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The advent of casinos has warranted the extensive application of statistics and probability to gambling.In this paper, we will consider a simple game between two players and attempt to apply this game todevelop a more intricate casino game.

Let us consider the situation in which two players, Ann and Bob, roll a single die once. If Ann rolls agreater number than Bob, Ann will win. However, if Ann rolls the same number as Bob, she will lose.

To find out the probability that Ann will win, we can construct the following table. Each row in thistable represents a unique combination of rolls for Ann and Bob that result in a victory for Ann.

Anns Number Bobs NumberProbability of

Combination Occurring

2 1!

!!

!

!!

!

!"

3 1!

!!

!

!!

!

!"

3 2

!

!!

!

!!

!

!"

4 1!

!!

!

!!

!

!"

4 2!

!!

!

!!

!

!"

4 3!

!!

!

!!

!

!"

5 1!

!!

!

!!

!

!"

5 2 !!!!

!!

!

!"

5 3!

!!

!

!!

!

!"

5 4!

!!

!

!!

!

!"

6 1!

!!

!

!!

!

!"

6 2!

!!

!

!!

!

!"

6 3!

!!

!

!!

!

!"

6 4!

!!

!

!!

!

!"

6 5!

!!

!

!!

!

!"

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Summing up the probabilities in the rightmost column, we arrive at the value!"

!"!

!

!". Using the table,

we have shown that P(Ann wins) =!

!". Consequently, P(Bob wins) = !!

!

!"!

!

!"!

Now, let us consider a different game. In this game, Ann can roll her die twice and take the greater

number. Bob rolls once. If the greater of Anns rolls is larger than Bobs single roll, Ann will win.Using a table, we can calculate the probability that a given number is the greatest number that Annrolled. The following table represents all possible combinations of Anns dice rolls.

First Roll Second Roll Greater Number First Roll Second Roll Greater Number

1 1 1 4 1 4

1 2 2 4 2 4

1 3 3 4 3 4

1 4 4 4 4 4

1 5 5 4 5 5

1 6 6 4 6 6

2 1 2 5 1 52 2 2 5 2 5

2 3 3 5 3 5

2 4 4 5 4 5

2 5 5 5 5 5

2 6 6 5 6 6

3 1 3 6 1 6

3 2 3 6 2 6

3 3 3 6 3 6

3 4 4 6 4 6

3 5 5 6 5 6

3 6 6 6 6 6

We are concerned with the Greater Number column. We know that there are 36 possible ways to rolla die twice. This is because there are 6 possible outcomes for a single roll. Because there are two rolls,the amount of possible combinations is (6)(6) = 36. Knowing this, we can find the probability of agiven number being the largest number that Ann rolled. For example, according to the table, the

number 3 is the greater number in 5 cases. Thus, the probability of 3 being the greater number is!

!". In

this manner, we can draw up a table of probabilities.

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NumberProbability that the number is the greater

number of two rolls

1!

!"

2!

!"

3!

!"

4!

!"

5!

!"

6!!

!"

With this information, we can calculate the probability that Ann will win this game. We can do this inthe same manner in which we calculated the probability that Ann will win the first game. That is, wecan use a table to list all of the combinations of Anns greatest number and Bobs greatest number thatresult in Ann winning the game. This table is found on the next page.

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Anns Greatest Number Bobs NumberProbability of

Combination Occurring

2 1!

!"!

!

!!

!

!"#

3 1!

!"!

!

!!

!

!"#

3 2!

!"!

!

!!

!

!"#

4 1!

!"!

!

!!

!

!"#

4 2!

!"!

!

!!

!

!"#

4 3!

!"

!

!

!

!

!

!"#

5 1!

!"!

!

!!

!

!"#

5 2!

!"!

!

!!

!

!"#

5 3!

!"!

!

!!

!

!"#

5 4!

!"!

!

!!

!

!"#

6 1!!

!"!

!

!!

!!

!"#

6 2!!

!"!

!

!!

!!

!"#

6 3!!

!"!

!

!!

!!

!"#

6 4!!

!"!

!

!!

!!

!"#

6 5!!

!"!

!

!!

!!

!"#

Summing up the probabilities in the rightmost column, we have:

! !""!"#$ ! !! ! ! ! ! ! ! ! ! ! ! !!!!!"#

!

!"#

!"#

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Therefore, ! !"#!"#$ ! !! !"#!"#!

!"

!"#!

While this table method works for situations in which the players are allowed to roll only a fewtimes, the same method would become increasingly unwieldy for a greater number of rolls. For thisreason, we will derive a formula that calculates the probability that Ann would win the game.

Let the random variable Grepresent the greatest number that is attained after one roll of the die. Theprobability that G is less than or equal tox, wherexis a number between one and 6, inclusive, is:

! ! ! ! !!

!

Where ! ! ! ! !!

The reason for this is that there arexnumbers that are greater than or equal to one, but less thanx.For

example, let us choosex to be 5. If we want the inequality ! ! !to hold, Gcan be 1, 2, 3, 4, or 5.

There is a 5/6 chance for G to be any of these numbers, and so

! ! ! ! !!

!

Now, let us consider the same probability, for nrolls of the die. That is, what is the probability that Gis less than or equal toxif we are allowed to roll the die ntimes? Let us define a success as the event

where ! ! !. We have shown that the probability of a success is simply:

! ! ! ! !!

!

If the inequality ! ! !is to hold true for nrolls, there must be nsuccesses:

! ! ! !!!"#!!"##$ ! !!

!

From this, we can deduce the probability that G=xfor nrolls.

! ! ! !!!"#!!"##$ ! ! ! ! !!!"#!!"##$ ! ! ! ! ! ! !!!"#!!"##$ It is easy to see that the above equation is true. For brevitys sake, the condition that there are nrolls

will be implicit in the following proof. Let us consider the right hand side of the above equation:

! ! ! ! ! ! ! ! ! ! ! !

! ! ! !!"! ! ! ! ! ! ! ! ! ! ! !! !!! ! ! !! ! !!! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! ! ! ! ! ! !

! ! ! ! ! ! ! ! ! ! ! ! ! ! !

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b-axis represents how many times Bob is allowed to roll the die, and the p-axisrepresents theprobability that Ann will win.

In this graph, aand bwere allowed to range from 1 to 20. This choice was made because the amountof rolls by each player cannot be negative. While the maximum amount of rolls can theoretically beinfinite, 20 was arbitrarily chosen to create a presentable graph. As expected, when we move parallel

to the b-axis, in direction that bincreases,pbecomes smaller. This is because as we allow Bob to rollthe die more times, Ann has a lesser chance of victory. Similarly, if we move from any point in theincreasing direction of a, the value ofpincreases because there is a larger chance for Ann to win.

With the summation programmed in Mathematica under the function name p(a,b), it is convenient tocalculate the probability that Ann will win, given the amount of rolls allowed for both Ann and Bob.For example, if we want to know the probability that Ann will win if she is allowed to roll the dietwelve times while Bob is allowed to roll the die five times, we evaluatep(a,b)at (12,5):

Mathematicas N operator allows us to find the decimal approximation.

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Evaluating the function at (1,1), (2,1), and (2,2), we arrive at the same results as we did when wemanually expanded the summation.

Our findings can be applied to create a casino game in which the casinos bank assumes Bobs role.

That is, we can create a game where the casino and the player each roll a die. If the value on thecasinos die is greater than or equal to the value on the players die, the casino wins. Because thecasino has essentially taken Bobs role, we can refer to the casinos greatest number rolled asB, andthe number of rolls allowed as b. Furthermore, the players maximum number rolled isAand thenumber of rolls allowed is a.

If both the casino and player are allowed a single roll, the probability that the player will win is!

!"!as

previously calculated. Let us define new variables. It costs Cdollars for the player to play, andDis thepayoff, or the dollars paid by the casino if the player wins. Note that if the player wins, the playerreceives his original bet of C dollars as well as the D-dollar payoff. Let us define the banks payoff asX. The banks expected payoff per roll,E(X)is:

! ! ! ! !"#$%&!"#$ ! !"#$!"!"#$ ! ! !"!"#$!"#$ !!!"#$%%!"!"#$%&!

! ! !!

!"! !

!

!"!

The casino wants the expected payoff to be positive. That is, ! ! ! !!This would mean that:

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!

!"! !

!

!"! ! !

!

!"! !

!

!"!

! !!

!!

For example, if it costs $10 to play, we have the following inequality:

!" !!

!!

!" ! !

If the casino chooses to pay the winner anything less than $14, it will make a profit in the long run.

The casino wants to charge enough money to make a reasonable profit, but at the same time, the gamemust be attractive to players. With external research, it was found that the casinos expected gain onthe money the players pay is called the casino edge. For Blackjack, the casino edge is approximately8% (Blackjack Probability). For our game, we can target an edge of 10%. That means our expectedpayoff for each roll should be:

! ! ! !"#!"#$ ! !"#!"!"#$ ! !! ! !" ! !"Substituting this forE(X),and $10 for C, we have:

! ! !!

!"

! ! !

!"

!

!" !!

!"!"#!

!

!"!

! !!"

!"!

!

!"!

!

!"

!"!

!

!

!"!

! !!"

! ! !""!!"

Therefore, we can see that to provide a reasonable game, the casino can maintain a cost to payoff ratioof 10:11.6, or equivalently, 1:1.16.

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Now, let us create a new game. We preserve the basic rules; the casino wins if it rolls a highest numbergreater than or equal to the players highest number. This time, however, we allot the casino three dicerolls. The reason for this is psychological; even if the player has 7 rolls, his chance of winning is still:

Yet, the player is allowed over twice as many rolls as the casino is. To a non-mathematician, thereseems to be an extremely large chance to win. Furthermore, it is more difficult to calculate theprobability of winning with the said distribution of rolls; in the case where both the player and casinohad a single roll, any sensible player could have determined their expected payoff. In this case, thecalculations are much more difficult and based on the summation we have derived.

We can allow the player to choose how many times he is allowed to roll. If we set the target edge ofthe casino to 10%, we can calculate what the payoff ratio should be for each number of rolls the playercan choose. Again, let us use the previously defined variables C and D. Furthermore, let PCbe the

probability that the casino wins and PPbe the probability that the player wins.

! ! ! ! !"#$%&!"#$ ! !"#$!"!"#$ ! ! !"#$%&!"#$ !!!"#$%%!"!"#$%&!! ! ! !!!! ! !!!!

This time, however, we know thatE(X)for each roll should be the edge (10%) multiplied by the cost toplay.

!!!! ! !!!! ! !!!!

!!

!! ! !!!! ! !!!!

!!!! !!!!

! ! !!

!!!!

! ! !!! !!!

!

! !

!!! !!!

!!

! ! !!!

!!

!

!

!!

Therefore, the cost to payoff ratio is:

!!!

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!!!!!! !!!

!!

!!!!! !!!

!!

Knowing this, excel was used to construct a table with payoff ratios for a dice game where acan be amaximum of 7 and bis always 3. Note that Mathematica was used to calculate Pp, using the functionP(a,b)that we previously defined.

!"#$%& () *(++, )(& -+./%&

0.1-2 3 -0.451 -6 3 78-2

!! ! !!!

!!

Cost to

Payoff

Ratio,

rounded to

the nearest

0.05

Cost to

Play, with a

Payoff of

$50

! #$!%& #$'() &$!%(&!*%+* !,&$!- .!($#&( #$('! #$%!+ ($(#('&)+%- !,($(# .(($%(

* #$*-( #$)&' !$--)'!'!'( !,!$-- .*($(-

& #$* #$-+% !$(**(-#)( !,!$(- .$##

- #$& #$-) !$#&-&-&-&- !,!$#- .&%$)(

) #$&)' #$-*( #$+(*#%)+(* !,#$+# .--$-)

% #$&'+ #$-!! #$'&+#%+' !,#$'- .-'$'(

This table can be used in many different ways to create various games. A particularly devious game

would progress as follows:

1. The payoff is set at some number. For simplicitys sake, lets call it $50. The player can chooseto purchase up to four dice, according to the payoff table above. For example, the cost to payoff

ratio associated with rolling three dice is 1.55. Therefore, the player should pay!"#

!!!!! !"#to

purchase 3 dice. The player then rolls the amount of dice he purchased. If his greatest numberis greater than the greatest number on the casinos dice, he wins.

2. If the player wins, he receives the money he bet, along with the promised $50 payoff.3. If the player loses, he may ask for a hit. That is, he may choose to play again, but only for a

$100 pot. The cost to purchase dice would be adjusted accordingly. However, upon losing thefirst dice roll, the player unlocks the next tier of permissible dice rolls; if he chooses to hit, he

can choose to purchase up to six dice while the casino rolls three again.4. If the player wins this time, he receives $100, along with the money he bet on the $100 pot.5. If the player loses, he can hit one more time. This time, he must play for a $200 pot. On the

second hit, he unlocks 7thdie and can choose to roll 7 dice if he wants to.6. If he wins, the player receives $200, as well as the money he bet on the $200 pot.

This game is deceptive and sneaky. If the gambler loses the first roll, he will immediately feel theurge to win his money back. At the same time, the casino will immediately present him the

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opportunity to do so with better odds than the previous gamble; the more money the gambler haslost, the better the chance becomes to make all that money back. Furthermore, the odds of agambler losing three times in a row are actually quite low. Let us assume that the gambler choosesto roll 4 dice, then 6 dice, then 7 dice.

! !"#$%&!"#$!""!!!""!"#$% ! !!!"#!!!!"#!!!!"" ! !!!"#This will ensure that the morale at the table is never too low. Yet, at the same time, each roll isdesigned for the casino to have a 10% edge. In the long run, the casino will make money.

As we have seen, an application of statistics to a simple dice game allowed us to construct a muchmore interesting and intricate scenario.

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Works Cited

"Blackjack Probability and Blackjack Odds."Blackjack Strategy. Web. 12 Feb. 2013.