If i roll 3 dice, what is that chance that at least one of them is a number between 1 and 3?

Question

ok, i feel like a dumbass for not getting that one....but how about the same deal, but at least one of them has to be either a 1 or 2?

Answer 1

On one die, the probability of getting a number between 1 and 3 (inclusive) is 3/6 or 50%. The probability of getting a number not between 1 and 3 is also 50%.

Since this question has so many possibilities, it is simpler if you solve the inverse: find the probability that *none* of them is a number between 1 and 3.

That probability is 50% * 50% * 50% = 12.5%

So, the probability that at least one is a number between 1 and 3 is (100% - 12.5%) = 87.5%.

Answer 2

A lot of times with questions like this, it's easier to compute the chance that the event WON'T happen than that it will.

The event WON'T happen if all 3 dice show either 4, 5, or 6. For each die, the probability of that happening is 3/6, or 1/2. So the probability of it happening for all three dice is 1/2 × 1/2 × 1/2:

3/6 × 3/6 × 3/6 = 1/2 × 1/2 × 1/2 = 1/8

So the chance that AT LEAST ONE of them is a number between 1 and 3 is 1 minus the probability that NONE of them have numbers between 1 and 3:

1 - 1/8 = 7/8 = 0.875, or 87.5%.

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For your followup question, if at least one has to be a 1 or 2, then in order for it NOT to happen, all three dice would have to show 3, 4, 5, or 6. The probability of that happening is 4/6 for each die:

4/6 × 4/6 × 4/6 = 2/3 × 2/3 × 2/3 = 8/27

So the chance that AT LEAST ONE of them is a number between 1 and 2 is 1 minus the probability that NONE of them have numbers between 1 and 2:

1 - 8/27 = 19/27 = 0.703703703, or about 70.37%.

Answer 3

Okay, for one die, P{1,2,3} = 3/6 = 1/2, P{4,5,6} = 3/6 = 1/2.

Let's take your question and ask the opposite.

What is the chance none of the three is 1, 2 or 3?
That is easy:
P{4,5,6} x P{4,5,6} x P{4,5,6} = 1/2 x 1/2 x 1/2 = 1/8

Now since either one or more dice come up 1-3 OR none do, their probabilties add to 1, so the answer is 1 - 1/8 = 7/8.

Answer 4

since you already have the first one answered, the second part is the same way, but using 2/3 for "losing". So 2/3 * 2/3 *2/3 = 8/27. and 1-8/27 = 19/27.

so you have a 19/27 chance of having a 1 or 2

Answer 5

Each die has a 50/50 chance of being 3 or less. To not get at least 1 in that range, the chance is (.5) cubed, or 1 in 8. So the chance to get at least 1 die 3 or less is 7 out of 8, or 87.5%

Answer 6

chance that one die roll gives you a number between 1 and 3 is 50%. (There are six sides, you "win" if you get any one of three of them)
Of the 8 possible outcomes of a roll of three dice, (Y = got a 1-3, N = rolled 4-6), only one of them gives a role with no "winner".
So the chance is 7/8 that at least one die will have 1-3.


N N N
N N Y
N Y N
N Y Y
Y N N
Y N Y
Y Y N
Y Y Y

Answer 7

The chance that the first one is NOT between 1 and 3 (I assume you mean 1,2,or 3...in the truest sense, only the number 2 is between 1 and 3, but I'm guessing you mean 1-3 inclusive) is 3/6, or 0.5. The chance that the second one is also not 1-3 is also 0.5. Same for the third one. So the chance that all three are not 1-3 is (0.5)(0.5)(0.5), or 0.125. So the chance that at least one IS between 1 and 3 is 1-0.125, or 0.875 (87.5%)

Answer 8

Each die has a 50% chance of NOT being in that range.
So the chance of ALL dice being NOT in that range is (1/2)*(1/2)*(1/2), or (1/8)
Hence, the chance that at least one is in that range is 1 - (1/8), or (7/8).

Answer 9

there is 18 possibilities and 3 numbers on each dice so 3by3 is 9 and 9 over 18 is on half

Answer 10

0.875 or 87.5%