I hope you all had a nice time yesterday celebrating any fathers in your lives. I can trace my love of puzzles back to my dad, who was quizzing me about Zeno’s paradox and the Fibonacci sequence before I was potty trained. This was us nearly every night of my childhood:
To honor our pops, I’ve posed three kinship puzzles below. The first one is an old chestnut that caused a hearty debate between my dad and me. I’d prefer not to share which one of us had it right.
Did you miss last week’s puzzle? Check it out here, and find its solution at the bottom of today’s article. Be careful not to read too far ahead if you haven’t solved last week’s yet!
Puzzle #47: All in the Family
1. A man was looking at a photograph and said, “Brothers and sisters I have none, but that man’s father is my father’s son.” Who is in the photograph?
2. A girl has an equal number of brothers and sisters. But each of her brothers has only half as many brothers as sisters. How many kids are in the family?
3. A family picnic had the following people in attendance:
1 grandfather, 1 grandmother, 2 fathers, 2 mothers, 4 children, 1 brother, 2 sisters, 2 sons, 2 daughters, 3 grandchildren, 1 father-in-law, 1 mother-in-law, and 1 daughter-in-law, but there were only 7 people present. How is this possible?
Clarifications: if somebody is counting for a certain relation, then the person with whom they have that relation must also be present. In other words, you can’t say a picnic with one person present contains a father, son, grandfather, and brother, just because that person is somebody’s father and somebody’s son, etc. But you can say that a father, his daughter, and his son constitute one father, one son, one daughter, one brother and one sister, because all of those relationships are in attendance.
I’ll be back Monday with the answers and a new puzzle. Do you know a cool puzzle that you think should be featured here? Message me on X @JackPMurtagh or email me at gizmodopuzzle@gmail.com
Solution to Puzzle #46: Coming Up Aces
Last week’s puzzles asked simple-sounding questions with highly counterintuitive answers.
Shuffle a normal face-down deck of 52 playing cards and then flip one card at a time face up.
Which card is more likely to come immediately after the first ace appears: the king of spades or the ace of spades? In other words, you’ll flip over cards until you see an ace of any suit. Is the next card more likely to be the king of spades or the ace of spades, or do they have equal likelihood?
The king of spades and ace of spades are equally likely to follow the first ace. Many people suspect that the king would be more likely because, after the first ace, only three more aces remain, whereas four kings might remain. Shout-out to Eugenius for cracking this one even when all of the other comments guessed that the king would be more common.
Here’s a good way to think about it: remove the ace of spades from a deck and shuffle the remaining 51 cards. If we were to reinsert the ace of spades, there are 52 available positions from the very top of the deck to the very bottom, but only one of those 52 would result in the ace of spades residing precisely after the first ace in the deck. The exact same reasoning applies to the king of spades. Only one of the 52 available positions would put the king of spades just after the first ace. Both cards have a 1/52 chance of following the first ace.
If you’d like to build up more intuition, imagine a three-card deck with an ace of spades (As), king of spades (Ks), and ace of clubs (Ac). There are six ways to arrange these cards:
- As Ac Ks
- As Ks Ac
- Ks As Ac
- Ks Ac As
- Ac Ks As
- Ac As Ks
Arrangements 4 and 6 show the ace of spades immediately after the first ace (the ace of clubs), while arrangements 2 and 5 show the king of spades immediately following the first ace, so both have a ⅓ chance of occurring.
Reshuffle the same deck and start flipping again. This time, before flipping, you must guess when the first black ace will appear. Which position in the deck is most likely, or are they all the same?
The first black ace is most likely to appear at the very top of the deck. Shout-out to mischlep for providing a rigorous analysis of the problem.The probability that the first black ace is at the top of the deck is 2/52, because there is a 1/52 chance that the ace of spades ends up there plus the 1/52 chance that the ace of clubs ends up there. For every other position, the probability that some black ace ends up there is also 2/52, but the probability that it is the first black ace begins to shrink, because you have to factor in the possibility that a black ace has already occurred, whereas any black ace at the top of the deck is guaranteed to be the first one.
To see how this operates in an actual calculation, the probability that the first black ace occurs in the second position of the deck equals the probability that some black ace appears in the second position times the probability that the other black ace did not already appear in the first position. This works out to 2/52 (probability that some black ace appears in second position) times 50/51 (once one black ace is in the second position, the other black ace has 51 possible remaining spots and 50 of them are not the top of the deck). Since 50/51 is less than 1, this probability is smaller than the probability that the first black ace is on top. These numbers continue to shrink as you go down the deck all the way to a zero percent chance that the first black ace occurs on the bottom.
Interestingly, the same argument runs backwards, so the second black ace is most likely to be the last card in the deck.