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E[N+1] denotes expected number of dice at N+1 stage.
E[N+1]= E[N]*E[1]
This is because, the process can be modeled as each die at the N stage starting a new process similar to the parent process.
Hence, the expected number of dice for each die in stage N is E[1], where E[1] denotes the expected number of dice after the first throw.
E[1] = (4+5+6)/6 = 2.5
Hence, the expected number of dice at the N+1th round of rolls will be (2.5*E[N]). This implies that E[N] = 2.5^N
Note that this is a sort-of intuitive argument (the expected number of dice could be non integral).
To make the argument concrete, consider Xn to be the random variable representing the number of dice at Nth stage.
Now, Xn+1 = Random variable representing number of dice at n+1 th stage.
Xn+1 = summation(X1)[Xn times]
An important (but reasonable) assumption is that Xn and X1 are independent.
Hence, E(Xn+1) = E(Xn)*E(X1)

Hence, expected number of dice after the Nth round of rolls is (E(X1))^N = (2.5)^N

I think total expected number at the end of Nth round of rolls is 2*((2.5)^N-1)/3.
the expected no of dice rolled in nth round is (2.5)^(N-1) if we consider roll of 1st dice as 1st round.

I think total expected number at the end of Nth round of rolls is 2*((2.5)^N-1)/3.
the expected no of dice rolled in nth round is (2.5)^(N-1) if we consider roll of 1st dice as 1st round

it shud be (3.5)* (2.5)^n-1

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is it (3.5)*(2.5)^N-1 ?..

@Everyone..
We all see the 2.5^(N-1) part..

I think it should be 3.5*() but its not correct somehow..

@Unknown, How did you come up with 2/3*()?

Thanks.

If I have read the problem correctly the total number FROM the i-th round itself should be 3.5*(2.5)^(i-1) as everyone has posted. This then gives us the total AFTER the n-th round should be \sum_{i=1}^n 3.5*(2.5^(i-1)) which equals 3.5*(1-2.5^n)/(1-2.5)=7/3*(2.5^n-1).

Note: For n=1 we get 7/3*(2.5^1-1)=7/2 and 1/6*(1+2+3+4+5+6)=7/2! Also, for n=2 we get 7/3*(2.5^2-1)=49/4 and 1/6*[1+2+3+(4+4*7/2)+(5+5*7/2)+(6+6*7/2]=49/4!

Let me know what you think, James.

I guess question have asked the total expected no. of rolls of dices at the end of the N-th round. So I just added all the expected no. dices rolled from round 1 to round N
1+2.5^1+2.5^2.....+2.5^N-1=
2*(((2.5)^N)-1)/3
I don't know if my view was correct or not.

@Unknown..
The question is a multiple choice question from the placement test. The option 3.5*() was not there. So, I guess your solution is correct.

@Others.. sorry for the ambiguity.

Thanks.

Hi...
Shouldn't the constant in unknown's solution be equal to 3. If {1,2,3} comes after the first roll then the number of dice remains 1 whereas it becomes {4,5,6} if the respective figure comes on the roll. Ergo, E(1) = (1+1+1+4+5+6)/6 = 3.
Please correct me if I am wrong.
Thanks!!! :)