So you put beakers into edges (it is 30 of them), not vertices. The rest is the same.

It will not work

That looks like it should work, except there are 30 beakers and not 20.

Right, my mistake. So you put beakers into edges (it is 30 of them), not vertices. The rest is the same.

How about using dodecahedron?

It has 20 vertices, so let's put one beaker into each vertex.

It also has 12 faces, but let's remove two opposite faces.

Now each out of remaining 10 faces represents a test.

To perform each test we use five beakers that are in vertices of the face representing this test.

If result of the test is positive (rat dies), we paint the face representing test black (a color of death).

If result of the test is negative (rat lives), we paint the face white (a color of life).

After all tests are performed our dodecahedron (without two faces) will be painted in black and white.

Will it be always possible to determine poisoned beakers from the way dodecahedron is painted?

It would be more interesting to ask what is the minimal length of the boards provided we have two boards of the same length.

The other is that an infinite number of balls were added, and an infinite number were removed and the answer would then be zero.

The argument is not that infinity number of balls were removed, and it is not about 'infinity - infinity' tricks.

The argument is that each particular ball was removed at specific moment. You can provide a formula that takes the number of the ball and gives the exact time it is removed. Each particular ball is inserted once and then removed once before noon, thus no ball can remain inside the box at 12:00.

My previous answer was wrong (too fast), so let me correct myself and explain solution.

To obtain the answer we need to find:

A = total number of ways it is possible do deal four players two hole cards each (out of 52-card deck)

B = as above but with additional condition: each player gets an ace.

Then B/A will be our probability.

Both A and B are easy and straightforward to calculate (though I made strange mistake calculating A in my first answer).

Let's call our four players: SB, BB, CO, BTN.

To find A we have to choose 2 cards out of 52 for SB, then 2 cards out of remaining 50 for BB, and so on:

A = Comb(52,2) * Comb(50,2) * Comb(48,2) * Comb(46,2)

A = 52*51*0.5 * 50*49*0.5 * 48*47*0.5 * 46*45*0.5

To find B we have to choose suit of the ace for SB, BB, CO and BTN (4*3*2*1) and choose remaining hole card for each player (48*47*46*45):

B = 4*3*2*1 * 48*47*46*45

So the answer is:

B/A = (4*3*2*1 * 48*47*46*45) / (52*51*0.5 * 50*49*0.5 * 48*47*0.5 * 46*45*0.5)

(You can paste above expression in google to evaluate it, if you need the number.)

(4*48 * 3*47 * 2*46 * 1*45) / (48*47*0.5 * 46*45*0.5 * 44*43*0.5 * 42*41*0.5)

You "don't have enough time" to simply add "Where n is the number of cuts?" Wow you must be one busy person.

It is not that I'm a busy person, it just didn't occur to me that there could be any doubt about what n stands for.

Anyway, my apologies to everyone confused.

The formula is based on the fact that with every N-th cut you will be able to cut at most N pieces (can anybody prove this?)

You start and finish the n-th cut at some two points at the edge of the pizza.

Ont it's way this cut crosses each of n-1 previous cuts.

When the n-th cut crosses first previous cut, a new piece appears. When it crosses second cut we have second piece, and so on.

When we cross last previous cut we have (n-1) new pieces, and finally we reach the edge of the pizza which produces one more piece.

This makes total of n new pieces with n-th cut.

...

Thanks, k-man. You have explained it all perfectly.

n*(n+1)/2 + 1

1. If you can weigh x and y then you can weigh x+y.

2. If you can weigh x and y then you can weigh x-y (for x>y).

3. If you can weigh x then you can weigh x/2.

And that's all.

Problem of finding 7 out of 5 is equivalent to finding 1 out of 5, because:

7 = (5+1)+1 (if you know how to weigh 1, you know how to weigh 7)

and

1 = (7-5)/2 (if you know how to weigh 7, you know how to weigh 1)

And problem of finding 1 out of 5 is equivalent to finding 1/5 out 1 (solution for one works for another, just with another starting pile).

Suppose that the last problem is solvable. Then there is a finite number of operations 1.-3., that produce 1/5 out of 1.

But how can finite number of additions, subtractions and divisions by 2 of number 1 produce a (irreducible) fraction with denominator 5?

It just can't. Clearly only powers of 2 are possible in the denominator of this fraction (in reduced form).

This shows that the original problem in also not solvable.

6. Weigh another 5 lbs sugar. Place the 4 3/8 lb of sugar on one side of the scale and enough of the 5 lb pile, leaving 7/8 lb.

Wrong.

The player who moves first has a winning strategy:

his first move is made diagonally, then he keeps repeating his opponent's moves (if his opponent moves right, he moves right, etc.).

Why this works? Suppose chessboard squares have horizontal and diagonal coordinates ranging from 1 to 8, so that the Kings starts at square (1,1) and finishes at square (8,8). Observe that the above strategy ensures, that after every move of the first player the King is on the square with both coordinates being even numbers, and every move of his opponent the King is on the square with at least one coordinate being odd. The final square of the King journey (8,8) has both coordinates even, so only the first player can get there.

The player who moves first has a winning strategy:

his first move is made diagonally, then he keeps repeating his opponent's moves (if his opponent moves right, he moves right, etc.).

The fundamental group of a plane with n holes is the free group of rank n. The rest is pure simple algebra.

If you have infinite amount of money, then you cannot win any more, since infinity plus your eventual winnings is still infinity.

In other words: what's the point of gambling when you have all the money in the world, can't loose it and can't get richer?

If above argument is not convincing for you, than accept that it is not possible to have infinite amount of money

I you have a finite amount of money, you can afford only a finite number of consecutive bets. Say you have $2^n-1 so you can afford n consecutive bets.

In this case there is a 2^-n chance to loose all the money and 1-2^-n chance to win $1 in one "session". Therefore the expected value of one session is:

EV = +$1*(1-2^-n) -$(2^n-1)*2^-n = $1 - $1 = $0

This explains why you should not expect to get rich playing this game.

And you should to expect to loose in real life casino with this strategy since you are not going to get even-money bet (EV=0) there.

for n=2:

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

for n>2: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

There is no way to put number 6 in 3rd column.

Will just one isosceles trapezoid do?

What you actually ask is to arrange given vectors into an ordered (rooted) tree that has the highest probability of construction according to given rules and vectors out of all possible ordered trees.

But observe, that when creating set of vectors according to given rules it is equally likely to produce vector W from vector V and to produce vector V from vector W. Therefore changing the root of the tree to any other node will not alter the probability of the tree, but will alter the relation of precession.

Finding the (non-ordered) tree with highest probability is definitely doable, but choosing the root is just guessing.

Maybe asking about non-ordered tree would make the contest more interesting.

I agree with you that these are almost certainly the right answers, but that's different from a rigorous proof.

I totally agree. This is why I wrote "sketch of a proof" instead of "proof".

Besides, I've left a case when there is one large circle (L) and a number of small circles,

small circles are pair-wise externally tangent to each other and each small circle is internally tangent to the big circle.

I just said, that "it is easy to see" that at most 3 small circles are possible, so I just want to complete it.

Suppose we have 4 or more inner circles. Let's number them 1, 2, 3, 4... in the following way:

number 1 is assigned freely, and subsequent numbers are assigned subsequently clockwise according to tangent points to the circle L.

Since circle 2 is tangent to circle 4 and both are tangent to the circle L, they divide interior of circle L in two parts.

But circles 1 and 3 are in different parts, so these circles cannot be tangent.

This contradiction shows that 4 or more small circles are not possible.

I suppose we're searching for a proof that this is really the largest number of circles possible in such a configuration (and that there are no other configurations with the same number)? Or is it just that these are the best solutions anyone has ever found?

There are two types of tangency: internal and external.

In case when only external tangency exists, 4 is a maximum: we start with 3 circles, then the only option is to put the fourth one "in the middle", and that's it.

Now suppose there are two internally tangent circles L (larger) and S (smaller). Now all other circles have to go inside L (if not, they will not be able to be tangent with S) and have to be externally tangent with S (if not, they will not be able to be tangent with L).

So in this case we have one "big" circle and a number of "small" circles, small circles are pair-wise externally tangent with each other and each small circle is internally tangent with the big circle. In this case at most 3 small circles are possible which is easier "to see", than to prove.

(I'm not suggesting that the prove is hard.)

So 4 is the maximum and only 2 configurations are possible.

PS I believe this reasoning can be extended for 3-dimensional case (mutually touching spheres).

The OP intended to require the pair-wise points of tangency be distinct.

I see. Then you must be asking about http://en.wikipedia.org/wiki/Descartes%27_theorem

The simple part is a special case with one of the circles replaced by a straight line.

Maybe the harder problem requires clarification (for me)...

It is clearly required for each pair of circles to be tangent, which is not the case in your example. For example the top-left circle is not tangent with bottom-right circle.