Logic Puzzles

Don't try to calculate the probability that two people share a birthday directly. Instead, calculate the probability that no two people share a birthday.

For each new person who enters the room, we are trying to find the odds that their birthday doesn't match anyone already there.

Just 23 people.

With 23 people in the room, the probability that at least two share a birthday is roughly 50.7%. By 70 people, it climbs above 99.9%.

This feels fundamentally wrong because we instinctively compare against our own birthday and assume there is a 1 in 365 chance. Instead, we should be comparing every possible pair of people in the room. With 23 people there are 253 unique pairs, and each one is an independent chance for a match.

I wrote a Python script that verifies this analytically and through simulation.

The math (roughly) is as follows:

Work out the probability that no two people share a birthday, then subtract from 1.

Each new person entering the room must have a birthday that avoids everyone already there:

• Person 1: 365/365 (no constraint)
• Person 2: 364/365 (must miss person 1's birthday)
• Person 3: 363/365 (must miss persons 1 and 2)
• Person n: (365 - n + 1) / 365

Multiply these together:

P(no match) = (365 × 364 × 363 × ... × (365 - n + 1)) / 365^n

For n = 23 this comes out to roughly 0.4927, so:

P(at least one match) = 1 - 0.4927 = 0.5073

If the dish is full at 60 minutes and the bacteria double every minute, it was half full exactly one minute before that. Seems obvious now, right?

Start by figuring out who the brakeman is. Work through each name and see which ones can be eliminated.

The salary clue is key: $20,000 divided by 3 is not a whole number. What does that tell you about Mr. Jones's relationship to the brakeman?

1. Smith beats the fireman at billiards, so Smith cannot be the fireman. He must be either the brakeman or the engineer.
2. The brakeman lives halfway between Detroit and Chicago. He also has a neighbor who earns three times as much as he does.
3. The neighbor cannot be Mr. Robinson, because Mr. Robinson lives in Detroit. The neighbor also cannot be Mr. Jones, because Mr. Jones earns $20,000 and $20,000 is not divisible by 3. So the brakeman's neighbor must be Mr. Smith.
4. With Mr. Smith established as the brakeman's neighbor (living near the halfway point), and Mr. Robinson in Detroit, the passenger in Chicago must be Mr. Jones. Clue 6 says the Chicago passenger shares a last name with the brakeman, so the brakeman's last name is Jones.
5. Smith is not the fireman (step 1) and not the brakeman (that's Jones). Smith must be the engineer.
6. Robinson is the only name left, so Robinson is the fireman.

The brakeman is Jones, the fireman is Robinson, and the engineer is Smith.

Use the inclusion-exclusion principle. When you add up all three groups, students who study two languages get counted twice, and students who study all three get counted three times. You need to subtract and add back accordingly.

The club has 39 members total. 14 study exactly one language.

Total = French + German + Spanish - (French & Spanish) - (French & German) - (Spanish & German) + (all three) Total = 21 + 20 + 26 - 12 - 10 - 9 + 3 = 39

• Only French: 21 - 12 - 10 + 3 = 2
• Only German: 20 - 10 - 9 + 3 = 4
• Only Spanish: 26 - 12 - 9 + 3 = 8

2 + 4 + 8 = 14

Think about what any islander, knight or knave, would say if asked whether they are a knight or a knave. Does that tell you anything about what the first person must have said? If you are still struggling, consider building out a truth table. Another hint: Assume that every islander will want you to think they are telling the truth.

The first and second are knights. The third is the knave.

Any islander, asked whether they are a knight or a knave, will claim to be a knight. A knight says so truthfully; a knave lies and says it anyway. So whatever the first person said, they definitely claimed to be a knight. The second person described that correctly, which means the second person told the truth and is therefore a knight.

The third person says the second is lying. But the second is a knight (as established), so the third person is wrong. That makes the third person the knave.

Work by assumption. Pick a suspect, assume they are guilty, and check how many statements come out true. If it matches the constraint, you have your answer.

Part 1 (only one statement is true): Gus is guilty.

If Gus did it: Archie is wrong (false), Dave is wrong (false), Gus's claim that he is innocent is wrong (false), and Tony is right that Dave lied (true). That gives exactly one true statement.

Part 2 (only one statement is false): Dave is guilty.

If Dave did it: Archie is right (true), Dave's claim that Tony did it is wrong (false), Gus is right that he didn't do it (true), and Tony is right that Dave lied (true). That gives exactly one false statement.

Assume for a moment that your forehead is clean. What would the other two travelers see? What would each of them be able to deduce from that?

The traveler reasoned as follows:

Suppose my forehead is clean. Then each of the other two sees one smudged face (the third person) and one clean face (mine). Both still raise their hands, because they each see at least one smudge. But now each of them should be able to reason: "The person I see with a smudge raised their hand because they see a smudge somewhere. The only smudge they could be seeing is mine. So I must be smudged." Each would immediately know and drop their hand.

But neither of them dropped their hand right away. That means the assumption is wrong: my forehead cannot be clean. It must be smudged.

The first traveler to complete this chain of reasoning dropped their hand.

The prisoner's argument relies on eliminating days one at a time from the end of the week. But each elimination step assumes the prisoner has already successfully ruled out all later days. Is that assumption actually solid?

The prisoner's reasoning is flawed, though pinpointing exactly why is surprisingly tricky and has been debated by logicians for decades.

The key problem is that the backwards elimination chain is not as airtight as it seems. The Saturday argument is however valid: if it's Friday night and no hanging has occurred, Saturday is no longer a surprise. But the Friday argument depends on Saturday already being ruled out. That elimination only holds if the prisoner can be certain Saturday will not be used, and that certainty requires the judge's statement to be known to be true in advance.

The judge's statement is a prediction about a future event. The prisoner can know it is true only after the fact, not before. Each step of the backwards induction works from the assumption that the prisoner has already proven the judge cannot use the remaining days, but that proof depends on steps that haven't been established yet. So the argument is circular.

In practice, the judge can carry out the sentence, and the prisoner will be surprised because the prisoner's logic collapses under its own self-referential weight.

A direct question like "which door leads to freedom?" will not work, because you cannot trust the answer without knowing who you are talking to.

Think about how you might route the question through both guards at once, so that their opposing tendencies cancel each other out.

Ask either guard: "If I asked the other guard which door leads to freedom, what would they say?" Then take the opposite door.

It works regardless of which guard you ask because:

• If you ask the truth-teller, they honestly report what the liar would say. The liar would point to the wrong door, so the truth-teller tells you: the wrong door.
• If you ask the liar, they lie about what the truth-teller would say. The truth-teller would point to the right door, so the liar tells you: the wrong door.

Both guards point to the wrong door, so you can safely go through the other one.

Alternative formulations that work by the same double-negation logic:

• "Which door would the other guard say leads to the hangman?" Then go through the door they point to (no inversion needed this time).
• "If I asked you which door leads to freedom, what would you say?" Then go through the door they indicate. The liar is forced to lie about their own lie, which cancels out and lands them on the truth.
• "If you were the other guard, which door would you point to?" Then take the opposite door.

Work out separately which days each creature could make the statement "Yesterday I was lying" without contradicting their own schedule. Then find the day that appears on both lists.

Remember: the statement can be consistent two ways. Either the creature is telling the truth (in which case yesterday really was a lying day), or the creature is lying (in which case yesterday was actually a truth-telling day).

It is Thursday.

Work out which days each creature can plausibly say "Yesterday I was lying":

The lion (lies Mon/Tue/Wed, tells the truth on Thu/Fri/Sat/Sun)

• If he was telling the truth today, yesterday must have been a lying day (Mon/Tue/Wed), making today Tue/Wed/Thu. The lion tells the truth only on Thu from that set. Thursday is a candidate.
• If he is lying today, yesterday must have been a truth day (Thu/Fri/Sat/Sun), making today Fri/Sat/Sun/Mon. The lion lies only on Mon from that set. Monday is a candidate.

The lion can say this consistently on **Thursday or Monday.

The Unicorn (lies Thu/Fri/Sat, tells the truth on Sun/Mon/Tue/Wed):

• If he is telling the truth today, yesterday must have been a lying day (Thu/Fri/Sat), making today Fri/Sat/Sun. The unicorn tells truth only on Sun from that set. Sunday is a candidate.
• If he is lying today, yesterday must have been a truth day (Sun/Mon/Tue/Wed), making today Mon/Tue/Wed/Thu. The unicorn lies only on Thu from that set. Thursday is a candidate.

The unicorn can say this consistently on Sunday or Thursday.

The only day that works for both is Thursday. On Thursday, the lion is telling the truth (Wednesday was a lying day), and the unicorn is lying (Wednesday was actually a truth-telling day for the unicorn).

Think about which box gives you the most information from a single fruit. One of the three labels, by being wrong, tells you something very specific about what that box cannot contain.

Pick a fruit from the box labeled "Apples and Oranges."

Since every label is wrong, this box cannot contain both. It holds either only apples or only oranges. One fruit tells you which.

From there, the other two boxes resolve by elimination:

• If you draw an apple: that box holds only apples. The box labeled "Apples" can't hold apples (wrong label) and can't hold the mixed contents (those belong to your identified box), so it must hold only oranges. That leaves the box labeled "Oranges" as the mixed box.
• If you draw an orange: that box holds only oranges. The box labeled "Oranges" can't hold oranges and can't hold the mixed contents, so it must hold only apples. The box labeled "Apples" is the mixed box.

One fruit is enough because the "Apples and Oranges" box is the only one whose wrong label constrains it to a single fruit type, and once it is identified, the remaining two boxes have only one valid assignment.

Unpack the phrase from the inside out, substituting "today + N days" at each step.

It is Saturday.

Working from the inside out, letting today = T:

1. "tomorrow" = T + 1
2. "the day before tomorrow" = T + 1 - 1 = T
3. "two days after [that]" = T + 2
4. "the day before [that]" = T + 2 - 1 = T + 1

So T + 1 = Sunday, which means today is Saturday.

The farmer does not have to move items in one direction only. Taking something back across is allowed, and may be necessary.

One solution (there are two):

1. Take the chicken across. Leave it on the far bank.
2. Return alone.
3. Take the fox across.
4. Bring the chicken back.
5. Take the grain across.
6. Return alone.
7. Take the chicken across.

Taking the chicken back may feel like moving backwards, but it is the only way that you can keep the fox and grain separated while you move the grain across.

The second solution swaps steps 3 and 5. Take the grain first, then the fox. It is identical otherwise.

He cannot tell the pills apart by sight, but he does not need to. Think about whether there is a way to guarantee the right dose without identifying any individual pill.

He breaks each of the four pills in half and takes one half of every pill.

The four halves he swallows add up to exactly one full blue pill and one full red pill, regardless of which half came from which pill. The other four halves are discarded.

List every combination of three whole numbers whose product is 36, then calculate the sum for each. The fact that you knew the house number but still could not determine the ages is itself a clue. What does that tell you about the house number?

The daughters are 2, 2, and 9.

Start by listing every combination of three positive integers with a product of 36, along with their sums:

Every sum is unique except 13, which appears twice: (1, 6, 6) and (2, 2, 9). Since you knew the house number but still could not determine the ages, the house number must be 13. Any other number would have pointed to a unique answer.

That leaves two candidates. The final clue resolves it: the only part that we care about is that the man has an eldest daughter, meaning one daughter is strictly older than the others. (1, 6, 6) has two daughters tied for oldest, so there is no single eldest. (2, 2, 9) has a clear eldest.

The daughters are therefore 2, 2, and 9.

You do not need to identify a single coin correctly. Think about what happens algebraically when you separate a group of 5 and flip every coin in it, regardless of which specific coins ended up in that group.

Pick any 5 coins, put them in a separate group, and flip all 5.

Say you happened to grab k heads. After flipping, your group has (5 - k) heads. The group left on the table also has (5 - k) heads. Both match no matter the value of k.

Try it out!

The last prisoner in line can see all 99 hats in front of them. They cannot save themselves with certainty, but they can encode something in their answer that gives every other prisoner the information they need. What single property of a set of hats can be conveyed in one word?

99 are guaranteed to survive. The last prisoner is a coin flip.

The trick is that before lining up, everyone agrees that the last prisoner will count the red hats he/she can see. If the count is odd, they say "red." If it's even, they say "black." Their own hat is just a guess, they are using their turn to pass a message.

Every prisoner afterword knows whether the red hats ahead add up to an odd or even number. So each prisoner must simply count the red hats they can see in front, then add this number to the number of "red" answers they have already heard. If the total matches the signal, their own hat is black. If it doesn't, their hat is red. One of those two is always right.

The last prisoner can't use the trick themselves because there is nobody behind them to start it.

Start with what you can pin down absolutely. Clues 8 and 9 fix two positions immediately. Clue 4 restricts where the green and white houses can go. Build a grid with houses 1 through 5 across the top and attributes down the side, then fill in what you can derive from each clue, revisiting earlier clues as new cells are filled.

The German owns the fish.

Working through the grid:

• From clue 9: Norwegian is in house 1.
• From clue 14: house 2 is blue.
• From clue 8: house 3 has milk.
• From clue 4: green/white must be houses 4/5 (green immediately left of white).
• From clue 5: green house (4) drinks coffee.
• Houses 1 and 3 are neither green, white, nor blue. House 3 is not red (Brit is red, and we can rule that out shortly). That leaves house 1 as yellow, house 3 as red.
• From clue 1: Brit is in house 3.
• From clue 7: Norwegian (house 1) smokes Dunhill.
• From clue 11: horse owner is next to Dunhill smoker (house 1), so horse is in house 2.
• From clue 3: Dane drinks tea. Tea is not in house 3 (milk) or 4 (coffee), and house 1 is Norwegian. Dane is in house 2 or 5. House 5 has white house. Checking remaining beverages: house 1 gets water or beer. From clue 15: Blends smoker is next to water drinker. From clue 12: BlueMaster smoker drinks beer. Working through: Dane is in house 2 (drinks tea).
• From clue 13: German smokes Prince. German is not in house 1 (Norwegian), 2 (Dane), or 3 (Brit). German is in house 4 or 5.
• Remaining nationality is Swede, in house 4 or 5. From clue 2: Swede keeps dogs.
• From clue 6: Pall Mall smoker keeps birds. From clue 12: BlueMaster smoker drinks beer. Beer is in house 1 or 5. House 1 drinks water or beer; from clue 15 the Blends smoker neighbors the water drinker. If house 1 drinks beer, the BlueMaster is in house 1, but house 1 smokes Dunhill. Contradiction. So house 1 drinks water and house 5 drinks beer.
• From clue 12: house 5 smokes BlueMaster. From clue 13: German smokes Prince, not BlueMaster, so German is in house 4. Swede is in house 5.
• From clue 15: Blends smoker neighbors water drinker (house 1), so Blends is in house 2.
• From clue 10: Blends smoker (house 2) neighbors cat owner, so cats are in house 1 or 3.
• From clue 6: Pall Mall smoker keeps birds. Remaining smokes for houses 3 and 4: Pall Mall and Prince. German (house 4) smokes Prince, so house 3 smokes Pall Mall and keeps birds.
• From clue 10: cats in house 1 or 3. House 3 has birds, so cats are in house 1.
• Swede (house 5) keeps dogs. House 4 (German) has the fish.