False coin, math problem.

[Drag]

There are 12 coins, one of them is a fake. It is either heavier or lighter than the rest. We have a balance in our disposal.

How many tries we need to find the fake coin and determine if it is heavier or lighter than the true coin?

[nightwood]

3

[Drag]

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3

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That's false. If we knew for sure that it is heavier, then we could do it in 3 tries.

And obviously, we are interested in the solution and not in the decaration.

[Andy Ross]

I think it's max 4, but we always have a 50% shot at 3, so stochastically it's 3.5.
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Firstly there's no point weighing 6 vs. 6 (tells us nothing)

TRY1: Take any 6, and balance them. We then have 6 normal coins (if the balance is off, then it's the 6 we didn't select.)

TRY 2: Take 3 normal coins, and weigh them against three unknowns. If the balance is off (50% chance), then we know that these three unknowns are candidates, and also whether the answer is lighter of heavier. Goto 3a. If they're measured equal, then the other three unknowns are the only candidates. Goto 3b.

TRY 3a: Weigh two of the candidates against each other. If they're equal, it's the other one. If they're not equal, then we know which one of those on the balance it is. (3 steps, but 50% of not getting this route)

TRY 3b+4: We do TRY2 on this group to gauge heavier/lighter, followed by 3a. (4 tries, 100%)
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[nightwood]

[ QUOTE ]
[ QUOTE ]
3

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That's false. If we knew for sure that it is heavier, then we could do it in 3 tries.

And obviously, we are interested in the solution and not in the decaration.

[/ QUOTE ]

It's correct [img]/images/graemlins/tongue.gif[/img] I'm too lazy too type out the answer b/c it's a bit long-winded but please read for yourself

here

[Henry17]

It is 3. This is a very old problem

[Drag]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
3

[/ QUOTE ]

That's false. If we knew for sure that it is heavier, then we could do it in 3 tries.

And obviously, we are interested in the solution and not in the decaration.

[/ QUOTE ]

It's correct [img]/images/graemlins/tongue.gif[/img] I'm too lazy too type out the answer b/c it's a bit long-winded but please read for yourself

here

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You are right, it's 3. I did it several years ago, so forgot the answer.
BTW there are at least 3 different ways to get this answer.

[jay_shark]

You may want to prove a more challenging problem .

Prove that the minimal number of weight trials necessary to determine the one counterfeit coin in a collection of n coins is [log(base3)(n-1/2)] + 1 , where [,] is the largest integer in the number.