Basic Probability Problem

[mikexz]

I am reviewing math for the MCAT and the problem is finding the prob. of getting a sum of 6 after rolling 2 dice.

I know that the solution is 5/36 but it doesn't intuitively make sense.

When determining the number of possible outcomes, if we get a result such as (1,2) then we still consider (2,1) as a different result??? yet when we have (3,3) why do we not consider the other (3,3) as a different result? doesn't it matter which number appears on which die? Why can we disregard this?

Perhaps the problem lies in a misunderstanding between permutations and combinations?

this reminds me of the Monty Hall problem where despite the outcome being the same regardless of which goat is chosen, they are considered different outcomes.

thank you for your help

 

[Deleted member 4993]

I am reviewing math for the MCAT and the problem is finding the prob. of getting a sum of 6 after rolling 2 dice. <<< For this excersize - assume one die is colored blue and the other red. Now go through the excersize.

Red 2 + blue 1 will be different from blue 2 and red 1

However red 3+blue 3 is indistinguishable from blue 3 + red 3

I know that the solution is 5/36 but it doesn't intuitively make sense.

When determining the number of possible outcomes, if we get a result such as (1,2) then we still consider (2,1) as a different result??? yet when we have (3,3) why do we not consider the other (3,3) as a different result? doesn't it matter which number appears on which die? Why can we disregard this?

Perhaps the problem lies in a misunderstanding between permutations and combinations?

this reminds me of the Monty Hall problem where despite the outcome being the same regardless of which goat is chosen, they are considered different outcomes.

thank you for your help

 

[galactus]

There are 6^2=36 different combinations when rolling two dice. Out of those we have ((3,3), (4,2), (5,1)). But the (4,2) and (5,1) can be (1,5) and (2,4)

For a total of 5 ways we can sum to 6.

If we want to be fancy, we can use a generating function. \(\displaystyle \left(\sum_{k=1}^{6}x^{k}\right)^{2}\)

Expand out and we get \(\displaystyle x^{12}+2x^{11}+3x^{10}+4x^{9}+5x^{8}+6x^{7}+5x^{6}+3x^{4}+2x^{3}+x^{2}\)

Look at the coefficient of the x^6 term. It is 5. That tells us that the number of ways of rolling two dice and summing to 6 is 5. Which we knew anyway.

This function is rather overkill when dealing with only two dice. But if it were, say, 5 dice and we wanted the number of ways they sum to 24, then it is handy.

 

[mikexz]

thanks for the replies

I get it now but the problem still bugs me intuitievley because if I was to look at the chances of strictly getting the same number on both dies vs. not getting the same number, I would think that since there is two possibilities of getting 3 on die A and 3 on die B and vise-versa the chances of getting the same number on both will be greater than getting 2 different numbers

so e.x. chance of getting (3,3) vs. (1,2)

 

[soroban]

Hello, mikexz!

Find the prob. of getting a sum of 6 after rolling 2 dice.

I know that the solution is 5/36, but it doesn't intuitively make sense.
Your intuition is off . . .

\(\displaystyle \text{Do you agree that there are: }\:6^2 = 36\text{ possible outcomes?}\)

\(\displaystyle \text{Here they are: }\)

. . . \(\displaystyle \begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\)


\(\displaystyle \text{You seem to think that }\,\boxed{3}\;\boxed{3}\,\text{ is different from }\,\boxed{3}\;\boxed{3}\)
. . \(\displaystyle \text{where the two dice have switched places.}\)

\(\displaystyle \text{They are }not\text{ different outcomes.}\)
. . \(\displaystyle \text{The outcome is: "Both dice have a 3."}\)
. . \(\displaystyle \text{How many different ways can you say that?}\)