Information Gain and Entropy

I recently read this question regarding information gain and entropy. I think I have a semi-decent grasp on the main idea, but I'm curious as what to do with situations such as follows:

If we have a bag of 7 coins, 1 of which is heavier than the others, and 1 of which is lighter than the others, and we know the heavier coin + the lighter coin is the same as 2 normal coins, what is the information gain associated with picking two random coins and weighing them against each other?

Our goal here is to identify the two odd coins. I've been thinking this problem over for a while, and can't frame it correctly in a decision tree, or any other way for that matter. Any help?

EDIT: I understand the formula for entropy and the formula for information gain. What I don't understand is how to frame this problem in a decision tree format.

EDIT 2: Here is where I'm at so far:

Assuming we pick two coins and they both end up weighing the same, we can assume our new chances of picking H+L come out to 1/5 * 1/4 = 1/20 , easy enough.

Assuming we pick two coins and the left side is heavier. There are three different cases where this can occur:

HM: Which gives us 1/2 chance of picking H and a 1/4 chance of picking L: 1/8 HL: 1/2 chance of picking high, 1/1 chance of picking low: 1/1 ML: 1/2 chance of picking low, 1/4 chance of picking high: 1/8

However, the odds of us picking HM are 1/7 * 5/6 which is 5/42

The odds of us picking HL are 1/7 * 1/6 which is 1/42

And the odds of us picking ML are 1/7 * 5/6 which is 5/42

If we weight the overall probabilities with these odds, we are given:

(1/8) * (5/42) + (1/1) * (1/42) + (1/8) * (5/42) = 3/56.

The same holds true for option B.

option A = 3/56

option B = 3/56

option C = 1/20

However, option C should be weighted heavier because there is a 5/7 * 4/6 chance to pick two mediums. So I'm assuming from here I weight THOSE odds.

I am pretty sure I've messed up somewhere along the way, but I think I'm on开发者_C百科 the right path!

EDIT 3: More stuff.

Assuming the scale is unbalanced, the odds are (10/11) that only one of the coins is the H or L coin, and (1/11) that both coins are H/L

Therefore we can conclude:

(10 / 11) * (1/2 * 1/5) and

(1 / 11) * (1/2)

EDIT 4: Going to go ahead and say that it is a total 4/42 increase.

You can construct a decision tree from information-gain considerations, but that's not the question you posted, which is only the compute the information gain (presumably the expected information gain;-) from one "information extraction move" -- picking two random coins and weighing them against each other. To construct the decision tree, you need to know what moves are affordable from the initial state (presumably the general rule is: you can pick two sets of N coins, N < 4, and weigh them against each other -- and that's the only kind of move, parametric over N), the expected information gain from each, and that gives you the first leg of the decision tree (the move with highest expected information gain); then you do the same process for each of the possible results of that move, and so on down.

So do you need help to compute that expected information gain for each of the three allowable values of N, only for N==1, or can you try doing it yourself? If the third possibility obtains, then that would maximize the amount of learning you get from the exercise -- which after all IS the key purpose of homework. So why don't you try, edit your answer to show you how you proceeded and what you got, and we'll be happy to confirm you got it right, or try and help correct any misunderstanding your procedure might reveal!

Edit: trying to give some hints rather than serving the OP the ready-cooked solution on a platter;-). Call the coins H (for heavy), L (for light), and M (for medium -- five of those). When you pick 2 coins at random you can get (out of 7 * 6 == 42 possibilities including order) HL, LH (one each), HM, MH, LM, ML (5 each), MM (5 * 4 == 20 cases) -- 2 plus 20 plus 20 is 42, check. In the weighting you get 3 possible results, call them A (left heavier), B (right heavier), C (equal weight). HL, HM, and ML, 11 cases, will be A; LH, MH, and LM, 11 cases, will be B; MM, 20 cases, will be C. So A and B aren't really distinguishable (which one is left, which one is right, is basically arbitrary!), so we have 22 cases where the weight will be different, 20 where they will be equal -- it's a good sign that the cases giving each results are in pretty close numbers!

So now consider how many (equiprobable) possibilities existed a priori, how many a posteriori, for each of the experiment's results. You're tasked to pick the H and L choice. If you did it at random before the experiment, what would be you chances? 1 in 7 for the random pick of the H; given that succeeds 1 in 6 for the pick of the L -- overall 1 in 42.

After the experiment, how are you doing? If C, you can rule out those two coins and you're left with a mystery H, a mystery L, and three Ms -- so if you picked at random you'd have 1 in 5 to pick H, if successful 1 in 4 to pick L, overall 1 in 20 -- your success chances have slightly more than doubled. It's trickier to see "what next" for the A (and equivalently B) cases because they're several, as listed above (and, less obviously, not equiprobable...), but obviously you won't pick the known-lighter coin for H (and viceversa) and if you pick one of the 5 unweighed coins for H (or L) only one of the weighed coins is a candidate for the other role (L or H respectively). Ignoring for simplicity the "non equiprobable" issue (which is really kind of tricky) can you compute what your chances of guessing (with a random pick not inconsistent with the experiment's result) would be...?