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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.

aditya gupta
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  • Could you show us the working for how you got ${\frac{1}{11}}$? – Riemann'sPointyNose Aug 12 '20 at 18:46
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    @Riemann ya by conditional probability. [1/6*1/6]/(1/6+1/6-1/36] – aditya gupta Aug 12 '20 at 18:53
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    This sounds just like the boy-girl problem. Have you heard of that? Have you seen any of the thousands of other posts about the boy-girl problem scattered around here and elsewhere on the internet such as wikipedia? – JMoravitz Aug 12 '20 at 19:00
  • To clarify also, you write in the title about asking if the other number was a six but in the body you write about asking if the other number was a four. You do recognize that these are two different questions with two different answers, yes? – JMoravitz Aug 12 '20 at 19:01
  • As for the answer of $\frac{1}{11}$ for asking if the other die also showed a $4$ and an answer of $\frac{2}{11}$ for asking if the other die showed a $6$ (or equivalently a $1,2,3,$ or $5$ respectively), yes that is correct given the natural interpretation of the problem as written. Oddities can occur however if you question the source of the knowledge that one of the numbers rolled was a $4$ and why this specific bit of knowledge was chosen to be shared, for instance if the person happened to choose one of the dice at random to reveal regardless of number, or if the largest gets revealed. – JMoravitz Aug 12 '20 at 19:05
  • Consider the sample space and you'll see more easily. – Andrew Chin Aug 12 '20 at 19:17

2 Answers2

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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.

BruceET
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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.

user247327
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    The declaration that "one of the dice is a 4" means it could be any of (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4). You did not correctly enumerate the possibilities given the information. – Jake Mirra Aug 12 '20 at 20:19