Brainteasers/Math problems - Page 15

First, if black kills gray, he dies to white. Simple. So the strategy of shooting at gray is infinitely inferior to purposefully missing, since purposefully missing guarantee's gray's turn comes whereas shooting at gray only gives a 2/3 chance of gray's turn coming. So black must either shoot at white or purposefully miss.

Let's say white's turn somehow comes around. If he shoots gray, there is a 2/3 chance he survives. If he shoots black, there is a 1/3 chance he survives. (This can be seen trivially by examining the possibilities). So white WILL shoot gray and gray will die if his turn comes.

So gray must shoot white and kill him or he will certainly die. If gray gets a chance to shoot at white, there is a (2/3) chance he kills him. Then its a duel with him and black with black getting the first shot. If black misses then gray hits, it's (2/3)(2/3). Or if gray misses once, it's (2/3)(1/3)(2/3)(2/3) he wins. And so on up to infinite misses from gray. Factoring out one of the (2/3) we can express the probability gray survives if he shoots at white (remember that if gray misses, he dies) as 4/9 times the sum from n = 1 to infinity of 2^n / 3^(2n + 1). This equals 8/21 chance of gray's survival.

If gray misses white (1/3 of the time) then white has a 2/3 chance of survival (shooting at gray). So white has a 2/9 chance of survival.

So, since the probabilities of survival equal one, these are the probabilities of survival if black shoots at white and misses or if he wastes his turn:

white: (14/63)
gray: (24/63)
black: (25/63)

That's right, if black wastes his turn or misses white, his probability of survival becomes the HIGHEST of the three!

So we must consider the probabilities if black shoots at white.

The probability is P(black kills white) * P(black kills gray when gray is allowed to shoot first) + P(black misses) * 25/63

Gray's chance of missing is 1/3. So if gray misses then black kills, it's (1/3)(1/3). If gray then black misses then gray misses then black kills, it's (1/3)(2/3)(1/3)(1/3). Then the next term in the sum is (1/3)(2/3)(1/3)(2/3)(1/3)(1/3).

Factoring out 1/3 and considering the possibilities of black killing on his nth shot, taking the sum from 1 to infinity:

1/3 * sum from n = 1 to infinity of 2^(n-1) / 3^(2n - 1)

That equals 1/7.

So, black's chance of survival if he shoots at white:

(1/3)(1/7) + (25/63)(2/3) = 59/189 (.312169 repeating)
25/63 = .396825 repeating.

Thus black should deliberately miss

lol i actually drew it on paper its 45

The first line intersects 0 lines. Then, the next line intersects 1, the next intersects 2...

Every Nth line intersects (N-1) lines at unique intersection points. So for a system of N lines, we take the arithmetic sum 0 + 1 + 2 + 3 + ... + N-1, which by the formula for arithmetic series equals N(N-1)/2 total unique intersections. When N = 10, the intersections number 45. (10 * 9 / 2)

Everyone doing the Mr. White, Mr. Grey, Mr. Black thing is working way too hard. Once you realize that one of your options is to intentionally miss the first shot, you can easily see that your other two options are strictly worse without actually computing the percentages.

With math, the goal is to be as lazy as possible IMO.

Think about the situations backwards to figure out the criteria for satisfying pirates on earlier votes. Let's use the format #n for pirates, xG for their respective gold split, a simple (y/n) for if they vote yes on the proposed situation, and label pirates from captain P10 and the lowest rank pirate P1.

Starting with 2 pirates remaining: P2, 100G, y :: P1, 0G, n. 50% means this passes.

3 pirates remaining: P3, 99G, y :: P2, 0G, n :: P10, 1G, y. 66% means this passes; the logic indicates that pirate 1 cannot get a better deal at this point, so 1g is enough for pirate 3 to buy his vote to save his life; at this point, pirate 2 cannot be bribed but is mathematically irrelevant.

4 pirates remaining: P4, 99G, y :: P3 0G, n :: P2, 1G, y :: P1, 0G, n. 50% means this passes. Our incredibly logical P2 realizes that there's no way to get past the p3 situation, so can be bribed for his vote for only 1 gold.

5 pirates remaining: P5, 97G, y :: P4 0G, n :: P3, 1G, y :: P2, 0G, n :: P1 2G (or 1G depending), y. 60% means this passes. This may seem tricky, but it's the only logical way to justify the voting--and we see a pattern emerge. P5 can guarantee getting P3's and P1's votes for their respective bribes because they can't do any better in any later situations. The reason I say it might be able to be 1G or 2G for P1 is because the pirates that are getting paid out alternate, and so every iteration puts each set of pirates at risk of getting nothing--thus any situation where the get 1G is better than them getting 0G, and logically they should accept any situation where the get 1G: adding 1G to their pay out just works to clinch the deal with the deal for the pirate deciding because none of the of the other pirates would be able to do better.

We can see based on this patterning that none of the pirates will be killed, and it all comes down to the captain's distribution.

It will either look like for the guaranteed ideal satisfaction of the even pirates:
P10, 90G, y :: P9, 0G, n :: P8, 1G, y :: P7, 0G, n :: P6, 2G, y :: P5, 0G, n :: P4, 3G, y :: P3, 0G, n :: P2, 4G, y :: P1, 0G, n.

or it will just be 96,0,1,0,1,0,1,0,1,0 if we're accounting for pirates' desire to minimize risk.

Sound travels faster than nerve impulse.
http://www.braingle.com/brainteasers/19579/madadian-assassin.html

4 captives are buried in the ground. There are 3 people on one side of a solid brick wall, and one person on the other. They are all facing the brick wall. The people on the left are in a line, so they can see the people in front of them. Person 1 can see person 2 and 3. Person 2 can see person 3. But person 3 and 4 can only see the brick wall in front of them. The people who are holding the captives place a star on each of the hats on their heads. 2 red, and 2 blue. They tell all 4 captives this, but do not tell them what order the stars are in. They say that if one of the captives can guess his color star, they will let him free. But if he's wrong, they all die. The captives can not talk to each other. They can only see what's ahead of them, including the other peoples stars, but not their own. The brick wall is completely solid and tall/wide. There is nothing reflective, and they cannot turn their heads to look at the people behind them. Which one of these people know what star they have, and how do they know?

Let's rephrase the problem. Given a 3-regular graph that has a Hamiltonian circuit, we want to show it has to have at least one more.

First some definitions:

Let a T-coloring of the graph be defined as a set of three disjoint classes of edges (which we will call T-classes) such that every edge in the graph belongs to one of the three classes and also such that every vertex is connected to one edge from each of the three classes. A T-coloring shall be unordered.

Let a S-subset be the union of simple cycles (A simple cycle is a cycle with no repeated vertices) such that each cycle contains an even number of edges and such that every vertex in the graph belongs to one of the cycles. Let n(S) be the number of cycles in S. When n(S)=1, we get a Hamiltonian cycle. For each edge in N, let its coefficient in X(S) be 1 if that edge belongs in S and 0 otherwise. X(S) shall be calculated mod 2.

Now onto the proof. We will prove that the sum of X(H) for all Hamiltonian circuits in N = 0 (mod 2). If there was only one Hamiltonian circuit, then that value will obviously not be 0.

For every T-coloring, we can associate 3 different S-subsets to it. This is what I mean by associate: take two T-classes and let their union form a S-subset. The third T-class is comprises of the edges that do not belong in that S-subset. It is easy to see that X(S_1)+X(S_2)+X(S_3) for the three subsets associated with the particular T-coloring = 0 (mod 2), as each edge occurs in two of the subsets. Call this result (1).

For every S-subset of N, there are 2^(n(S)-1) different T-colorings that associate with it. That is because in each cycle of the S-subset, the two T-classes that form it will alternate, and so each cycle can be colored in two ways. Note it is 2^(n(S)-1) not 2^(n(S)) because the T-coloring is unordered.

Now, the sum of 2^(n(S)-1) * X(S) over every S of N is equivalent to summing (1) over all T-colorings which equals 0 (mod 2). 2^(n(S)-1) * X(S) is naturally 0 (mod 2) for all S for which n(S) >1. Thus it follows that sum{2^(n(S)-1) * X(S)} = 0 for when n(S) = 1, i.e. sum{X(H)} = 0. Q.E.D.

This is a classic result in graph theory.

Feel free to ask for any clarifications.

#2 knows. If 2 and three had the same color, then 1 would know and would shout out his (the opposite of what both of them have). but because 1 is silent, 2 and 3 are different. So 2 can look at three and say the opposite.

You have a jar filled with 100 black marbles and a jar filled with 100 white marbles. You can rearrange the marbles any way you like as long as all of the marbles are in jars. After sorting, you randomly select one jar and then randomly select one marble from the jar. If the marble is white you win. What are your maximum odds for winning?

You are blind folded and presented with 100 coins. You are told that 50 are heads and 50 are tails. How can you split them into two piles so that the two piles contain an equal amount of heads? You can not distinguish the orientation of the coins by touch.

You have two cups with 10 teaspoons of tea in one and 10 teaspoons of coffee in the other. You take one teaspoon from the tea cup and put it in the coffee cup, then take a teaspoon of the coffee-tea mix and put it in the tea cup. Is there more coffee in the tea cup, or tea in the coffee cup?

I think that the 2 guy knows his hat is blue. Given he knows that #3 has a red hat, but #1 hasn't said his color hat (if he saw 2 red hats he would know his hat is blue), then its safe to assume he has a blue hat. #4 seems irrelevant to the riddle at all as he cannot make any statement due to lack of information.