These 17 problems are adapted from the chapter “Quesiti, problemi e paradossi” in Giovanni Vittorio Pallottino’s Il caso e la probabilità. Le sorprese di una strana coppia, “Sfide e giochi matematici” series, Hachette, pp. 165–175. The questions are paraphrased and the solutions rewritten step by step while preserving the book’s data and results, except for one card-count error noted in problem 3. Try each question before opening its solution.
We assume a standard 52-card deck, fair dice and coins, and independent throws where stated. To express a probability as a percentage, we explicitly multiply its fraction by 100%.
1. A face card from a deck
What is the probability that a card drawn from a well-shuffled deck is a king, queen or jack?
Solution
There are three face cards in each of four suits, hence 12 favourable cards out of 52. P = 12/52 = 3/13.
2. Three specified die rolls
On three successive rolls of a die, we want 1, 2 and 3 in that order. What is the probability?
Solution
Each specified result has probability 1/6 and the rolls are independent. P = (1/6)³ = 1/216. A different order would be a different event.
3. An ace or a heart
We draw one card. What is the probability that it is an ace, a heart, or both?
Solution
There are 4 aces and 13 hearts, but the ace of hearts is counted twice. The favourable cards number 4+13−1=16, so P = 16/52 = 4/13. The photographed book solution says there are 12 hearts and gives 15/52; a standard 52-card deck has 13 hearts, so we correct that error here.
4. Two twin girls
A mother is expecting twins. What is the probability that both are girls (a) with no further information, (b) knowing at least one is a girl, (c) knowing the first-born is a girl? Assume independent, equally likely sexes.
Solution
The ordered outcomes are GG, GB, BG and BB. (a) One of four: 1/4. (b) Knowing only that at least one is a girl leaves GG, GB and BG: 1/3. (c) If the first-born is a girl, only GG and GB remain: 1/2. Part (b) means “at least one”; observing a randomly chosen twin to be a girl would be a different information model.
5. Two aces without replacement
We draw two cards in succession, without returning the first. What is the probability that both are aces?
Solution
The first ace has probability 4/52. Afterwards, 3 aces remain among 51 cards. P = (4/52) × (3/51) = 1/221.
6. Mutually exclusive events
A and B cannot occur together. Given P(A)=1/4 and P(B)=1/6, find the probability of (a) A or B, (b) neither A nor B, (c) both.
Solution
(a) The events are disjoint: P(A∪B)=1/4+1/6=5/12. (b) The complement is 1−5/12=7/12. (c) Their intersection is impossible: P(A∩B)=0.
7. Six dice
We roll six dice. What is the probability that they show (a) the same number on every die or (b) six different numbers?
Solution
(a) The first die is unrestricted; each of the other five must match it: (1/6)⁵=1/7776. (b) After the first die, the numbers of permissible faces are 5, 4, 3, 2 and 1 out of 6: (5/6)(4/6)(3/6)(2/6)(1/6)=5/324.
8. Two exact sequences
In six die rolls, compare the chance of the exact sequence 123456 with that of the exact sequence 314265.
Solution
Each specified face has probability 1/6, whether or not the sequence looks orderly. P(123456)=P(314265)=(1/6)⁶=1/46656. The set of all ordered-looking sequences would be a different event.
9. Matching cards
We shuffle two 52-card decks independently and pair cards in corresponding positions. How many identical pairs should we expect on average?
Solution
At each of the 52 positions, the chance of a match is 1/52. Linearity of expectation does not require the matches to be independent: E=52×(1/52)=1 pair. An average of one does not mean every shuffle produces exactly one.
10. Steps near a cliff
A person stands one step from a cliff. Each minute they step towards it with probability 1/3 or away with probability 2/3. What is the chance of falling within three minutes?
Solution
They may fall on the first step, probability 1/3. Otherwise the only way within three steps is away, towards, towards: (2/3)(1/3)(1/3)=2/27. The cases are disjoint, so P=1/3+2/27=11/27. Falling precisely on step two is impossible.
11. A jury with a random vote
A majority jury has two members who each vote correctly, independently, with probability p; the third votes by a fair coin toss. A single judge rules correctly with probability p. Which is more reliable?
Solution
If both informed members are right (p²), the majority is right. If exactly one is right [2p(1−p)], the coin settles the vote correctly with probability 1/2. Hence P(jury correct)=p²+2p(1−p)(1/2)=p. Under the stated assumptions, the two systems are equally reliable.
12. Why do buses arrive in pairs?
Buses leave the terminus at regular intervals but are often seen close together at a stop. What mechanism could amplify a small delay?
Solution
A delayed bus finds more passengers waiting and spends longer boarding them; the bus behind meets fewer passengers and catches up. This feedback creates bus bunching. Regular departures alone do not give a numerical probability or guarantee that it always happens: we would need a model of traffic, passenger arrivals and boarding times.
13. Is the two-dice game fair?
You lose €6 if the sum of two dice is neither 2 nor 12, but gain €100 net if it is 2 or 12. What is your average net result per game?
Solution
The 36 ordered die outcomes are equally likely; only (1,1) and (6,6) win. Thus P(win)=2/36=1/18 and P(loss)=17/18. Expected net gain is 100(1/18)−6(17/18)=−€1/9, about −€0.11 per game. It is not a fair game for the player.
14. Neighbours in a theatre row
Eight boys and seven girls occupy 15 seats in a row at random. How many adjacent pairs of seats containing one boy and one girl should we expect?
Solution
For any fixed neighbouring pair, the probability of boy–girl or girl–boy is (8/15)(7/14)+(7/15)(8/14)=8/15. There are 14 adjacent pairs. By linearity of expectation, E=14×(8/15)=112/15≈7.47 pairs.
15. A coin on squared paper
A coin 20 mm in diameter lands on a square grid with 40 mm sides. If its centre is uniformly distributed relative to the grid, what is the chance that the whole coin lies inside one square?
Solution
The centre must stay at least 10 mm from every edge. Inside a 40 mm square, the safe central square has side 40−2×10=20 mm. Comparing areas gives P=20²/40²=1/4. As a percentage: (1/4)×100%=25%.
16. Breaking a stick into three
Choose two independent uniformly distributed cut points on a stick. What is the probability that the three pieces can form a triangle?
Solution
Set the stick length to 1 and order the cuts as x<y. The pieces are x, y−x and 1−y; each must be shorter than 1/2. Thus x<1/2, y>1/2 and y<x+1/2. These inequalities enclose a small right triangle of area 1/8 inside the ordered-cut region of area 1/2. Hence P=(1/8)/(1/2)=1/4. Equalities describe degenerate triangles and have probability zero.
17. Three doors
One of three doors hides a car and the other two hide goats. You choose a door. The host, who knows the prizes, always opens another door showing a goat and offers you a switch. Should you stay or switch?
Solution
Your first door has the car with probability 1/3; staying wins only then. With probability 2/3 you initially chose a goat; the host removes the other goat, so switching wins. P(win by switching)=2/3, compared with 1/3 by staying. The rule that the host always reveals a goat is essential.
Takeaway: before dividing favourable by possible cases, ask whether the cases are equally likely, whether trials depend on one another, and whether new information changes the sample space.