bubbled

Here is my best reasoning:

There is a well known gambling hustler from a time gone by name Titanic Thompson. He was a great golfer, but lived during a time when you could make more money hustling amateurs than playing on tour. One of his classic hustles was he'd go down to Florida during the winter and play for weeks with a group of players and play right-handed (he was a very good right-handed golfer). After beating them out of a fair amount of money, he'd offer to play them double or nothing left-handed. Of course, he was a natural lefty. Here's the hustle that will form my puzzle. He once bet a group of businessmen that he could hit a golfball a mile. He went to the Grand Canyon and simply hit the golf ball off the edge. He told them that he felt bad and then offered to bet double of nothing that he could hit a ball over 500 yards horizontally. They took the bet. This is a time when given the hickory clubs of the time, it was hard to hit a ball over 200 yards. How did he win?

Jasen, you are on the right track, but you can't move the boxes. Imagine that the boxes are wooden with a hinged lid. The boxes are nailed to the table in position. You can open the boxes and look in, but you can't touch the names inside and when you've finished looking in your 50 boxes, you have to close all the boxes. Also, for Rainman, the problem is definitely achievable. The best strategy is not on the order of 1 / 2*100

The percentage bonanova is looking to get above 30% is the probability that the ENTIRE group of 100 prisoners will all find their names. Without a powerful group strategy, each prisoner will be exactly 1/2 to find their name, thus the chances the entire group will succeed would be 1 / (2^100).

Here's one where you only flip the coin once:

Based on this exacting 1000 time simulation of the best night of your high school life, you got very lucky!! BTW, this code works fine. from random import randint def he_gets_the_bottle(): if randint(0,9) == 1: return True def it_points_to_uma(): if randint(0, 9) == 1: return True def bonanova_gets_a_kiss(): if he_gets_the_bottle() and it_points_to_uma(): return True kisses = 0 for i in range(1000): if bonanova_gets_a_kiss(): kisses += 1 print kisses

Just for giggles I ran a simulation for 1,000,000 games. Here's my output and Python code:

I believe I posted a very similar problem years ago involving a roulette wheel and a ball with paint on it. It can be solved with just logic. No math is needed:

Clarification:

A general solution:

Should be:

My guess:

I think:

OK. Here's my attempt at a formal inductive proof:

Here's my answer:

It appears that there's a general solution for how many barrels/bottles (B) of wine you can test given a number of soldiers (N) and a number of testing time periods (P). B = (P + 1)^N. If the poison killed in a day and you had 7 days and 3 soldiers, you could test 512 barrels. Cool.

Beautifully done. I had never considered going with a base-3 approach. That now answers a question I couldn't figure out. Given N soldiers, why is the maximum number of barrels one can investigate in two time periods equal to exactly 3^N?

If you read the question closely, you will see you can't use smaller slices of time than 24 hours. A person is guaranteed to die within 24 hours, but when they die, is unknown. I assure you, you will need all 5 soldiers.

I think I posted this one before. But it's certainly similar to the OQ. The king has 240 barrels of wine, one of which has been poisoned. After drinking the poisoned wine, one dies within 24 hours. The king has 5 soldiers whom he is willing to sacrifice in order to determine which barrel contains the poisoned wine. How does he achieve this in 48 hours?

Here's another approach:

Well done k-man. The explanation I heard from the lecturer was very similar:

Express (M choose N) + (M choose N-1) as a single (X choose Y) in terms of M and N. The problem is solvable using algebra, but the result can be explained using logic. I'm looking for that approach.

bonanova, it certainly looks like your formula is right. I wrote some new quicker code using suggestions from CaptainEd. I then simulated 100,000 trials for each even number from 4-100 bullets. I then compare the simulation to the result predicted by your formula. The results track the formula very well:

I agree with bubbled about Jasen's statement: Hidden Content Bubbled, I agree that you have to recognize that bullets disappear, but I disagree that you have to simulate each second; merely calculate collision timestamps Hidden Content OK, I'm trying to wrap my head around you statement of calculating collision times, without simulating the the actual firing of the bullets. How would you calculate the result of this 4-bullet sequence. b_0 = v 0.9; b_1 = v 0.1; b_2 = v 0.2; b_3 = v 0.99. Unless you account for the firing of the bullets and the timing thereof, it might appear that b_3 annihilates b_2 and b_0 and b_1 survive. But taking into account the firing phase, b_2 annihilates b_1, and later b_3 annihilates b_0. I agree with bubbled about Jasen's statement: Hidden Content Bubbled, I agree that you have to recognize that bullets disappear, but I disagree that you have to simulate each second; merely calculate collision timestamps Hidden Content OK, I think I figured out what you meant. 1. I create n bullets all at once. 2. I store them in a list of the form [(n-1, V_n-1), (n-2, V_n-2),...,(0, V_0)] 3. For each pair in order starting from the n-1 bullet, I calculate time of collision. Example for n-1 bullet and n-2 bullet: t = (((n -1) - (n -2)) * V_n-2) / (V_n-1 - V_n-2). The two velocities can't be equal (division by zero) which is very unlikely. In the example, the ((n-1) - (n-2)) term is obviously 1, but in later iterations it might be higher. 4. If t < 0 ignore. 5. If t > 0, then actual collision time is t + n-1. 6. Find min of all actual collision times and remove those two bullets. Then repeat the process with remaining bullets until all bullets gone or there are no collisions left. I'd be afraid to try to optimize further. CaptainEd, do you have a way of removing all bullets will eventually collide in one pass?

I'm going with: