2 dice are rolled, and it is revealed that one of the numbers rolled was a 4. P(other number rolled was a 4)=?

Question

The above question was present on brilliant with slight change as written below. I was able to correctly solve it and get the probability=2/11. Question on brilliant: Two fair dice are rolled, and it is revealed that (at least) one of the numbers rolled was a 4. What is the probability that the other number rolled was a 6?

Note: You are not told which of the numbers rolled is a 4.

I am getting the answer to the question I have asked as 1/11 (both 4's). Is that correct? Just wanted to confirm.

Answer 1

If the two throws are independent, then the probability of a "6" on one throw is 1/6 no matter what the other throw was.

If you need a detailed analysis there are 36 results of two 6 sided dice: {(1, 1), (1, 2), (1, 3), ... (6, 4), (6, 5), (6, 6)}.

Of those, exactly 6 have a "4" in the first place: {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)} one of which has a "6" in the second place. Therefore, the probability that when rolling two dice, if one die is a "4", the other is a "6" is 1/6.

Answer 2

You already have the correct answer $1/11$ and have responded to @Riemann in your comment, giving the correct method.

I'm a great fan of simulation, especially to verify my answers to such problems, guarding against a lapse in logic. So here's my simulation in R, which is consistent with your $1/11.$

I simulated rolling two dice a million times, counting the number of 4s at each roll of a pair of dice. This gives me a vector nr.4 with a million entries (each of which may be 0, 1, or 2). The simulation gives the answer $0.0913 \pm 0.0006,$ consistent with $1/11 = 0.0909.$

set.seed(812)  # for reproducibility
nr.4 = replicate(10^6, sum(sample(1:6, 2, rep=T)==4))
table(nr.4)/10^6
nr.4
       0        1        2 
0.695074 0.277077 0.027849 

mean(nr.4[nr.4>=1] == 2)
[1] 0.09133036     # aprx P(both 4|one or more) = 1/11
1/11
[1] 0.09090909
2*sd(nr.4[nr.4>=1] == 2)/1000
[1] 0.0005761576   # aprx 95% margin of sim error for answ

Notes on R code (if interested):

• If you repeat this bit of code starting with my set.seed statement, you will get exactly my results; for your own fresh simulation, pick a different seed, or omit the statement and R will pick an unpredictable seed.

• The procedure sample simulates rolling two dice.

• The logical vector nr.4 >= 1 has a million entries (TRUEs and FALSEs).

• I look at the number of instances with nr.4==2 which also have nr.4 >= 1. The bracket notation [ ] can be read as 'such that'. The mean of a logical vector is the proportion of its TRUEs.

• The last statement allows me to get a 95% margin of simulation error. With a million iterations one can usually expect two places of accuracy.