Name ___________Answers________________________

1. What is the probability of rolling a six with one die? Express the answer as a decimal, fraction, and percentage.

 

0.16667, 1/6, 16.67%.

For questions 2-4 consider the place bet on 4 in craps. The player bets that a 4 will be rolled before a 7. You may take it on faith (for now) that the probability of winning is 33 1/3%. The bet pays 9 to 5.

2. What are the true odds?

For a bet with probability of winning p the true odds are (1-p)/p to 1. So in this case the true odds are (1-(1/3))/(1/3) = (2/3)/(1/3) = 2 to 1. 

3. If the player bets $100 and wins, how much does he win?

$100*(9/5) = $180 

4. What is the house edge?

The expected value is Pr(win)*payoff odds – Pr(loss) = (1/3)*1.8 – (2/3) = -0.06667. So the house edge is 0.06667, or 6.67%. 

For questions 5-8 consider the 1-12 column bet in double zero roulette. The player bets that a 1 to 12 will be rolled on the next spin. So 12 numbers will win and 26 will lose. If the player bets $100 he stands to win $200.

5. What is the probability of winning, in decimal form?

0.3158 

6. What is the probability of winning, in fraction form?

12/38 

7. What is the probability of winning, in percentage form?

31.58%. 

8. What are the payoff odds?

2 to 1 

9. What are the true odds?

Again use (1-p)/p = (1-(12/38))/(12/38) = (26/38)/(12/38) = 26/12 = 13/6 = 13 to 6. 

10. What is the house edge?

The expected value is (12/38)*2 – (26/38) = -2/38 = -1/19 = -0.0526. So the house edge is 5.26%. 

 

For questions 17-21 consider the $10 bet in Big Six. There are 54 total positions on the wheel but only 4 of them win. If the player bets $100 he stands to win $1000.

11. What is the probability of winning, in decimal form?

0.0741 

12. What is the probability of winning, in fraction form?

4/54 

13. What is the probability of winning, in percentage form?

7.41%. 

14. What are the payoff odds?

10 to 1. 

15. What are the true odds?

P=4/54, payoff odds = (1-p)/p = (50/54)/(4/54) = 50/4 = 25/2 = 25 to 2. 

16. What is the house edge?

Expected value = (4/54)*10 – (50/54) = 40/54 – 50/54 = -10/54 = -0.1852. So the house edge is 18.52%. 

For the remainder of the questions consider a new game called "five-sided dice sic bo", in which 5-sided dice are used instead of 6-sided dice. You may assume that the probability of each side is equal. Three of these five-sided dice are thrown and the player bets on the outcome.

Following is a table showing the number of permutations and probability of each total. This can be arrived at a variety of ways but I would recommend Excel.

Total	Permutations	Probability
3	1	0.008000
4	3	0.024000
5	6	0.048000
6	10	0.080000
7	15	0.120000
8	18	0.144000
9	19	0.152000
10	18	0.144000
11	15	0.120000
12	10	0.080000
13	6	0.048000
14	3	0.024000
15	1	0.008000
Total	125	1.000000

 

17. A total of 3 pays 100 to 1. What is the probability of winning and expected value?

There is only 1 way to throw a total of 3 (1,1,1) or 3!/3! = 1

Probability of winning = 1/125.

Expected value = (1/125)*100 – (124/125) = -24/125 = -19.25%. 

18. A total of 14 pays 35 to 1. What is the probability of winning and the house edge?

There is only one combination totaling 14 (5,5,4) but it can be rolled 3 ways: (4,5,5); (5,4,5); and (5,5,4), or 3!/(2!*1!) = 3.

So there are 3 permutations of a total of 14.

Probability of winning = 3/125

Expected value = (3/125)*35 – (122/125) = -17/125 = -13.60%. So the house edge is 13.60%. 

19. A total of 5 pays 17 to 1. What is the probability of winning and expected value?

Ways to roll a total of 5:

(1,1,3): 3 permutations {(3,1,1); (1,3,1); (1,1,3)} or 3!/(2!*1!) = 3

(1,2,2): 3 permutations {(1,2,2); (2,1,2); (2,2,1)} or 3!/(2!*1!) = 3

So there are 6 permutations resulting in a total of 5.

Probability of winning = 6/125.

Expected value = (6/125)*17 – (119/125) = -17/125 = -13.60%. 

20. A total of 12 pays 10 to 1. What is the probability of winning and the house edge?

Ways to roll a total of 12:

(2,5,5): 3 permutations {(2,5,5); (5,2,5); (5,5,2)} or 3!/(2!*1!) = 3

(3,4,5): 6 permutations {(3,4,5); (3,5,4); (4,3,5); (4,5,3); (5,3,4); (5,4,3)}

(4,4,4): 1 permutations {(4,4,4)}

So there are 10 permutations of a total of 12.

Probability of winning = 10/125.

Expected value = (10/125)*10 – ((125-10)/125) = -15/125 = -12.00%. So the house edge is 12.00%.

21. A total of 7 pays 7 to 1. What is the probability of winning and the expected value?

22. A total of 10 pays 5 to 1. What is the probability of winning and the house edge?

Ways to roll a total of 10:

(1,4,5): 6 permutations {(1,4,5); (1,5,4); (4,1,5); (4,5,1); (5,1,4); (5,4,1)} or 3!/(1!*1!*1!) = 6

(2,3,5): 6 permutations {(2,3,5), (2,5,3), (3,2,5); (3,5,2); (5,2,3); (5,3,2) } or 3!/(1!*1!*1!) = 6

(2,4,4): 3 permutations {(2,4,4); (4,2,4); (4,4,2)} or 3!/(2!*1!) = 3

(3,3,4): 3 permutations {(3,3,4); (3,4,3); (4,3,3)} or 3!/(2!*1!) = 3

So there are 18 permutations of a total of 10.

Probability of winning = 18/125. 

Expected value = (18/125)*5 – (1-(18/125)) = 90/125 – 107/125 = -17/125 = -13.60%. So the house edge is 13.60%. 

23. A total of 9 pays 5 to 1. What is the probability of winning and the expected value?

Ways to roll a total of 9:

(1,3,5): 6 permutations: {(1,3,5); (1,5,3); (3,1,5); (3,5,1); (5,1,3); (5,3,1)} or 3!/(1!*1!*1!) = 6

(1,4,4): 3 permutations: {(1,4,4); (4,1,4); (4,4,1)} or 3!/(2!*1!) = 3

(2,2,5): 3 permutations: {(2,2,5); (2,5,2); (5,2,2)} or 3!/(2!*1!) = 3

(2,3,4): 6 permutations: {(2,3,4); (2,4,3); (3,2,4); (3,4,2); (4,2,3); (4,3,2)} or 3!/(1!*1!*1!) = 6

(3,3,3): 1 permutations: {(3,3,3)} or 3!/3! = 1

So there are 19 permutations of a total of 9.

Probability of winning = 19/125. Expected value = (19/125)*5 – (1-(19/125)) = 95/125 – 106/125 = -11/125 = -8.80%. 

 

24. The "Two of a Kind" wins if the two specific numbers both appear in the throw of the dice. If the player chooses 2 and 4 the bet will win if both a 2 and 4 appear, for example: (2,2,4) or (4,2,5). The bet pays 4 to 1. What is the probability of winning and the expected value?

The following table shows the possible combinations and the number of permutations each.

Dice	Permutations
2,4,1	6
2,4,2	3
2,4,3	6
2,4,4	3
2,4,5	6
Total	24

 

So the probability of winning is 24/125.

The expected value is (24/125)*4 – (101/125) = -5/125 = -4.00%. 

25. The "Double" bet wins if a specific number appears at least twice. If the player chooses 3 the bet will win if 2 or 3 threes appear in the throw of the dice, for example (3,6,3) or (3,3,3). The bet pays 8 to 1. What is the probability of winning and the house edge?

There are 16 winning permutations: 3 ways each of {(3,3,1); (3,3,2); (3,3,4); and (3,3,5)}, plus (3,3,3). So the probability of winning is 13/125 = 10.40%.

The expected value is 8*(13/125) – (1-(13/125) = 104/125 – 112/125 = -8/125 = -6.40%. So the house edge is 6.40%. 

 

26. The "Triple" bet wins if a specific number appears on all three dice. If the player chooses 5 the bet will win if and only if three fives appear. The bet pays 100 to 1. What is the probability of winning and the house edge?

The probability of winning is 1/125.

The expected value is (1/125)*100 – (124/125) = -24/125 = -19.20%. So the house edge is 19.20%. 

 

27. The "Any Triple" bet wins if any number appears on all three dice. For example, (1,1,1) or (4,4,4). The bet pays 20 to 1. What is the probability of winning and the expected value?