When n standard 6-sided dice are rolled, the probability of obtaining

Question

When n standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of S. The smallest possible value of S is

(A) 333

(B) 335

(C) 337

(D) 339

(E) 341

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yashu5279 · Answer

Probabilities of sums of N 6-sided dice rolls are symmetrical - as an example, the probability of rolling a sum of 2 with 2 6-sided dice is the same as probability of rolling a sum of 12 with them; probability of sum 3 is the same as probability of sum 11 in the same scenario.

1994/ 6 = 332.3. ==> You would need at least 333 6-sided dice to be able to roll 1994.
With 333 dice, the minimum value you could roll is 333 x “1” = 333. The maximum value is 333 x “6” = 1998.

Since 1994 = 1998 - 4, it has the same probability of being rolled with 333 dice as “333 + 4 = 337”.

chandan1405042 · Answer

could you please explain the last line.

f2h · Answer

if the dice are standard, then the probability of 6 for each die will be the same as the other numbers( 1/6)
if we have 332 of 6 and one 4, the probability will be (1/6)^332*(1/6)=(1/6)^333
and it is the same as 333 of 1 =(1/6)^333
what's my mistake? could anyone help me?

IanStewart · Answer

The probability, when rolling 333 dice, that you get a '1' every time, or that you get a '6' every time is indeed (1/6)^333. But if you want to know the probability that you get one '4', and get a '6' on all of the 332 other dice, the probability is much larger than that, because there are many ways that can happen. If we imagine we're rolling the dice in order, we get that result if we get any of these sequences of rolls:

4, 6, 6, 6, 6, 6, 6, 6, ... 6, 6
6, 4, 6, 6, 6, 6, 6, 6, ..., 6, 6
6, 6, 4, 6, 6, 6, 6, 6, ..., 6, 6
etc

and since any of the 333 dice can display the '4', there are 333 different sequences of rolls that give us one four, and the rest sixes. Each individual sequence occurs with a probability of (1/6)^333, so the probability we get one four and the rest sixes is 333*(1/6)^333. You correctly answered a slightly different question than the one you need to answer to solve this question: you worked out the probability that you get a 4 on your first roll, then get a 6 on every other roll, but that's only one of many ways to get a single 4 and the rest 6's.

It's not the best question to illustrate the concept, because the numbers are so large. But if you think about just rolling two dice, and ask "what is the probability the sum is 2" then the answer is 1/6^2, because you need to get '1' both times, and that can only happen in one way. But if you ask "what is the probability the sum is 3?" the answer is larger (you can't just find the probability of getting a 1 then a 2), because there is more than one way to get a sum of 3: we can roll a 1 then a 2, or we can roll a 2 then a 1, and the answer is 2/36 = 1/18.

It's not really relevant to your question, but the original problem asks about a sum of 1994, which we actually don't get with 332 sixes and one 4 (that would give a sum of 1996), and we can get a sum of 1994 in quite a few ways. yashu's solution using symmetry is the fastest way to solve the problem.

Kinshook · Answer

When n standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of S.

The smallest possible value of S is

1994 = 6*332 + 2

1998 = 6*333
333 = 1*333

Probability of getting 1994(1998-4) is same as probability of getting (333+4=337)

IMO C

When n standard 6-sided dice are rolled, the probability of obtaining

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When n standard 6-sided dice are rolled, the probability of obtaining

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