6=7 then 8L (three chips determined in 3 weighings)

if = then either 5 is heavy or 9 or 10 are light

3) Compare 9 vs 10; if 9>10 then 10L, if 9<10 then 9L if 9=10 then 5H (three chips determined in 3 weighings)

4) Compare 1 vs N to confirm if it was truely normal (one chip determined heavy in 1 weighing). Why gamble when you dont have to?

Average number of weighings is just less than 2.929 to determine which chip is heavy or light. So at $1000 per weighing and $25 to

play, each game costs $2954 on average to win $3000 or $46 per game won. If this reasoning is legit, you always beat the house.

(This method just now came to me and I havent yet tried with other initial weighings which may bring the slightly better

results of the OP.)

Hope this is on the right track. Awesome puzzle Tuck!

[plainglazed]

my last "solution" (fingers crossed) determines the weight 1/6 of the time in two weighings, 3/4 of the time in three weighings and 1/12 of the time in four weighings for an average of 2.917. At a cost of $1000 per weighing and $25 to play, you win $58 per game on average or $9860 after 170 games.

[Guest]

Plainglazed, I have a confession to make. My original incarnation of this puzzle had you tricking the house into letting you use your own "lucky" coin during the game which you knew to be real. But then, a solution very similar to yours popped into my head and I was convinced I had it without using an extra coin so the puzzle went ahead without it. Only now, reading your solution, do I realize the mistake I made in my own solution. It's with great shame that I have to admit that I don't think it can be done without this 13th coin after all...

I'm very sorry to anybody who has spent significant time working on this one but that time is not necessarily wasted. I know for a fact that it can be done with a 13th coin which is known to be real.

Anyways, a short word on what we seem to have overlooked in our solutions:

The problem arises in step 3:

3) Compare 9 vs 10; if 9>10 then 10L, if 9<10 then 9L if 9=10 then 5H (three chips determined in 3 weighings)

when 9=10 the fake could be either 5 or 1. If we don't use another weighing to distinguish between them, we have a 50% chance of guessing wrong and losing 6025$. Certainly not worth it. Which means there are 2 cases here that will require 4 weighings.

My solution was basically the same as yours:

The coins are 1234567890AB

12345=67890, then A vs N. If A=/=N then it's A (2 cases in 2). If A=N then it's B (2 more cases in 2).

12345>67890, then 1236 vs 45NN. If 1236=/=45NN then we can find the fake in 1 more weighing (6 cases in 3).

If 1236=45NN, then 7 vs 8. If 7=/=8 then we found our fake (2 more cases in 3). And here is where I also thought I would only have 1 case left for 4 but it if 7=8 then it is in fact: (2 cases in 4)

Now we do the same for 12345<67890. (8 more cases in 3 and 2 more cases in 4)

Totals: 4 cases in 2

16 cases in 3

4 cases in 4

Not good enough...

Again, I apologize.

-Tuck

[Guest]

adding your own coin, now that is cheating, and there is one other thing your are missing out on, you never want to use more than two weighs for then you would be loosing. You would just not guess if you could not figure it out in two weighs

[Guest]

adding your own coin, now that is cheating, and there is one other thing your are missing out on, you never want to use more than two weighs for then you would be loosing. You would just not guess if you could not figure it out in two weighs

sure, but i believe we assume you have enough money to ride out the losing rounds and that the favorable odds will eventually win you money

[Guest]

Wait wait wait!! I take it back. I take it all back! The puzzle CAN be solved as written after all. Lightning struck my brain on the way to work this evening. The solution isn't a tenth as elegant as I would like it to be but it exists.

Feel free to solve it with the extra coin as well if you like but it can be done without it after all.

[plainglazed]

You're right, of course, regarding my last post. Should have been more alert or realized I wasnt pre-coffee. Though just weighing the last "reserved" chip does complete all the possibilities, it does so by defining the ninth of the ten as well; as you pointed out. That is of course if you also rectify a mistake I made in the wrong answer above: 2) 2,6,7 vs 3,8,9 (not 2,3,4 vs 6,7,8); if > 2,8,or9; if < 3,6,or7; if = 4,5,or10.

As for the solution given an extra coin, I believe it is not so obvious and a worthy puzzle in itself. Give it some thought before reading the below spoiler if you disagree.

Label the coins 1-12 and the coin known to be normal and coins determined to be normal label N:

1) Compare N,1,2,3 vs 4,5,6,7

if N,1,2,3 > 4,5,6,7

2) Compare 1,2,3,4,5,6 vs N,N,N,N,N,N

if = then 7 is light

if > then one of 1,2,3 is heavy

3) Compare 1 vs 2; if 1>2 then 1H, if 1<2 then 2H, if 1=2 then 3H

if < then one of 4,5,6 is light

3) Compare 4 vs 4; if 4<5 then 4L, if 4>5 then 5L, if 4=5 then 6L

if N,1,2,3 < 4,5,6,7 then the same method as above applies

if N,1,2,3 = 4,5,6,7 then one of 8,9,10,11,12 is heavy or light. Quit there.

One coin is defined in two weighings (+$975). Six coins are defined in three weighings (6 x -$25 = -$150).

So seven times out of twelve you average winning $825 (7 x $825 = $5775). And five times out of twelve you lose $1025 (5 x $1025 = $5125).

Your average winnings in twelve games would then be $650. You beat the house?

Now, Tuck, are you absolutely certain this can be done without the extra known coin?

Edited September 26, 2009 by plainglazed

[Guest]

I like it! I especially like backing out if it's one of those other 5 and still making a profit. I did it slightly differently:

Coins are 1234567890AB

If 1234 =/= 567R then do it the same as you did (2 cases in 2 and 12 cases in 3)

If 1234 = 567R then 890 vs RRR.

If 890 =/= RRR then 1 more weighing finds it (6 cases in 3)

If 890 = RRR then A vs R.

If A = R then it's A, if not then it's B (4 Cases in 3 Weighings)

So in 2 out of 24 We get it in 2 and the rest in 3. (1000*2 - 25 * 24)/24 = 58.33$ per game average.

Anyways, the solution for just 12 coins does exist but I don't like it, it's not very imaginative like your solution for 13. It relies on the fact that you don't need to know if the coin is lighter or heavier like the last step in my solution.

Though, now I think about it, though your solution doesn't quite make as much money, it doesn't rely on that little loophole and so is, in my opinion, more efficient.