Fair Dice Rolled

  • Thread starter Redhead711
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  • #1
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I am studying for a test tomorrow and this question is on the review sheet and I can't figure it out.

Suppose a fair die is rolled ten times. Find numerical values for the expectations of each of the following random variables.
a) The sum of the numbers in ten rolls;
b) The sum of the largest two numbers in the first three rolls;
c) The maximum number in the first five rolls.

I believe I figured out part a to be 35.

However, the next two parts of the equation I am completely stuck on. If someone could give me some detailed instructions on what I need to do it would really help me out for the test. Thank you so much.
 

Answers and Replies

  • #2
509
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Well, for c you can just count up the cases and the probability for each case: first find the probability of the largest number = 1, then largest number = 2, etc.

I can't think of any simple way to do b. You could consider the probability that the smallest value = 1, and then find the expected sum when you know that the other two are in the range 1-6, then you could consider the probability that the smallest value = 2, and then find the expected sum when you know that the other two are in the range 2-6, and so on.
 
  • #3
38
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How was the test? I was thinking about your problem and I realized you were asking "order statistic" questions. To find the probability distribution function for part (c) you find P(d(1)<=k,d(2)<=k,...,d(5)<=k) = [P(d<=k)]^5=(k/6)^5. Where {d(i)<=k} is at least i dice out 5 dice that are <=k. The expectation for the maximum then become Sum(k * [(k/6)^5-((k-1)/6)^5], k=0,1..6). However I seemed to be stuck on how to find the joint distribution P(d(2)<=k and d(3)<=m) where d(3) is the max out of 3 dice and k<=m. Actually I would be interested in knowing how to determine
P(d(N-1)<=k and d(N)<=m) for N dice.
 
  • #4
10
0
I ended up getting a B on the test. I am glad, I am a senior and this is the last math class I need before I graduate. I finally got what I think is the answer to the problem:

In one roll the expected value is

(1/6)(1 + 2 + 3 + 4 + 5 + 6)

= 21/6 = 3.5

So in 10 rolls the expected sum = 35

>b) The sum of the largest two numbers in the first three rolls;

We must first find the expected value of the largest number when a
die is rolled 3 times.

If y = largest number when a die is rolled 3 times

the c.d.f., F(y) of Y

F(1) = (1/6)^3
F(2) = (2/6)^3
F(3) = (3/6)^3
F(4) = (4/6)^3
F(5) = (5/6)^3
F(6) = (6/6)^3


the pdf, f(Y) of Y

f(1) = (1/6)^3 = 1/216 = 1/216
f(2) = F(2) - F(1) = (2^3 - 1^3)/216 = 7/216
f(3) = F(3) - F(2) = (3^3 - 2^3)/216 = 19/216
f(4) = F(4) - F(3) = (4^3 - 3^3)/216 = 37/216
f(5) = F(5) - F(4) = (5^3 - 4^3)/216 = 61/216
f(6) = F(6) - F(5) = (6^3 - 5^3)/216 = 91/216

The expected value for the largest number is

= (1/216)[1 + 2x7 + 3x19 + 4x37 + 5x61 + 6x91]

= (1/216)[1071] = 4.9583

and so the expected SUM of the two largest = 2 x 4.9583

= 9.9166


>c) The maximum number in the first five rolls.

Using same reasoning as in part (b)

f(1) = (1/6)^5 = 1/7776 = 1/7776
f(2) = F(2) - F(1) = (2^5 - 1^5)/7776 = 31/7776
f(3) = F(3) - F(2) = (3^5 - 2^5)/7776 = 211/7776
f(4) = F(4) - F(3) = (4^5 - 3^5)/7776 = 781/7776
f(5) = F(5) - F(4) = (5^5 - 4^5)/7776 = 2101/7776
f(6) = F(6) - F(5) = (6^5 - 5^5)/7776 = 4651/7776

The expected value for the largest number is

= (1/7776)[1 + 2x31 + 3x211 + 4x781 + 5x2101 + 6x4651]

= (1/7776)[42231]

= 5.4309
Thanks
 
  • #5
OlderDan
Science Advisor
Homework Helper
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I was reading here to refresh my memory or learn something I never knew about things probabilistic, and I'm trying to figure out the logic of the approach to part b). If I get that, then I will think about c). It seems to me the result posted for b) is too big for the following reason.

I believe the result for a) is correct. The expectation for each roll is 3.5, and so for 10 rolls it would be 35. The expectation for 3 rolls would be 3*3.5 = 10.5, and the smallest number that can be rolled on one die is 1. It seems to me the largest expectation for the sum of the two largest dice has to be no greater than the expected sum for 3 dice minus the minimum on one die or 10.5 - 1 = 9.5.

I happen to have an old Excel file I used to illustrate frequency distributions for sums of multiple dice throws, so I quick added a column to calculate the sum of the two largest dice for all 216 possible outcomes for 3 rolls and took the average. I came up with 8.458, which seems a more reasonable result to me than the one posted. I'd like to see a more elegant approach than my brute force method, or at least confirm that I did the right thing to get that result.
 
Last edited:
  • #6
38
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OlderDan, I agree with your 8.4583333 (9-13/24) number. I have been looking into the Joint Probability Mass Function for "Order Statistics" and I am not yet clear on how to clean up the brute force technique for a discrete random variables (dice).
 
  • #7
644
1
Nice work Older Dan!
As for rigorous mathematical footing, it isnt very hard either (since we are considering just three rolls ofcourse)

Let X1,X2,X3 be R.Vs associated with the first three rolls
Let Y = max(X1+X2,X2+X3,X1+X3)
-------Y---------|---------P(Y=i)-----------|
-------2---------|----------1/216-----------|
-------3---------|----------3/216-----------|
-------4---------|----------7/216-----------|
-------5---------|---------12/216-----------|
-------6---------|---------19/216-----------|
-------7---------|---------27/216-----------|
-------8---------|---------34/216-----------|
-------9---------|---------36/216-----------|
------10---------|---------34/216-----------|
------11---------|---------27/216-----------|
------12---------|---------16/216-----------|

These probabilities arent very hard to calculate at all!
E(Y)
= 2(1/216)+3(3/216)+...+12(16/216)
= 8.458333333

-- AI
 
  • #8
644
1
Redhead,
Now that i check your answer to c, i wonder if thats correct either?
I am getting a very low expectation compared to yours (1.4623)

Let X1,X2,X3,X4,X5 be RVs associated with the first five rolls.
Let Y = max(X1,X2,X3,X4,X5)
-----------Y-----------|----------P(Y=i)-----------|
-----------1-----------|----------1/6^5-----------|
-----------2-----------|-------(2^4*5)/6^5-------|
-----------3-----------|-------(3^4*5)/6^5-------|
-----------4-----------|-------(4^4*5)/6^5-------|
-----------5-----------|-------(5^4*5)/6^5-------|
-----------6-----------|-------(6^4*5)/6^5-------|

-- AI
 
  • #9
38
0
TenaliRaman, Redhead's part c is correct. If Y=max(X1,X2,X3,X4,X5) then P(Y<=k) = P(X1<=k & X2<=k & X3<=k & X4<=k & X5<=k) where & is used as "and". Since the RVs are independent and identical you have the following for 5 dice
P(Y<=k) = P(X1<=k)*P(X2<=k)*P(X3<=k)*P(X4<=k)*P(X5<=k)=[P(X<=k)]^5=(k/6)^5.
The probability mass function can then derived as p(Y=k)=[P(X<=k)^5-P(X<=k-1)^5]=(k/6)^5-((k-1)/6)^5. If you take the example from part b of the three dice, where Y3=max(X1,X2,X3). You will quickly see that p(Y3=k)=P(X<=k)^3-P(X<=k-1)^3
=(k/6)^3-((k-1)/6)^3.
 
  • #10
644
1
LittleWolf said:
If you take the example from part b of the three dice, where Y3=max(X1,X2,X3). You will quickly see that p(Y3=k)=P(X<=k)^3-P(X<=k-1)^3
=(k/6)^3-((k-1)/6)^3.
As much as i agree with your analysis and thanks for pointing my error out(i knew i must have screwed up with probability calculations there). However the above quoted statement is baffling me. I believe for part b, Y3 = max(X1+X2,X1+X3,X2+X3).

-- AI
 
  • #11
38
0
I meant take the example of 3 dice and consider Y3=max(X1,X2,X3) then check if
P(Y3=k)=P(Y3<=k)-P(Y3<=k-1)=(k/6)^3-((k-1)/6)^3. I did not mean Y3 to be the
max(X1+X2,X1+X3,X2+X3).
 

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