# Thread: Dice interpretation problem

1. ## Dice interpretation problem

I understand how the question arrived at the conditional probabilities, however couldn't the question also be interpreted this way?

Since the question never stated that the 2 dice were different, what if we assume they were not distinct and also we can assume that the 2 dice is thrown at the same time, thus $\displaystyle P(S = 4 | M=3) = \frac{P( S=4 \cap M = 3)}{P(M=3)}$

Now to have a sum of 4 and a max of 3, all we need is to roll a 1 and a 3, there is no order here since we are throwing the dice at the same time and the dice are NOT distinct. Thus the numerator becomes 1/6 * 1/6

To get a maximum of 3, all we need is to roll a 1 and a 3 or 2 and a 3 or 3 and a 3 (again there is no order since we are throwing the dice at the same time)

Thus the denominator is 3*(1/6 * 1/6)

Thus $\displaystyle P(S = 4 | M=3) = \frac{1}{3}$?

Would the above also be a valid interpretation?

2. ## Re: Dice interpretation problem

I can give a quicker explanation.
$\displaystyle M=3$ is the space $\displaystyle \{(1,3),~(3,1),~(2,3),~(3,2),~(3,3)\}$

Now it is clear that $\displaystyle P(S=5)|M=3)=\frac{2}{5}.$

3. ## Re: Dice interpretation problem

Thanks Plato, however if the 2 dices are rolled at exactly the same time (and the 2 dices are assumed to be identical) then wouldn't (1,3) be the same as (3,1) ?

4. ## Re: Dice interpretation problem

Originally Posted by usagi_killer

I understand how the question arrived at the conditional probabilities, however couldn't the question also be interpreted this way?

Since the question never stated that the 2 dice were different, what if we assume they were not distinct and also we can assume that the 2 dice is thrown at the same time, thus $\displaystyle P(S = 4 | M=3) = \frac{P( S=4 \cap M = 3)}{P(M=3)}$

Now to have a sum of 4 and a max of 3, all we need is to roll a 1 and a 3, there is no order here since we are throwing the dice at the same time and the dice are NOT distinct. Thus the numerator becomes 1/6 * 1/6

To get a maximum of 3, all we need is to roll a 1 and a 3 or 2 and a 3 or 3 and a 3 (again there is no order since we are throwing the dice at the same time)
Because, as you say, "there is no order here", you must also consider "a 3 and a 1" and "a 3 and a 2".

Thus the denominator is 3*(1/6 * 1/6)

Thus $\displaystyle P(S = 4 | M=3) = \frac{1}{3}$?

Would the above also be a valid interpretation?

5. ## Re: Dice interpretation problem

Originally Posted by usagi_killer
Thanks Plato, however if the 2 dices are rolled at exactly the same time (and the 2 dices are assumed to be identical) then wouldn't (1,3) be the same as (3,1) ?
Absolutely not. If you roll two dice the outcome space is a set of 36 ordered pairs.
If you roll three dice the outcome space is a set of $\displaystyle 6^3$ ordered pairs.
If you roll n dice the outcome space is a set of $\displaystyle 6^n$ ordered pairs.

It makes no difference if you roll n dice at once or one die n times the outcome space is the same.

6. ## Re: Dice interpretation problem

Thanks to both, I understand now

7. ## Re: Dice interpretation problem

Hi again, I just have an extension to this problem:

I was wondering in order to compute E(S) using the method they described, is it really as tedious as the following? (Obviously I could just apply the definition of $\displaystyle E(S) = \sum_{s = 2}^{12} s \times p_S(s)$ straight away, but here I wanted to try their exercise)

So in order to compute E(S) using the method they described we have to work out E(S|M=y) first which is given by $\displaystyle E(S|M=y) = \sum_{s = 2}^{12} s \times p_{S|M}(s|y) = \sum_{s = 2}^{12} s \left(\frac{p_{S,M}(s,y)}{p_{M}(y)}\right)$

Now we have do that for $\displaystyle y = 1, \cdots, 6$: which means we have to manually find $\displaystyle \sum_{s = 2}^{12} s \left(\frac{p_{S,M}(s,y)}{p_{M}(y)}\right)$ for $\displaystyle y = 1, \cdots, 6$

Then we have to manually find $\displaystyle p_M(y)$ for $\displaystyle y = 1, \cdots, 6$

Then after all that sub back into $\displaystyle E(S) = \sum_{y=1}^{6} p_M(y) E(S|M=y)$

I did all of the above and ended up with so much working, is there any shortcuts rather than applying the definitions of each expression directly?