Question #418

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?

Expert's answer

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.

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.

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