Probability of rolling at least one snake eyes (pair of two ones) with four dice, rolled 3 times

Question

Rolling a pair of dice has a 1/36 chance of getting a pair of ones. But if I have four dice, at first I assumed it will still be 1/36 since the other two dice won't matter, but then the other two dice both have the chance to get a 1, so my possible results should be [1,1,x,x],[1,x,x,1],[x,x,1,1],[x,1,1,x],[1,x,1,x],[x,1,x,1],[1,1,1,x],[1,1,x,1],[1,x,1,1],[x,1,1,1],[1,1,1,1], so its 11/1296? but why has it become smaller than 1/36? I'm doing something wrong? Also how would I compute it for 3 rolls

Sorry, if there is an obvious answer for this, but its been years since I last took basic probability

Answer 1

The eleven options listed are not equally likely. There are $5^2=25$ ways to roll $[1,1,x,x]$, but only one way to roll $[1,1,1,1]$. Hence, what you need is $$\frac{25\cdot 6+5\cdot4+1}{6^4}=\frac{171}{1296}\approx 0.132$$ To deal with three rolls, the simplest way to compute it is to compute the probability that you FAIL to roll snake-eyes (three times), then subtract this from $1$. Your answer is $$1-\left(\frac{1125}{1296}\right)^3=\frac{1032859}{2985984}\approx 0.346$$

Answer 2

Your $x$s can have any one of five values, so for example $[1,1,x,x]$ is really $25$ options, not $1$. The total number of successful rolls of four dice is $$6\times25+4\times5+1=171$$ and the probability is $$\frac{171}{1296}\ .$$ The chance of failing on one roll of four dice is $$1-\frac{171}{1296}=\frac{1125}{1296}\ ,$$ the chance of failing on three rolls is $$\Bigl(\frac{1125}{1296}\Bigr)^3$$ and the chance of succeeding at least once in four rolls is $$1-\Bigl(\frac{1125}{1296}\Bigr)^3=0.3459\ .$$

Answer 3

You have 2 pairs of dice (4 dies) and you roll the dice once (1) and 3 times (2) and are looking for snake eyes (two 1’s). What is the probability (P) of rolling snake eyes exclusively (3) and inclusively (4). Note that there will be 4 answers at the end.

Answer: (can this be explained step by step and not just plugged in to an equation?) First of all the rolling of four dies can be modeled as xxxx where each x is a base 6 number system (i.e. they are numbered 1 to 6) and four x’s represent four columns. The possible combinations are 6^4 or 1296 possibilities (To find the possibilities you take base^(# of columns)).

P(no 1’s) + P(one 1) + P(two 1’s) + P(three 1’s) + P(four 1’s) = 1 ; (All the possibilities mean P = 1)

P(no 1’s) = (cases without 1's)/(# of possibilities)=(5/6)^4= 625/1296 ; The P of not rolling a 1 on a die is 5/6, two dies is (5/6)*(5/6), four dies (5/6)^4. P(four 1’s) = 1/1296 (“1111” only occurs once). We eliminated the cases without 1’s but we still have the same amount of 1’s in our system which is 1296/6(in the leftmost column(6^3 column)) and 1296/6 in the next column (6^2 column) and 1296/6 in the third column (6^1 column) and 1296/6 in the last column (6^0 column). Each column treats numbers in a number system equally. So we have (1296/6)4 = 864 1’s total. We have only identified P(four 1’s) so far which had 1 case which means it used up 4 1’s and we have 860 left.

P(one 1) = (4C1)*(5^3) = (4!/1!3!)(125) = 4(125) = 500 (there are 500 cases with only one 1). For the case where you have “one 1” there are (4C1) combinations that can occur. (4C1) = 4 which are 1xxx, x1xx, xx1x, and xxx1. Since all columns have the same quantity of 1’s and x cannot be a “1” there are 3 columns of x’s with 5 choices (2,3,4,5,6) (you cannot use 1) so we multiply 4 by 5^3.

P(two 1’s) = (4C2)*(5^2) = (4!/2!2!)(25) = 6(25) = 150 (there are 150 cases with only two 1’s). For the case where you have “two 1’s” there are (4C2) combinations that can occur. (4C2) = 6 which are 11xx, x11x, xx11, 1x1x, 1xx1, and x1x1. Since all columns have the same quantity of 1’s and x cannot be a “1” there are 2 columns of x’s with 5 choices (2,3,4,5,6) (you cannot use 1) so we multiply 6 by 5^2.

P(three 1’s) = (4C3)*(5^1) = (4!/3!1!)(25) = 4(5) = 20 (there are 20 cases with only three 1’s). For the case where you have “three 1’s” there are (4C3) combinations that can occur. (4C3) = 4 which are 111x, x111, 1x11, and 11x1. Since all columns have the same quantity of 1’s and x cannot be a “1” there is 1 column of x’s with 5 choices (2,3,4,5,6) (you cannot use 1) so we multiply 6 by 5^1.

P(no 1’s) + P(one 1) + P(two 1’s) + P(three 1’s) + P(four 1’s) = 1 ; (All the possibilities mean P = 1) 625/1296 +500/1296 +150/1296 +20/1296 +1/1296 =1 (works!) 500(1) + 150(2) +20(3) +1(4) = 864 (the number of 1’s) (works!)

1. Answer: P(two 1’s) = 150/1296 = 11.57%
2. Answer: P(two 1’s) + P(three 1’s) + P(four 1’s) = 171/1296 = 13.19% If we now roll the 4 dies 3 times what is the probability of getting snake eyes? We are not looking for the probability of both to occur (P(A)^3) but for at least one to occur which is the complement of nothing occurring (1 – (p(A)’)^3)
3. Answer: (1 – (1 - 150/1296)^3) = 0.3086 = 30.86%
4. Answer (1 – (1 - 171/1296)^3) = 0.3459 = 34.59%