Ben

Here's a rather simple problem, but it's fairly cute and open to a whole gamut of possible extension problems.

Consider three bins. Each bin contains many balls. In one of the bins, each ball weighs 2, in another bin each ball weighs 3, and in the last each weighs 5 [units are arbitrary]. We don't know which bin is which.

Find a way to take determine which bin is which with only one weighing. You may take as many balls from as many bins as you wish.

Bonus: since that should be fairly easy, can you find a description of ALL such strategies?

My two-year Y!A anniversary, celebrated again with a new question:

For a few months now I've been developing a chess variant which involves four spatial dimensions. I've finally gotten around to typing up a few notes on it and making a couple of diagrams if you're interested:

http://my.ilstu.edu/~bmreini/Chess-basic...

My main question is, does the idea of 4D chess interest you? If so, then please have a read through the pdf. I've left quite a bit out really, particularly toward the end, so if you notice something interesting or have a comment, make a note of it here. [That's what my best answer criteria is for now.]

You'll notice I don't pick a solution for the slippery king; which one do you like? Do you have another suggestion? My main objective over the next few months is to decide on one central ruleset, which will be dubbed "TessChess" -- as in tesseract chess; "chesseract" is already taken by another supposedly 4D game -- and any other notable rules will be variants thereof in my mind.

Thanks!

[I've asked this in both the Mathematics section and the Board Games section, to get an impression on whether this is more a chess-players game or a mathematician's one.]

My two-year Y!A anniversary, celebrated again with a new question:

For a few months now I've been developing a chess variant which involves four spatial dimensions. I've finally gotten around to typing up a few notes on it and making a couple of diagrams if you're interested:

http://my.ilstu.edu/~bmreini/Chess-basicmath.pdf

My main question is, does the idea of 4D chess interest you? If so, then please have a read through the pdf. I've left quite a bit out really, particularly toward the end, so if you notice something interesting or have a comment, make a note of it here. [That's what my best answer criteria is for now.]

You'll notice I don't pick a solution for the slippery king; which one do you like? Do you have another suggestion? My main objective over the next few months is to decide on one central ruleset, which will be dubbed "TessChess" -- as in tesseract chess; "chesseract" is already taken by another supposedly 4D game -- and any other notable rules will be variants thereof in my mind.

Thanks!

[I've asked this in both the Mathematics section and the Board Games section, to get an impression on whether this is more a chess-players game or a mathematician's one.]

During our holiday celebration, my family started a game of zilch. Zilch is a dice game played with six dice, and is somewhat similar to yahtzee. I'm interested in the probability of getting a zilch in the first roll, with the six dice.

In the first roll, the following situations score:

any die 1

any die 5

three or more of a kind

three pair

straight (1-6)

If you fail to get any of those, you score 0 and lose your turn (you "zilch"). Assuming fair dice, what is the probability of zilching?

Bonus: What about with 5, 4, 3, 2, or 1 die? The same scoring rules apply, except obviously you can no longer get a straight or three pair, nor can you get a triple with 1 or 2 dice.

[In the game, you set aside at least one die that earns you points from each roll and have the option of rolling the remaining dice. You add all points you earn to your score; however, if you zilch at any point in your turn, you get no points whatsoever for that turn. So knowing how likely it is to zilch is fairly important. 1 and 2 dice are easy, 3, 4, 5 a bit harder, and I think all 6 is a bit tricky. I have an answer for 6, but want to make sure I've got it right.]

Here's the follow-up I promised to:

http://answers.yahoo.com/question/index;_ylt=AjUPi...

You'll want to refer to an old question of mine:

http://answers.yahoo.com/question/index;_ylt=Ah9xL...

http://tech.groups.yahoo.com/group/mathforfun/mess...

Basically, we have a coin which, given that we've tossed it n times and gotten X heads, the probability for heads in the next toss is 1-X/n (and p=1/2 in the first toss).

It's been shown that, for n tosses, X has a distribution that follows

Pr(X=x) = E(n-1,x-1)/(n-1)!

for n>=2, x=1, 2, ..., n-1 and is zero otherwise. Here E(n, k) is an Eulerian number:

http://en.wikipedia.org/wiki/Eulerian_numbers

However, that may not be too useful here, as I care about sequences.

In sync with the previous few coin questions here, I want to know this: If we flip the coin until the combination THT or TTH shows up, which sequence is more likely to be the end of our coin flipping?

This is a follow-up to the question:

http://answers.yahoo.com/question/index;_ylt=Ah6X0...

which is, in turn, a follow-up to

http://answers.yahoo.com/question/index;_ylt=Anr2n...

Suppose we flip a coin until either the sequence THT or TTH shows up. If we want to have equal probabilities for either sequence to end our coin flipping, how should we weight the coin? That is, if the probability in any toss of heads is p, what value of p creates equal probabilities (of 1/2) of ending with THT or with TTH?

[I have two more follow-ups in mind; let me know if you like this enough to see some more...]

I came across a rather interesting identity while messing around with one of my favorite functions (which will be left unidentified, for now). The identity, in text, is

sum{sum{ (-2)^n*x^(2n+2k+1)/[k! (2n+1)!!] }}

= sum{x^(2m+1)/[(2m+1) m!]}

the summations are taken wrt the appropriate variables n, k, and m, and all run from 0 to infinity.

image version:

http://www.ilstu.edu/~bmreini/sumid.bmp

I'd like to see someone give a direct proof of the identity. I want to know if my round-about discovery of this is actually better than trying to show it directly. Bonus points if anyone can deduce how I came up with the thing in the first place.

(Oh, and that is a double factorial in the LHS; that means the product of the odd integers up to 2n+1 in this case, NOT the factorial of the factorial.)

I'll try to put up some helpful hints up here every day or so...

Hmm, the "similar questions" presented by Yahoo seem to indicate this one might be simple for some of you. Ah well, here goes:

Let n be a positive integer such that 5n has 36 positive divisors and 3n has 40 positive divisors.

How many positive divisors can 15n have?

How many values of n are possible?

[I have a solution which I believe is correct, but it's far from elegant; as such, whoever gets the most elegant solution will get best answer.]

Suppose we have a coin that "tries" to be fair. To be more specific, if we flip the coin n times, and have X heads, then the probability of getting a heads in the (n+1)st toss is 1-(X/n). [The first time we flip the coin, it truly is fair, with p=1/2.]

A short example is in order. Each row here represents the ith coin toss, with associated probability and its outcome:

p=1/2: H

p=0: T

p=1/2: H

p=1/3: T

p=1/2: T

p=3/5: T

It certainly would seem that the expected value of X should be n/2. Is this the case? If so (and even if not), then think of this coin as following a sort of altered binomial distribution. How does its variance compare to that of a binomial [=np(1-p)]?

Does the coin that tries to be fair become unfair in the process? Or does it quicken the convergence to fairness?

(I haven't worked this one out, it just occurred to me. If the algebra becomes intractable, computer solutions are allowable.)

In celebration of my one year anniversary here at Answers, I thought I would put up my first question(s). This one comes from the 2007 Putnam competition (no cheating!). I had a pretty good idea where to start, but couldn't finish it up. One of the other test-takers from my university did manage to finish it.

Suppose you write down the numbers 1, 2, ..., 3k+1. What is the probability that the sum of your partial list is never divisible by 3, at any time in the process? (That is, with each new number you write down, the sum of your list is never divisible by 3.)

Cheers!