HotCore,

I am just being an old party pooper. I am thinking about what I would REALLY do, I would simply place one bullet in the center of one pan (eye balling it) and then start swapping out a single bullet at a time in the center of the other pan (again eye balling it) until I found the light one. See I seriously believe I could find the light one quicker this way.

I really do hate statistics. I know how to perform the math, but to me it is a tool, like ratchet, nothing more. I don't love ratchets either (at least most of the time) Oh yeah, that doesn't mean anyone can take my Snap ons and throw them around the shop!

ASS_CLOWN

The absoloute minimum is two weighings. Its not too likley, but it is possible.





Heres my hint; 5/5 1/1

I came up with two comparisons. Here is my logic.

The problem asks for the minimum number so you need to assume you will get lucky and draw the off-weight bullet the first or second time, but the trick is to insure that you have the off-weight bullet and that you are able to determine if it is heavier of lighter than the norm.

You select two bullets (A and B) and they are equal weight. Now you know what the standard weight (norm)is. You save one of the bullets (A) to use in the next comparison. The next bullet is C which is lighter or heavier than A. Two comparisons and you are done.

Now if the first two bullets are different, A is heavier than B or B is heavier than A, then you need to compare again. Select either A or B and compare with C. Lets say that A is heavier than B, and C and A are the same, then B is the off-weight bullet and it is lighter. If A is heavier than B and C is lighter than A, then C is the off weight bullet and it is lighter. The same logic is used if A is lighter than B. Again you have two comarisons and you are done.

'course this might help some too... http://www.jdawiseman.com/papers/easymath/weighing.html

(this has gone on too long... )

The obvious way to solve this problem is going to be a process of elimination. 6.5 Bandit is right. It can be done in two, for certian, no less.



A few things to consider. The question to the problem is "what is the MINIMUM number of weighings required"?



Also, we are already given the FACT that only one of the bullets are odd, so we need not wrestle with that on the scale.



Heres an example. With luck!! 10 bullets are placed on the scale, 5/5 and it ballances. We have just eliminated 10 possibilities of 12. Remove all but one bullet from the scale, place one of the remaining two on the opposing pan. If it happens to be the one then it will show, and you will also know if it is light, or heavy. That is the MINIMUM number of weighings and it can be accomplished several ways. 4/4 1/1, 3/3 1/1, 2/2 1/1. All other answers can be modified within the process according to..... the luck of the draw!



Cool brain wringer Hot core..

You guys all blew it. you dont need to weigh any of them on the scale. load all bullets into equal brass and powders. shoot all through a chrony. the one that goes the fastest is the lightest, or the slowest is the heaviest. And I did not touch your darn scale once.

I do regret wasting time on this irrelevant topic. What I remember from math probability is that there are 144 combinations.

I promise not to read this topic again so have at it.

OK folks, here is Part 2. There is absolutely no Luck or Guessing involved with this method. By the way, I found "coloring" the Bullets helped me see the "trick".

...

Now we go back and look at the First Comparison which was set up as follows backin Part 1:

[1] [2] [3] [4] ^{5} {6} {7} {8}

But, this time the Scale is "NOT" Balanced. That will mean the "different" Bullet is in:

[1] [2] [3] [4] ^{5} {6} {7} {8}

And, we know for sure that <9> <10> <11> <12> are all alike.

For the Second Comparison set it up as follows:

[1] [2] {5} <9> ^ [3] [4] <10> <11>

And I have {6} {7} (8} and <12> off to the side.

If the Scale is Balanced, I know the different Bullet is either {6} {7} (8} and determine which one is Heavier or Lighter just as we did back up in Part 1.

If the Right side of the Scale goes down, I know that either [1] [2] must be light and the Third Comparison:

[1] ^ [2] would tell me which one is Light.

If however the Right side of the Scale went Up after the Second Comparison, then either {5} is Heavy or [3] or [4] is Light.

For the Third Comparison we do:

[3] ^ [4]

If it is Balanced, then {5} is Heavy.

If either Side goes Up, that is the Light Bullet.

And you are done!

...

The "trick" is grouping things into "3s". No luck and no guessing is involved. This way works irregardless of the position you choose to put the "different" Bullet in the original sequence.

Pick any one as being Heavier or Lighter. Then go back and follow the Logic Steps as outlined above in Part 1 & 2 and you can prove it to yourself.

If there are other ways to do it where no luck or guessing is involved, please include your method and my hat will be off to you.

Glad to see some of you folks enjoyed it!

Put the house on 'FOUR' no less no more
Should know which one and if higher or lower everytime with no 'luck'

That brick would be two pounds, and a brick and a half would be three pounds, but you would need to load them and shoot them through a chrony to see if the sun was out.

Hot Core

I'm a little confused.

Please tell me how this scenario would work in either [2] was heavy or [5] was light.

/r