Three Good Puzzles

The word riddle and puzzle are generally used interchangeably but have slight differences. Riddles have more of a verbal history and connotation. They often intentionally phrase themselves in such a way to have a double meaning and attempt to mislead the victim, er, recipient.

I dislike the majority of riddles because they rarely have applications beyond themselves. Here are some examples:

What is in seasons, seconds, centuries and minutes but not in decades, years or days?
The letter ‘n’.

What has a head, a tail, is brown, and has no legs?
A penny.

What can travel around the world while staying in a corner?
A stamp

They’re fun little things but have limited applicability. The knowledge rarely transfers and you quickly forget about each one. After getting each answer you understand why it’s correct, but it’s not like you now know how to travel the world while staying in a corner.

Puzzles on the other hand are usually a bit more fair. They are generally solved by logical reasoning where each piece, when put together, leads to final answer. They test your ingenuity and knowledge and often grant some sort of insight into the world, and the solutions can be generalized and applied to various disparate problems. Here are three of some of my favorite puzzles. I’ll post some more in the future. You can highlight the area in black following each riddle to reveal the answer.

  1. Suppose you are shooting free throws and each shot has a 60% chance of going in (there is no “learning” effect and “depreciation” effect, all have the some probability no matter how many shots you take). Now there are three scenarios where you can win $1000.
    A: Make at least 2 out of 3.
    B: Make at least 4 out of 6.
    C: Make at least 20 out of 30.Which do you choose?

    You would choose “A: Make at least 2 out of 3.” Indeed, all the probabilities reduce to the same amount of 2/3, so why should this be the case? The answer is linked to the Law of large numbers. Try to imagine that instead of shooting 3 free throws that you instead shoot 1,000,000. Since each one has a 60% chance of going in, you’ll probably get around 600,000 in. However, you need to get 666,667 in (2/3). After 1,000,000 shots, each with a 60% chance of going in, it is extremely unlikely that you’d get an additional 66,667 above the average.

    As a more extreme example, can you imagine flipping a coin 1,000,000 times and getting heads 750,000 times? No, after a large enough amount of trials the variance will be minimal and the expected and actual amount of successes will be quite close. However, with a small amount of trials, you can just get plain lucky and beat the odds. There will be high variance with a small number of trials, and this gives you the greatest chance of winning. So where else would this apply? Well, if you’re gambling and the house always wins in the long term, the optimal strategy is to “go big or go home” and minimize the number of bets, because in the short-term variance will be much greater and you can beat the expected long-term returns.
  2. Linda is 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations. Which is more probable?A: Linda is a bank teller.
    B: Linda is a bank teller and is active in the feminist movement.

    The answer is A, although most people choose B. This is known as the Conjunction fallacy. The probability of two events occurring together is always going to be less than (or equal to) the probability of either one occurring alone. Let’s say the probability of Linda being a bank teller is 5% and the probability of her being a feminist is 99%. The probability of A would just be 5% and the probability of B would be .99 * .5 = .495, or 4.95%, still less than 5%. The linked Wikipedia article has additional examples and ways you can debias yourself in these sorts of situations.
  3. A king has 1,000 barrels of wine stored in his cellar. Unfortunately the guards found an assassin that had snuck in last night and poisoned exactly one of the barrels. Even 1 drop of poisoned wine would be enough to kill a man, but the poison takes 30 days before becoming fatal, plus or minus a few hours. The poison is colorless, odorless, and victims show exactly 0 signs of being poisoned in the 30 days prior to dying. There is absolutely no way to detect the poison.In 31 days time, there will be a grand feast in which the king will need as much of the wine as possible. He has 10 prisoners on which to test the wine. His first idea is to give each prisoner a drop from 100 barrels of wine, and then when 1 prisoner dies, throw out all 100 barrels of which the dead prisoner drank from. However his jester speaks up and explains a solution that will pinpoint exactly which barrel is poisoned using only the 10 prisoners.What was the jester’s solution?

    The easiest way to understand this answer is through binary. If you’re not already familiar, it’s rather quick to learn and all you need to know is how to count in it, so take some time for that first and continue here afterwards.
    The answer is to assign each prisoner a position in a 10 digit binary number. So, in the binary number 00000 00000, the first 0 refers to prisoner 1, the second 0 to prisoner 2, the third 0 to prisoner 3, etc. Next, you number each of the 1000 barrels, and convert each barrel to its binary equivalent. For example, barrel #5 would be 00000 00101, because 5 in binary is that number. So for this barrel you have prisoners #8 and #10 drink. You do this for each of the 1000 barrels and at the end of the 30 days, depending on the combination of dead prisoners you will know exactly which barrel is poisoned. For example, if the poisoned barrel was #5, then only prisoners #8 and #10 would be dead.
    The reason why this is one of my favorite riddles is because of how clean it is. It works because there’s exactly 10 prisoners and exactly 1,000 barrels. Because 2^10 is 1024, if there were 1025 barrels then you would not have enough prisoners to be sure. The numbers 10 and 1,000 just look like clean numbers that were chosen somewhat arbitrarily but are actually quite fine-tuned. Also, this combinatorial solution can be applied to other problems and represents an extremely dense way of conveying information.
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