Recently I used this one as a first question on the final exam for students taking a first course in probability. It is definitely not a new one but I have not seen it in this forum before. Enjoy!A die is constructed in the following way: each side is painted with a number chosen at random from the set {1,2,3,4,5,6}. The sides are painted independently of each other. It is thus possible for the die to have e.g. the number 6 on all sides or three sides with the number 1 and the remaining three sides with the number 4. A person throws such a die two times. What is the probability of obtaining a 6 in both throws?

When N=6 and p=1/6 the result is ~5.09%P.

P = [sum (i=1 to 6) [(1/6)^i * (6 over i) * (i/6)]]^2 = 12,97%

Paolos got it right. Of course there are other ways of seeing this. Any more suggestions?

Let the dice land showing side X and then side YChance that both are sixes = P(X is a six) * P(Y is a six given X is a six)The chance that side X is a six is 1/6The chance that side Y is a six given that X is a six is P(X=!Y)*P(Y=6) + P(X=Y) = (5/6)*(1/6) + (1/6) = 11/216 or about 5.093

QuoteOriginally posted by: timedsLet the dice land showing side X and then side YChance that both are sixes = P(X is a six) * P(Y is a six given X is a six)The chance that side X is a six is 1/6The chance that side Y is a six given that X is a six is P(X=!Y)*P(Y=6) + P(X=Y) = (5/6)*(1/6) + (1/6) = 11/216 or about 5.093That was my original reasoning as well.

= (1/6) * SUM(n=1 to 5) {[(n+1)/6] * 5Cn * (1/6)^n * (5/6)^(5-n)}= 0.0509

Blacksheep,I dont understand this: I think there are 6^6 ways of designing such a die. Number of ways in which this die can have Just 1 six = 5^6. So the probability of such a die is 5^6/6^6.Given such a die was constructed, the probability of throwing a six is (5^6/6^6)*1/6.Similarly, Prob of constructing a die with ONLY 2 six is 5^4/6^6, Hence Prob of getting a six from such a die is 5^4/6^6*2/6.Dont we need to do something like this? But I am getting a different answer. Why is my approach wrong? QuoteOriginally posted by: BlackSheepQuoteOriginally posted by: timedsLet the dice land showing side X and then side YChance that both are sixes = P(X is a six) * P(Y is a six given X is a six)The chance that side X is a six is 1/6The chance that side Y is a six given that X is a six is P(X=!Y)*P(Y=6) + P(X=Y) = (5/6)*(1/6) + (1/6) = 11/216 or about 5.093That was my original reasoning as well.

hi no expertcould if be that the 2 events are not indept.once the first throw shows a 6, the no. of possible die configuration dramatically reduce, ie we have to rule out all configurations that does not have a 6.

noexpert,I dont think your approach is wrong but the probabilities you are using are not correct. There are indeed 6^6 ways of constructing such a die. However, the number of ways in which this die will have exactly 1 six is (6 choose 1)*5^5=6*5^5: Choose one side which will be painted with a six (=(6 choose 1)=C(6,1)=6 ways). For each of the remaining 5 sides there are 5 possibilities (=5^5 ways). Similarly a die with exactly 2 sixes can be contructed in (6 choose 2)*5^4 ways: Choose two sides which will be painted with a six (=(6 choose 2)=C(6,2)=15 ways). For each of the remaining 4 sides there are 5 possibilities (=5^4 ways).Perhaps it is easier to note that " X= number of sixes on a die " has a binomial distribution with parameters 6 and 1/6.