Question

What is more likely: when throwing four dice receive a unit at least on one of them or with 24 tosses of two dice at least once get two units?

Answer 1

The probability of each digit in the throwing of a dice is 1/6. The probability of any digit except unit is 5/6. The results of throwing dice are independent. So the probability of getting figures different from unit on all 4 dice is (5/6)^4, hence the probability of getting at least one unit is p1 = 1–(5/6)^4.

When throwing two dice the probability of not getting 2 units is: (1/6)*(5/6)+(5/6)*(1/6)+(5/6)*(5/6) = 35/36. The probability of not getting 2 units in 24 tosses is (35/36)^24. The probability that 2 units will appear at least once is: p2 = 1–(35/36)^24.

P1 – P2 = (1–(5/6)^4) – (1–(35/36)^24) = (35/36)^24 – (5/6)^4 > 0 => p1 > p2. Thus, the probability of getting a unit when throwing four dice is bigger than getting 2 units with 24 tosses of two dice.