Frequency for rolling a die (without replacement)...and finding expected value

earthman

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Credit for this math problem goes to the Cirrus Test Prep company's book Praxis Mathematics Content Knowledge 5161. I took it from the book referenced above and I changed a few erroneous words.

This is a strange type of problem. It's kind of like rolling a 6 sided die, then rolling a 5 sided die that doesn't have the first result on it.

Here it is:

A bag contains 6 balls numbered 1 - 6. Two balls are removed from the bag, and the sum of the two balls is recorded. If this experiment is related 50 times, about how many times would the sum be 3? What would the average sum of the two balls be (Expected Value)?

here is my reasoning at this point:

If I was to take a random guess at the expected value of the sums, I would say 7 because as anyone who plays settlers of catan should know, it is the most common die roll of two dice. (Something like 11/36 chance.)

X is represents the sum of the two balls. The possible values of X are 3, 4, 5, 6, 7, 8, 9, 10, and 11. But, my question is, how many times can each result occur?

If I take a 1 and a 2, I get a three. What if I take a 2 and a 1? Should that also count as a possible outcome? This is important because I need to know the frequency of each result to calculate the probability distribution of the possible sums, P(x). I can then find the expected value. It's also important because I need to know the frequency to predict how many times the sum would be 3.

I think there are 30 different possible out comes. With the first draw, there are 6 different outcomes. With the second draw, there are 5 different outcomes. I I use the fundamental counting principal, wouldn't there be 30 different outcomes? I am arguing with a solution manual that claims only 15 different possible outcomes.

Possible results according to me:
1+2 = 3
1+3 = 4
1+4 = 5
1+5 = 6
1+6 = 7
2+1 = 3
2+3 = 5
2+4 = 6
2+5 = 7
2+6 = 8
3+1 = 4
3+2 = 5
3+4 = 7
3+5 = 8

And so on.... I get 30 because if there was replacement, there would be 36 possible results. The only results that are not possible are 1+1=2, 2+2=4, 3+3 = 6, 4+4=8, 5+5=10, and 6+6=12. Another way to do this is to multiply 6 times 5. You get 30.

If you conducted the experiment 50 times, I think you could get about how many times the sum would be 3 by multiplying the number of trials by the single probability of getting a 3. By my count, the probability of getting a 3 is (2/30). So the expected value of times getting a 3 or not getting a 3 after 50 trials is 3.333 times.

Now I am ready to find the expected value of the sums of all 50 trials.

Before I was asking, doesn't it matter whether you say the frequency is 1 or 2 for three? And, Doesn't it matter if the number of possible results is 30 or 15? Well it turns out to be two routes to the same answer.


E(X) = Sum from i=1 to 9 XiP(xi) = 3(2/30) + 4(2/30) + 5(4/30) + 6(4/30) + 7(6/30) +8(4/30) +9(4/30) +10(2/30)+11(2/30) =

6/30 + 8/30 + 20/30 + 24/30 + 42/30 + 32/30 + 36/30 + 20/30 + 22/30 = 210/30 = 7.

Interestingly, my original guess of 7 was the correct answer. But, I am still wondering whether this is a better way of explaining this problem. If the total number of possible out comes for choosing two balls numbered 1 to 6 from a bag was 5x6=30, then the frequency of each sum would be what I listed above. However, if you said well don't count the same sum twice, you would end up with 6C2 = 6!/4!2! = 15 possible sums. In that case, then the frequency of each sum would just be half as much as the frequencies that I found. You would still end up with an expected value of 7 for the function P(x), the sum of two balls numbered 1-6 being chosen out of a bag.

However, my question still stands. Does it matter whether you say that the total number of possible out comes is 30 or 15? In other words, is the range of P(x) from 1/15 to 1/5 or is it 2/30 to 6/30? I can see how mathematically maybe it doesn't make a difference. However, conceptually it feels like it should make a difference.

What do you all think? Also, could I have explained this problem better? I am still trying to wrap my mind around it. Do you have an alternate explanation?
 
Hello earthman,

Here is my first thought, on the counting of the possible outcomes. The ordering doesn't matter. The problem doesn't even say that you first draw one ball, and then the other: it says you take them out at the same time. But even if it did say that you drew them sequentially, getting first "1" and then "2" would not be a different outcome from first "2" and then "1". It would be just as if you drew them both at the same time, because ultimately you pick two of these unique balls and sum up their numbers, and both the scenarios I described give the same sum. There is only one ball with a "1" printed on it, and only one ball with a "2" printed on it, and both outcomes lead to these being the two balls that you have selected and added together.

The question is, how many different ways are there of selecting two objects out of a set of six? Since the six objects are unique, that will give you the number of unique pairings, and therefore the number of possible sums

The answer is given by what we call (6 choose 2), which is also written as:

\(\displaystyle \left(\begin{array}{c} 6 \\ 2 \end{array}\right) = \dfrac{6!}{2!4!} = \dfrac{6\cdot5}{2} = 15 \)

There are 15 unique pairings.

EDIT: if you need an explanation of why this is the equation for (6 choose 2), let me know.
 
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When you are counting it out brute force, what you are trying to do is list all possible unique pairings:

1+2 = 3
1+3 = 4
1+4 = 5
1+5 = 6
1+6 = 7
2+3 = 5
2+4 = 6
2+5 = 7
2+6 = 8
3+4 = 7
3+5 = 8
3+6 = 9
4+5 = 9
4+6 = 10
5+6 = 11

Notice that, once I'm done with the 1+X entries, I then ignore the number 1. The reason is because every possible pairing of a ball with the "1 ball" has already been counted, and if I now write things like (2+1), (3+1), etc. I am double counting those pairings.

The expectation value is the given by

\(\displaystyle \sum_x xP(x) \)

where each of the x values is a possible sum, and P(x) is the probability of that sum occurring. The probability is just given by the number of possible ways of getting that sum, divided by the total number of possible outcomes:

\(\displaystyle \sum_x xP(x) = \dfrac{1}{15}\left[ (1 \cdot 3) + (1 \cdot 4) + (2 \cdot 5) + (2 \cdot 6) + (3 \cdot 7) + (2 \cdot 8) + (2 \cdot 9) + (1 \cdot 10) + (1 \cdot 11) \right ] = 7\)

EDIT: I would say that your other method is conceptually wrong, but arithmetically equivalent, since you have double counted every single pairing, and hence you have also double counted the total number of possible outcomes
 
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