saw elsewhere the following two interesting problems:(1) two loaded dices: with 1, 2, ..., 6 coming up with various probability (could be different).is it possible that when we roll them both, the sum 2, 3,..., 12 comes up with the same probability?i.e., P(sum=S)=const., , S = 2, 3, ..., 12?(2) two fair dices: each face has a positive integer (could be same) on it.someone rolls these two dices and tells you that the sum follows the exact distribution as two normal fair dices,i.e., P(sum=S)=(6-|S-7|)/36, S = 2, 3, ..., 12.are these two dices just two normal fair dices?

(1) is impossible:Let p_i, q_i be the probabilities of the first/second die showing i. We need P(sum=2) = p_1 q_1 = p_6 q_6 = P(sum=12). We can assume that q_6 >= q_1 (if q_6 < q_1 and p_6 < p_1, we have p_6 q_6 < p_1 q_1).P(sum=7) >= p_1 q_6 + p_6 q_1 >= p_1 q_1 + p_6 q_1 = P(sum=2) + p_6 q_1. Hence p_6 q_1 = 0 and P(sum=2) = P(sum=12) = 0.

"two loaded dices" are too many!

1) two loaded dices: with 1, 2, ..., 6 coming up with various probability (could be different).is it possible that when we roll them both, the sum 2, 3,..., 12 comes up with the same probability?i.e., P(sum=S)=const., , S = 2, 3, ..., 12?If two loaded dices are the sameFor symmetry, we know P1=P6, P2=P5, P3=P4 P1+P2+P3 should be equal to 0.5P(sum=2)=P1^2=1/11 => P1=sqrt(1/11)P(sum=3)=2*P1*P2=1/11 => P2=1/2*sqrt(1/11)P(sum=4)=2*P1*P3+P2*P2=1/11 => P3=3/8*sqrt(1/11)P1+P2+P3=15/8*sqrt(1/11) is not equal to 0.5 : ContradictionIf two loaded dices are not the same, We have 12 variables for probability. Then there have 11+2(for totalprobability to equal to 1) equations.I would not try but the answer might be still impossible. (2) two fair dices: each face has a positive integer (could be same) on it.someone rolls these two dices and tells you that the sum follows the exact distribution as two normal fair dices, i.e., P(sum=S)=(6-|S-7|)/36, S = 2, 3, ..., 12.are these two dices just two normal fair dicesTo have S=2, 12 , We know 1 and 6 must present in two dicesWe can have 1-11 integer on each integer, but because sum(P(sum=S)| S=2,3,....,12)=1,we have to accept we only have 1-6 on the faces.1) If the two dices are the sameGenerally, we have only 6 equations for the distribution of each integer,appearantly, the answer is yes.2) If the two dices are not the sameGenerally, we have 12 variable for the distribution of each integerBut remember we still have 11+2(for totalprobability to equal to zero) equations. However,by the simple arguement, we only have 8 variables,and there have 9+2 equations. Apparently,the answer is still yes. The fair dicesare the only chance.See:For P(sum=2)=P1_1*P1_2= 1/36 -> only have 1 integer 1 on each as both P1_1 and P1_2>=1/6For P(sum=12)=P6_1*P6_2= 1/36 -> only have 1 integer 6 on each as both P6_1 and P6_2>=1/6

cm27874's answer to (1) is the simplest solution i know if the two dices are independent.btw, i guess you are considering the cases when the two dices might be correlated, quantyst?QuoteOriginally posted by: letiand2) If the two dices are not the sameGenerally, we have 12 variable for the distribution of each integerBut remember we still have 11+2(for total probability to equal to zero) equations. However, by the simple arguement, we only have 8 variables, and there have 9+2 equations. Apparently, the answer is still yes. The fair dices are the only chance.See:For P(sum=2)=P1_1*P1_2= 1/36 -> only have 1 integer 1 on each as both P1_1 and P1_2>=1/6For P(sum=12)=P6_1*P6_2= 1/36 -> only have 1 integer 6 on each as both P6_1 and P6_2>=1/6i am not following your simple argument...and we already know the dices are fair - you want to find whether there exists two dices other than {1,2,3,4,5,6} which has a same distribution of the sum as the normal ones. (hint: you might want to consider the generating function)----- ----- ----- ----- -----for (2), the answer is no. the two dices could also be: {1,2,2,3,3,4} and {1,3,4,5,6,8}.one-line proof that they yield the same sum-distribution as the two normal dices:

I almost forget the stochastic process as I took it several years ago. So my arguement is just based on simple logic .

QuoteOriginally posted by: wileyswcm27874's answer to (1) is the simplest solution i know if the two dices are independent.btw, i guess you are considering the cases when the two dices might be correlated, quantyst?If the two dice do not have to be independent, it should be possible to start with any two marginal distributions and link them by a (not uniquely defined) copula in order to produce equal probabilities for the sums.