Question

1. Many games require rolling 2 dice and adding the rolls together. Fill in the table below with the sum of the two die rolls. The first few cells have been completed as an example.

Sum of Die Rolls		First Die Roll
		1	2	3	4	5	6
Second Die Roll	1	2	3	4
	2	3
	3
	4
	5
	6

a. We assume die rolls are all equally likely. There are 36 possible outcomes (6x6) when we give them as ordered pairs like (2, 3), but when we look at adding them together, we get sums 2, 3, 4, 5, etc.

Complete this table with the sum of two dice, and the probability of each sum. (If you use decimals, use at least 3 digits, like 2.78%.)

Sum	Probability	Your Results (part c)
2 3 4 5	1/36 ≈ 2.78%

b. Which number is the most likely? Which are the next most likely?

c. Now we’re going to compare the theoretical distribution (part a) with some empirical data.

• Roll 2 dice, and record the sum. Do this 30 times and put the results in the table above. (If you don’t have dice handy, use the Excel command =RANDBETWEEN(1,6) to roll a die.)
• Write a couple sentences about how your results compare with the theoretical distribution.
• What does the Law of Large Numbers say about what should happen to your results if you roll thousands and thousands of times?

2. We are going to roll a die with 20 sides, numbered 1 – 20. Each number on a die is assumed to be equally likely, but let’s mix things up a bit here.

Let’s say A = the number is 1 – 10, B = the number is 11 – 12, and C = the number is 13 – 20

a. Let’s say you roll the die once. Give the probability of each outcome A, B, and C.

(Make sure P(A) + P(B) + P(C) = 1.)

b. Suppose you roll the die two times. Now you have sequences like AA, AB, etc. Complete the table with all the possible sequences, and the probability of each sequence.

Sequence	Probability
AA AB

c. Make sure the probabilities add up to 1.

d. What is the probability that you get a two-roll sequence with no A’s in it?

Answer 1

1.

Sum of Die Rolls		First Die Roll
		1	2	3	4	5	6
Second Die Roll	1	2	3	4	5	6	7
	2	3	4	5	6	7	8
	3	4	5	6	7	8	9
	4	5	6	7	8	9	10
	5	6	7	8	9	10	11
	6	7	8	9	10	11	12

a.

	Sum	Probability	Percentage
	2	0.0278	2.78%
	3	0.0556	5.56%
	4	0.0833	8.33%
	5	0.1111	11.11%
	6	0.1389	13.89%
	7	0.1667	16.67%
	8	0.1389	13.89%
	9	0.1111	11.11%
	10	0.0833	8.33%
	11	0.0556	5.56%
	12	0.0278	2.78%

b. The number is the most likely=7, Next most likely numbers=6,8 and so on.

c.

Serial no.	Die 1	Die 2	Sum
1	2	1	3
2	3	6	9
3	1	4	5
4	4	2	6
5	6	5	11
6	5	4	9
7	1	4	5
8	1	4	5
9	1	1	2
10	2	5	7
11	5	2	7
12	3	5	8
13	3	1	4
14	3	4	7
15	4	3	7
16	1	3	4
17	4	4	8
18	3	3	6
19	4	5	9
20	1	5	6
21	1	3	4
22	3	4	7
23	2	6	8
24	3	2	5
25	6	5	11
26	4	4	8
27	4	3	7
28	5	5	10
29	4	2	6
30	6	2	8

Outcome(Sum)	Frequency	Relative frequency	Percentage
2	1	0.0333	3.33%
3	1	0.0333	3.33%
4	3	0.1	10%
5	4	0.1333	13.33%
6	4	0.1333	13.33%
7	6	0.2	20%
8	5	0.1667	16.67%
9	3	0.1	10%
10	1	0.0333	3.33%
11	2	0.0667	6.67%
12	0	0	0%

The number is the most likely=7, Next most likely number=8 and so on. Then we see that the result of the number which is the most likely is matched with the theoretical result.

If we repeat this process 1000 times then

Outcome(Sum)	Frequency	Relative frequency	Percentage
2	35	0.035	3.5
3	51	0.051	5.1
4	87	0.087	8.7
5	120	0.12	12
6	135	0.135	13.5
7	174	0.174	17.4
8	140	0.14	14
9	101	0.101	10.1
10	79	0.079	7.9
11	57	0.057	5.7
12	21	0.021	2.1
Total	1000	1	100

so we get a same result as above.

2.a. P(A)=10/20=1/2, P(B)=2/20=1/10, P(C)=8/20=2/5

b. P(AA)=P(A)P(A)=1/4, P(AB)=P(A)P(B)=1/20, P(BA)=P(B)P(A)=1/20, P(BB)= 1/100, P(AC)=P(A)P(C)=1/5, P(CA)=1/5, P(BC)=1/25, P(CB)=1/25

c. P(AA)+P(AB)+P(BA)+P(BB)+P(AC)+P(CA)+P(BC)+P(CB)+P(CC)=1

d. The probability that you get a two-roll sequence with no A’s in it=P(AA)+P(AB)+P(BA)+P(AC)+P(CA)=1/4+2/20+2/5=3/4.