Probability: 10 difficult problems solved step by step

Count empty boxes using inclusion–exclusion.

1. There are 4⁸ equally likely assignments. The possible vectors of box counts are not equally likely.

2. A specified empty box leaves 3⁸ assignments. Subtract four such counts, add C(4,2)2⁸ to correct double subtraction, then subtract C(4,3)1⁸. All four boxes cannot be empty.

P = [4⁸ − C(4,1)3⁸ + C(4,2)2⁸ − C(4,3)1⁸]/4⁸
  = 40824/65536 = 5103/8192 ≈ 62.29%

Do not multiply four probabilities of nonempty boxes: those events are dependent.

Track the useful suffix, not disjoint groups of three.

1. Let w(s) be A’s chance given useful suffix s: ∅, H, HH, T or TH. Completing HHT gives 1 and THH gives 0. From HH the first subsequent T wins for A almost surely.

2. From T, H leads to TH and T stays at T. From TH, H loses and T returns to T:

w(HH)=1
w(T)=w(TH)/2+w(T)/2
w(TH)=w(T)/2
⇒ w(T)=w(TH)=0
w(H)=w(HH)/2+w(T)/2=1/2
w(∅)=w(H)/2+w(T)/2=1/4

3. A wins with probability 1/4 and B with 3/4. Indeed A can win only if the first two tosses are HH. If a T appears first, the first later HH completes THH. Equal probabilities for a fixed triple do not make this race fair.

The first vote must be A. Reflect a bad path up to its first tie.

1. There are C(12,7) orders, of which C(11,6) start with A.

2. For an order starting with A that later ties, interchange A and B through its first tie. That prefix has equal numbers of both, so the totals remain unchanged, but the order now starts with B.

3. This is a bijection: every order starting with B must reach a tie before ending two votes ahead for A, and reflection reverses the operation. Hence C(11,7) orders starting with A are bad.

C(12,7) = 792
C(11,6) − C(11,7) = 462−330 = 132
P = 132/792 = 1/6 ≈ 16.67%

Allowing ties would change the answer: strict leadership is essential.

The longest side must be shorter than the other three combined.

1. We need max Lᵢ<1/2. Equalities and coincident cuts have probability zero.

2. Ordered cuts have constant density 3!; converting cuts to lengths has determinant 1. Thus lengths are uniform on the three-dimensional simplex Lᵢ≥0, ΣLᵢ=1.

3. For L₁≥1/2, subtract 1/2 from L₁. The remaining simplex has total 1/2, so its three-dimensional volume is scaled by (1/2)³. The same holds for every piece. Two pieces cannot both exceed 1/2, and boundary intersections have probability zero.

L₁+L₂+L₃+L₄=1
P(Lᵢ ≥ 1/2) = (1/2)³ = 1/8
P(max Lᵢ < 1/2) = 1−4/8 = 1/2

The answer is 1/2. Cutting the stick in half first and then splitting each half is a different random experiment.

Find the probability of a specific order containing k white draws.

1. The denominators are 2,3,…,11. For an order with k whites, the white numerator factors are 1,…,k and the black factors 1,…,10−k, regardless of order.

2. Multiply the probability of one order by the C(10,k) possible orders:

P(WW…BB… in a fixed order) = k!(10−k)!/11!
P(K=k) = C(10,k) k!(10−k)!/11! = 1/11
P(K ≥ 8) = 3/11 ≈ 27.27%

3. K is uniform on 0,…,10, so the favourable values are 8, 9 and 10. This is not binomial: draws depend on the history. Colour symmetry does not imply independence.

Weight each possible p by the likelihood of the observation.

1. Given p, five heads have probability p⁵. Integrating against the initial uniform density gives the observation probability.

2. Five initial heads and a sixth head is exactly six initial heads. Use conditional probability:

P(HHHHH) = ∫₀¹ p⁵ dp = 1/6
P(HHHHHH) = ∫₀¹ p⁶ dp = 1/7
P(H₆ | H₁H₂H₃H₄H₅) = (1/7)/(1/6) = 6/7

3. The answer is about 85.71%. We are learning an unknown bias, not changing a known fair coin. If p were known to be 1/2, the answer would remain 1/2; redrawing p at every toss would also be a different model.

Condition on position k of the best candidate.

1. If the best appears at k>r, we select them exactly when the best among the preceding k−1 lies in the first r positions. This has probability r/(k−1), and ensures we have not stopped earlier.

2. Each k has probability 1/10. Sum from k=r+1 to 10 and put j=k−1. With r=0 we choose the first immediately.

Pᵣ = (r/10) Σⱼ₌ᵣ⁹ 1/j  (1 ≤ r ≤ 9)
P₀ = 1/10
P₃ = (3/10)(1/3+1/4+…+1/9) ≈ 0.39869048
P₄ ≈ 0.39825397

3. For r=1,…,9 the rounded probabilities are 0.282897, 0.365794, 0.398690, 0.398254, 0.372817, 0.327381, 0.265278, 0.188889, 0.100000. Reject 3. Its advantage over 4 is small but real. The optimization concerns the stated family without extra information about candidates.

Find the chance that you have not stopped after n draws.

1. N>n is equivalent to Sₙ≤1. In the n-dimensional unit cube this region is a simplex of volume 1/n!. Inductively, Vₙ(t)=tⁿ/n!, and Vₙ₊₁(1)=∫₀¹(1−u)ⁿ/n! du=1/(n+1)!, starting with n=1.

2. For a positive integer-valued N, expectation is the sum of P(N>n) from n=0: an outcome N=m contributes exactly m indicators.

Sₙ = U₁+…+Uₙ,  S₀=0
P(N>n) = P(Sₙ ≤ 1) = 1/n!
E[N] = Σₙ₌₀∞ P(N>n)
     = 1+1+1/2!+1/3!+… = e ≈ 2.71828

The mean is e, not 2: the final sum overshoots the threshold. Also P(N>n)=1/n! tends to zero, so stopping occurs almost surely.

Count sequences avoiding HHHHH rather than treating overlapping windows as independent.

1. Let aₙ count length-n strings with no five consecutive heads. For n<5, all 2ⁿ strings qualify.

2. For n≥5 every qualifying string contains T and ends in exactly one of T, TH, THH, THHH, THHHH. Removing that final block leaves a shorter qualifying string, giving the recurrence:

a₀=1, a₁=2, a₂=4, a₃=8, a₄=16
aₙ=aₙ₋₁+aₙ₋₂+aₙ₋₃+aₙ₋₄+aₙ₋₅  (n ≥ 5)
a₅=31, a₁₀=912, a₁₅=26784, a₂₀=786568
P = 1−a₂₀/2²⁰ = 32751/131072 ≈ 24.9870%

3. Subtract the avoidance probability from 1. The shortcut 1−(31/32)¹⁶ is invalid because the sixteen windows overlap. The recurrence counts each string exactly once.

Follow the cycles of the drawer-to-content permutation.

1. Participant i opens drawer i, then the drawer named by its content, and repeats. They succeed exactly when their permutation cycle has length at most 10.

2. Everyone succeeds if no cycle exceeds 10. For k>10 choose k elements, arrange their cycle in (k−1)! ways, and permute the rest in (20−k)! ways.

3. Two cycles longer than 10 cannot coexist, so the excluded events are disjoint:

P(a cycle of length k)
= C(20,k)(k−1)!(20−k)!/20! = 1/k  (k>10)
P(success) = 1−Σₖ₌₁₁²⁰ 1/k
           = 77107553/232792560 ≈ 33.1229%

Do not multiply twenty individual probabilities 1/2: this strategy makes their successes dependent. This is a twenty-participant version of the classical names-in-boxes puzzle.