Question

To play the 7-11 game at a gambling casino one must pay $1. if one rolls a sum of 7 or a sum of 11 one wins $5, since one paid $1 to play the net gain is $4. for all other sums rolled the net gain is -$1.

a. Play the game for 36 rolls of two dice. Record the sums you rolled in the chart provided.

Sum: 2 3 4 5 6 7 8 9 10 11 12

Tally:

Frequency:

b.What were your average winnings for the 36 times you played the game?

c. How many times should you have theoretically won $5?_____ What is P(7 or11)?______

d. How many times should you have theoretically lost $1?____ What is P(7 or 11)?_____

e. What should your theoretical net average winnings be if you play the game 36 times?

f. If the casino charges $2 per game and the payoff was still $5, what would your expected payoff for this game be?

g. What should the casino charge if the 7-11 game is to be a fair game?

Answer 1

a)

Sum : Frequency(Tally: First Dice-Second Dice)   

2	1 (1-1)
3	2 (1-2,2-1)
4	3 (1-3,2-2,3-1)
5	4 (1-4,2-3,3-2,4-1)
6	5 (1-5,2-4,3-3,4-2,5-1)
7	6 (1-6,2-5,3-4,4-3,5-2,6-1)
8	5 (2-6,3-5,4-4,5-3,6-2)
9	4 (3-6,4-5,5-4,6-3)
10	3 (4-6,5-5,6-4)
11	2 (5-6,6-5)
12	1 (6-6)

b) Only games with sum of 7 and 11 results in win. As per the above table, sum of 7 happens 6 times and sum of 11 happens 2 times. So in all 8 times one will be winner out of 36 games.

Net winnings = Win count * 5 - Amount paid to play ($1 for each of 36 games)

Net winnings = 8*5 - 1*36 = 40-36 = $4

c) Times one can won $5 = 8 Times as per the table for (a). Probability of 7 or 11 = 8/36 = 44.44%

d) Times one will lose $1 = 36 - 8 = 28.  Probability of 7 or 11 = 8/36 = 44.44%