2. Description of the Game
x = Winnings from Game
Sum of Two
Die
2
3
4
5
6
7
8
9
10
11
12
Probability of
Rolling Sum
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
x (in dollars)
3
0
3
0
3
0
3
0
15
20
25
Hello, welcome to the AP Statistics Casino! The game you will be playing today is Lucky Double Digits. The cost of play is $5.00 per
round. The object of the game is to either roll sum of 10, 11, or 12, but rolling any even number will get you $3 of your buy in money back.
As shown above, you, the player, will roll two dice per round. Your winnings depend on the sum you roll. The game is high risk/high
reward, because approximately 5/6 of the time you will lose some or all of your money, but 1/6 of the time you will either double, triple, or
quadruple your money. Good Luck!
3. Expected Value and Standard Deviation Theoretically
P(x=0) = probability of getting a sum of 3, 5, 7, or 9 = (2/36) + (4/36) + (6/36) + (4/36) = (16/36) = .444
P(x=3) = probability of getting a sum of 2, 4, 6, or 8 = (1/36) + (3/36) + (5/36) + (5/36) = (14/36) = .389
P(x=15) = probability of getting a sum of 10 = (3/36) = .083
P(x=20) = probability of getting a sum of 11 = (2/36) = .056
P(x=25) = probability of getting a sum of 12 = (1/36) = .028
Mean of x = E(x) = expected value of x
E(x) = .444(0) + .389(3) + .083(15) + .056(20) + .028(25) = $4.23
σ2x= p1(x1-μx)2 + p2(x2-μx)2 + ... + pn(xn-μx)2 = .444(0-4.23)2 +.389(3-4.23)2 +.083(15-4.23)2 +.056(20-4.23)2 +.028(25-4.23)2 = $44.17
σ=
σ2x =
44.17 = $6.65 = The average winnings deviate approximately $6.65 from the mean of $4.23.
Due to the $5.00 buy in cost, the house should make an average of $0.78 per game ($5.00 buy in - $4.23 average winnings).
However, the standard deviation of the amount the house will make is still $6.65. This is because standard deviation is not
affected when values are added or subtracted from every value of x.
4. Expected Value and Standard Deviation of Game Through
Simulation
Sum of Two Die
2
3
4
5
6
7
8
9
10
11
12
Number of Times
Rolled (out of 50)
0
3
2
7
9
6
2
7
5
7
2
Percent sum was rolled
(# of times / 50)
0
.06
.04
.14
.18
.12
.04
.14
.10
.14
.04
In our simulation, we played Lucky Double Digits 50 times, just without using any money. We recorded
the results in this data table.
Total Payout = 0(3) + 3(0) + 3(2) + 7(0) + 9(3) + 6(0) + 2(3) + 7(0) + 5(15) + 7(20) + 2(25) = $304 total
payout for 50 games
$304 total payout for 50 games / 50 games = $6.08 average payout per game = mean of x
$6.08 average payout per game - $5.00 buy in per game= $1.08 actual winnings from game
$1.08 actual winnings per game = $1.08 lost by House per game
x = .46(0-6.08)2 +.26(3-6.08)2 +.1(15-6.08)2 +.14(20-6.08)2 +.04(25-6.08)2
σx = $8.29
= $68.87
5. Explain your results
The results we got from our simulation were actually fairly surprising, but not unbelievable. The
results were fairly surprising, because the House ended up losing approximately $1.08 per game,
when theoretically the house should be making $0.78. However, due to the nature of game having
a very high standard deviation in comparison to the amount it “should” be making ($6.65
standard deviation compared to a $0.78 expected profit), the game is likely to have large
fluctuations in payouts. However, if we consider the total amount of people that would probably
be playing Lucky Double Double Digits per year, these fluctuations should approximately cancel
out due to the law of large numbers. However, the fact that the house did lose money in the
simulation brings up the issue of whether the House should design the game to have more house
advantage or not. The House could experiment with lowering the payout, increasing the buy in,
etc. However, I do not think any modifications should be made to the game that would give it
more house advantage. Doing so would drastically decrease the attractiveness of the game, and
less people would be interested in playing it. Overall, despite the house losing money in the
simulation, this game would definitely work in real life.
