GAM 470 Test #1

February 22, 2005

Name______________________________

All problems are worth 4 points each, except the last two, which are 6 points each.

Problems 1 to 6: An octahedron is an 8-sided die. Imagine a craps game in which two octahedrons are rolled.

What is the probability the two dice have a total of 3?

A 3 can be rolled two ways: {1,2} and {2,1}. There are a total of 8²=64 combinations. So the probability is 2/64 = 1/32 = 3.13%.

If a total of 3 pays 25 to 1 what is the expected value?

Pr(win)*win – Pr(loss) =

(1/32) * 25 – (31/32) =

-6/32 = -3/16 = -18.75%.

A place bet may be made that a total of 12 is rolled before a total of 9. What is the probability of winning?

Ways to roll as 12: {4,8}, {5,7}, {6,6}, {7,5}, {8,4}, total =5

Ways to roll a 9: {1,8}, {2,7}, {3,6}, {4,5}, {5,4}, {6,3}, {7,2}, {8,1}, total =8

The probability of rolling a 12 before a 9 is 5/(5+8) = 5/13 = 38.46%.

The place bet on a 12 before a 9 pays 3 to 2. What is the house edge?

Expected value = Pr(win)*win – Pr(loss) =

(5/13)*(3/2) – (8/13) =

15/26 – 16/26 =

-1/26

House edge = Expected Value * -1 = 1/26 = 3.85%.

The player may bet on over or under a total of 9. What is the probability the total is under 9?

The average roll per die is (1+2+3+4+5+6+7+8)/8 = 4.5. An easy way to find the sum of 1+2+…+n = n*(n+1)/2, which in this case is 8*9/2=36.

So the average of two dice is 4.5*2 = 9. The rest of the totals must obviously fall half and half above and below 9.

The probability of a total of 9 is 8/64 = 1/8 (see problem 3 for the 8 ways to roll a 9). So the probability of under 9 is (1-(1/8))/2 = (7/8)/2 = 7/16 = 43.75%.

The under 9 bet pays 1 to 1. What is the expected value?

Expected value = Pr(win)*win – Pr(loss) =

(7/16) * 1 – (9/16) = -2/16 = -1/8 = -12.5%.

Problems 7-12: Consider a Sic Bo game with four six-sided dice.

The player may bet on a four of a kind. The bet wins if all four dice are the same face. What is the probability of winning?

Winning combinations = 6

Total combinations = 6⁴=1296

Probability = 6/1296 = 1/216 = 0.46%.

Alternatively, the probability the next three dice match the first one is (1/6)³ = 1/216 = 0.46%.

The four of a kind bet pays 200 to 1. What is the house edge?

Expected value = Pr(win)*win – Pr(loss) =

200*(1/216) – (215/216) = -15/216 = -6.94%.

House edge = -1*Expected Value = +6.94%.

The player may also bet on a two pair. The bet wins if two different numbers appear, each twice. For example {3,3,5,5}. What is the probability of winning?

There are combin(6,2)=6!/(4!*2!) = 15 ways to choose 2 numbers out of 6. Then there are combin(4,2)=4!/(2!*2!) = 6 ways to order them: xxyy, xyxy, xyyx, yyxx, yxyx, yxxy.

Winning combinations = 15*6=90.

Total combinations = 6⁴=1296.

Probability = 90/1296 = 15/216 = 6.94%.

Alternatively, if the first die landed on x then following are the ways the next 2 rolls can result in a 2-pair:

x,y,y

y,x,y

y,y,x

It stands to reason the probability of all three of these would be equal. So it would be enough to work out the probability of just one way and multiply by 3. The probability of rolling x,y,y is (1/6)*(5/6)*(1/6) = 5/216. Multiplying by 3 the total probability is 15/216.

The two pair bet pays 12 to 1. What is the expected value?

Expected value = Pr(win)*win – Pr(loss) =

(15/216)*12 – (201/216) = -21/216 = -7/72 = -9.72%.
Consider a bet in which the player picks any number from 1 to 6. The player is paid according to the number of times his number appears according to the following table.

Appearances	Pays
0	Loss
1	Push
2	2 to 1
3	10 to 1
4	50 to 1

11. What is the probability of winning per roll?

Probability of 0 appearances: combin(4,0)*(1/6)⁰*(5/6)⁴ = 625/1296 = 48.23%.

Probability of 1 appearance: combin(4,1)*(1/6)¹*(5/6)³ = 500/1296 = 38.58%.

Probability of 2 appearances: combin(4,2)*(1/6)²*(5/6)² = 150/1296 = 11.57%.

Probability of 3 appearances: combin(4,3)*(1/6)³*(5/6)¹ = 20/1296 = 1.54%.

Probability of 4 appearances: combin(4,4)*(1/6)⁴*(5/6)⁰ = 1/1296 = 0.08%.

Probability of winning = (150+20+1)/1296 = 171/1296 = 13.19%.

12. What the house edge of the bet in problem 11?

Expected value = (625/1296)*-1 + (500/1296)*0 + (150/1296)*2 + (20/1296)*10 + (1/1296)*50 = (-625+0+300+200+50)/1296 = -75/1296 = -5.79%.

House Edge = -1*Expected Value = -1*-5.79% = +5.79%.
Problems 13-15: Consider a drawer with 6 red socks, 4 green socks, and 2 yellow socks.

13 You pull out three socks at random without replacement. What is the probability you pick one of each color?

Winning combinations = 6*4*2=48.

Total combinations = combin(12,3) = 12!/(9!*3!) = 220.

Probability = 48/220 = 24/110 = 12/55 = 21.82%.

Alternatively there are 3!=6 ways to order three socks of different colors. The probability of picking any one (red, green, yellow for example) is (6/12)*(4/11)*(2/10) = 48/1320 = 3.64%. Then multiply by 6 for all possible orders: 6*(48/1320) = 21.82%.

14 You pull out four socks at random without replacement. What is the probability you get a two pair, in other words two different colors of two socks each?

RRGG combinations: combin(6,2)*combin(4,2) = 15*6 = 90.

RRYY combinations: combin(6,2)*combin(2,2) = 15*1 = 15.

GGYY combinations: combin(4,2)*combin(2,2) = 6*1 = 6.

Total winning combinations = 90+15+6 = 111.

Total combinations = combin(12,4) = 495.

Probability = 111/495 = 22.42%.

15 You pull out six socks at random and without replacement. What is the probability you pull out exactly three reds?

Winning combinations = combin(6,3)*combin(6,3) = 20*20 = 400.

Total combinations = combin(12,6) = 924

Probability = 400/924 = 200/462 = 100/231 =43.29%.
Problems 16-17: Another sock drawer has 5 blue socks, 5 orange socks, and 5 purple socks.

16 You pull out three socks at random and without replacement. What is the probability you have at least 2 of the same color?

All three socks same color combinations: 3*combin(5,3)= 3*10 = 30.

Two socks same color combinations: There are 3 ways to pick the color with 2 socks and 2 left for the color with 1 sock. Since all sets are the same size the number of ways to pick 2 socks out of 5 for the 2-sock color is combin(5,2)=10. There are combin(5,1)=5 ways to pick the sock of one color. So the total 2 & 1 combinations are 3*2*10*5=300.

Total winning combinations = 30+300 = 330.

Total combinations = combin(15,3) = 15!/(12!*3!) = 455.

Probability = 330/455 = 60/91 = 72.53%.

Alternatively, the probability of getting one color of each is 5³/combin(15,3) = 125/455 = 25/91 = 27.47%. So the probability of NOT getting three different colors is 1-(25/91) = 60/91 = 72.53%.

17 You pull out five socks at random and without replacement. What is the probability you get a full house (3 of one color and 2 of another)?

There are 3 ways to pick the 3-sock color and 2 ways left to pick the 2-sock color. Note that you can't do combin(3,2) because the sizes are different. Then there are combin(5,3)=10 ways to pick 3 socks out of 5 for the 3-sock color, and combin(5,2)=10 ways to pick 2 socks out of 5 for the 2-sock color. So the total winning combinations are 3*2*10*10=600. The total combinations are combin(15,5) = 3003. So the probability is 600/3003 = 200/1001 = 19.98%.

Alternatively, one way to get a full house would be to pick BBBOO in that order. That probability is (5/15)*(4/14)*(3/13)*(5/12)*(4/11) = 0.00333. However there are 3 ways to pick the 3-sock color and 2 for the 2-sock color, so multiply by 6, resulting in 6*0.00333=0.01998. Then there are combin(5,3)=10 ways to order the 3 red socks out of the 5 drawn. So multiply again: 0.01998*10 = 0.1998.
Problems 18-22: You have a deck of cards consisting of 13 ranks of 5 suits each. Dealing five cards at random and without replacement what is the probability of the following hands?

18 A full house

Ranks: 13*12 = 156

Suits: combin(5,3)*combin(5,2)=10*10=100

Total full house combinations=156*100=15600.

Total combinations = combin(65,5) = 8,259,888.

Probability = 15600/8,259,888 = 7800/4129944 = 3900/2064972 = 1950/1032486 = 975/516243 = 325/172081 = 0.19%.

19 A straight (aces count high or low). Straight does not include straight flush.

Ranks: 10 possible spans (A-5) to (10-A)

Suits: 5^5-5 = 3125-5 = 3120.

Total straights: 10*3120 = 31200.

Probability = 31200/8259888 = 0.03777%.

20 A three of a kind

Ranks: 13 (for the 3 of a kind), plus combin(12,2) = 66 for the two singletons. 13*66=858.

Suits: combin(5,3)=10 for the three of a kind. 5 for each singleton. 10*5*5=250.

Total three of a kinds = 858*520=214,500.

Probability=214500/8259888 = 2.60%.

21 A flush. Flush does not include straight flush.

Ranks: combin(13,5)-10 = 1287 – 10 – 1277.

Suits: 5

Total flushes: 1277*5=6385.

Probability = 6385/8259888 =0.0773%.

22 A two pair

Ranks: combin(13,2)=78 for 2 ranks out of 13 for the two pairs. There are 11 ranks left for the singleton. 78*11=858.

Suits: combin(5,2)=10 for each pair. 5 for the sington. 10*10*5=500.

Total two pairs = 858*500 = 429,000.

Probability 429,000/8259888 = 5.19%.

Problems 23-24: The slot machine Test Anxiety has the following pay table.

Payline	Pays
3 clocks	500
3 coffee cups	100
3 books	50
3 pencils	40
3 notes	30
3 chairs	20
3 calculators	5
Any 2 clocks	5
Any 1 clock	2

Following is how each symbol is weighted on each reel.

Symbol	Reel 1	Reel 2	Reel 3
coffee cup	2	1	1
book	2	2	1
pencil	2	2	2
notes	3	2	2
chair	4	4	4
calculator	4	6	7
clock	1	1	1
total	18	18	18

Fill in the table below to determine the hit frequency and the expected return.

Payline	Pays	Probability Combinations	Return Combinations
3 clocks	500	1	500
3 coffee cups	100	2	200
3 books	50	4	200
3 pencils	40	8	320
3 notes	30	12	360
3 chairs	20	64	1280
3 calculators	5	168	840
Any 2 clocks	5	51	255
Any 1 clock	2	867	1734
Total		1177	5689

What is the hit frequency? (6 points)

1177/18³ = 1177/5832 = 20.18%.

What is the expected return? (6 points)

5689/18³ = 5689/5832 = 97.55%.