Course Name: Quant Interview Questions Preparation, Section No: 3, Unit No: 6, Unit type: Quiz
Which of the following events is most likely?
A) At least one 6, when 6 dice are rolled
B) At least 2 sixes when 12 dice are rolled
C) At least 3 sixes when 18 dice are rolled
D) All have same probability
Can you please tell me how is the answer C?
Cause all of the dice have same probability, you can vouch for randomness tho that C has the least one…
Hi,
We are looking at this query and will update on this.
Hi Shubham
Thank you for raising this.
The answer key marked C as correct, but the mathematically accurate answer is A
Solution:
P(exactly k sixes in n rolls) = ⁿCₖ × (1/6)ᵏ × (5/6)ⁿ⁻ᵏ
P(at least k sixes) = 1 − [ P(0 sixes) + P(1 six) + … + P(k−1 sixes) ]
OPTION A : At least one 6 in 6 rolls
P(0 sixes) = ⁶C₀ × (1/6)⁰ × (5/6)⁶ = 1 × 1 × (5/6)⁶ = 0.3349
P(at least one 6) = 1 − 0.3349 = 0.6651
OPTION B : At least two 6s in 12 rolls
P(0 sixes) = ¹²C₀ × (1/6)⁰ × (5/6)¹² = 1 × 1 × (5/6)¹² = 0.1122
P(1 six) = ¹²C₁ × (1/6)¹ × (5/6)¹¹ = 12 × (1/6) × (5/6)¹¹ = 0.2692
P(fewer than 2 sixes) = 0.1122 + 0.2692 = 0.3814
P(at least two 6s) = 1 − 0.3814 = 0.6187
OPTION C: At least three 6s in 18 rolls
P(0 sixes) = ¹⁸C₀ × (1/6)⁰ × (5/6)¹⁸ = 1 × 1 × (5/6)¹⁸ = 0.0376
P(1 six) = ¹⁸C₁ × (1/6)¹ × (5/6)¹⁷ = 18 × (1/6) × (5/6)¹⁷ = 0.1353
P(2 sixes) = ¹⁸C₂ × (1/6)² × (5/6)¹⁶ = (18×17/2) × (1/36) × (5/6)¹⁶ = 153 × (1/36) × (5/6)¹⁶ = 0.2306
P(fewer than 3 sixes) = 0.0376 + 0.1353 + 0.2306 = 0.4035
P(at least three 6s) = 1 − 0.4035 = 0.5966
Hence A has the highest probability, not C and we will correct the answer key.
This question is based on the historical de Méré problem, and there’s a common misconception that as you scale up dice and thresholds proportionally, probability stays the same or increases. It actually decreases slightly each time, which is precisely the interesting lesson of this puzzle.