Introduction- In the real world events cannot be predicted with total certainty. The best way in which we can do this is by saying that how likely is an event to happen, using the idea of probability. Probability is the branch of mathematics, which studies the possible outcomes of given events together with the outcomes it also studies the relative likelihoods and distributions. In Lehman terms, the word probability means the chance that a particular event {or set of events) will occur expressed on a linear scale from 0 {impossibility) to 1 {certainty), also expressed as a percentage between 0 and 100%.

Considering a game with two players, Ann and Bob. Ann has a red die and Bob a white die. They roll their dice and note the number on the upper face. Ann wins if her score is higher than Bobs and Bob wins if the scores are the same. This can be summarized as follows: Number of Ann is greater than the number of Bob = Ann wins Number of Ann is less than equal to number of Bob = Bob wins Now considering the probabilities if both players roll their dice once each. To find the probabilities of both plays we can use a probability tree diagram, which is shown below:

The above tree diagram shows the probabilities and the winner of each game where the color red represents Anns wins and bobs wins are represented by the color blue. For the following results to be obtained we make an assumption that both player are rolling the same, fair die each once. The above probability results can also be obtained using a space diagram that is shown below: Possible outcomes 1 2 3 4 5 6 1 Bob wins Bob wins Bob wins Bob wins Bob wins Bob wins 2 Ann wins Bob wins Bob wins Bob wins Bob wins Bob wins 3 Ann wins Ann wins Bob wins Bob wins Bob wins.

Bob wins 4 Ann wins Ann wins Ann wins Bob wins Bob wins Bob wins 5 Ann wins Ann wins Ann wins Ann wins Bob wins Bob wins 6 Ann wins Ann wins Ann wins Ann wins Ann wins Bob wins The above space diagram shows the probabilities and the winner of each game where the color red represents Anns wins and bobs wins are represented by the color blue. For the following results to be obtained we make an assumption that both player are rolling the same, fair die each once. The probabilities for the above events presented in the space as well as the tree diagram are calculated below.

As we already know that the general formula which is as follows: We use the above formula to calculate the probabilities for occurring events. The calculations are as follows: As we have made an assumption that the die being used is a fair die the total number of outcomes (nT) = 36 And the number of ways in which each event can occur (nW)= 6 So, Thus, P (1) P (2) P (3) P (4) P (5) P (6) Thus we can calculate the sum of the above probabilities as 1, which is a condition for a fair die to be rolled.

We can conclude that as the above probabilities are of the first throw by Ann and that as bob is allowed to throw the die once this means that as they throw the same fair die the probabilities for each number will remain same. If we reconsider the situation given to us where, Number of Ann is greater than the number of Bob = Ann wins Number of Ann is less than equal to number of Bob = Bob wins By looking at the space diagram we can conclude that the number of times Ann wins is 15 and as the total number of outcomes is 36 the probability that Ann wins is.

Since we already know that the sum of the probabilities is equal to 1 so, we can calculate bobs probability by using the formula If a game with two players, Ann and Bob. Ann has a red die and Bob a white die that are fair. They roll their dice and note the number on the upper face. Ann wins if her score is higher than Bobs then her probability of winning is and Bob wins if the scores are the same thus bobs probability of winning is Now we consider the same game but where Ann can roll her die a second time and will note the higher score of the two rolls but Bob rolls only once. All the assumptions made above still apply to the game.

To advance towards solving the problem presented to us above we first find out the probabilities of Ann of rolling the die twice as her highest score using a space diagram shown below: Anns possible number of outcomes 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6 2 6 3 6 4 6 5 6 6 The color green represents Anns first roll and the color orange shows Anns second roll the highest score chosen from Anns rolls is shown in the table below in the color red and highlighted by the color yellow. 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2

6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6 2 6 3 6 4 6 5 6 6 Using this space diagram above we can calculate Anns probabilities using the same formula we used in the first case as we assume that Ann is using a fair Die she can only obtain the outcomes of 1,2,3,4,5, or 6. To calculate the probability we use the data from the table and the formula, P (when Anns highest score = 1) this is because the total number of Outcomes = 36 and the highest score =1 is only obtained once as we can see in the table above. P (when Anns highest score = 2) this is because the total number of

Outcomes = 36 and the highest score =2 is only obtained thrice as we can see in the table above. P (when Anns highest score = 3) this is because the total number of Outcomes = 36 and the highest score =3 is only obtained five times as we can see in the table above. P (when Anns highest score = 4) this is because the total number of Outcomes = 36 and the highest score =4 is only obtained seven times as we can see in the table above. P (when Anns highest score = 5) this is because the total number of Outcomes = 36 and the highest score =5 is only obtained nine times as we can see in the table above.

P (when Anns highest score = 6) this is because the total number of Outcomes = 36 and the highest score =6 is only obtained eleven times as we can see in the table above. Using the probabilities above we can now construct a new tree diagram adapted from the first case and making an assumption that bob rolls a fair die and can only obtain outcomes of 1 or 2 or 3 or 4 or 5 or 6. Possible outcomes 1 2 3 4 5 6 1 Bob wins Bob wins Bob wins Bob wins Bob wins Bob wins 2 Ann wins Bob wins Bob wins Bob wins Bob wins Bob wins 3 Ann wins Ann wins Bob wins Bob wins Bob wins Bob wins 4 Ann wins.

Ann wins Ann wins Bob wins Bob wins Bob wins 5 Ann wins Ann wins Ann wins Ann wins Bob wins Bob wins 6 Ann wins Ann wins Ann wins Ann wins Ann wins Bob wins Using the tree diagram and the space diagram above we can calculate the probability of the events in which Ann wins, this can be presented in the table below: Anns highest outcomes from two rolls Bobs Outcomes Probability of the event (Ann wins) 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 Using the above found probabilities in the table the sum of these probabilities gives the probability when Ann wins the game.

P (Ann wins) = Since we already know that the sum of the probabilities is equal to 1 so, we can calculate bobs probability by using the formula If we consider the same game where Ann can roll her die a second time and will note the higher score of the two rolls but Bob rolls only once and all the assumptions made above still apply to the game then Anns probability of winning the game as compared to Bob increases as shown above. Now considering a situation in which both players can roll the die twice and all the assumptions made above still apply to the game.

As the highest score obtained from the two rolls is to be chosen as a result we can make an assumption that Bobs possible outcomes are same as Anns presented in the table below. 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6 2 6 3 6 4 6 5 6 6 Using the table above we create another table, which tells us the outcomes in which Ann wins the game in the situation provided to us. 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6.

2 6 3 6 4 6 5 6 6 The color red represents the outcomes Ann needs to get to win the game when Bob gets the outcomes represented by the color blue. As we have already calculated the probability of Ann getting Specific outcomes in the earlier situation provided to us, we can say that Bobs probability of getting the specific outcomes is same as Anns. So the probability of Ann winning the game in this situation can be shown with the help of the table below: Anns highest outcomes from two rolls Bobs Outcomes Probability of the event (Ann wins) 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 Using the above found probabilities in the table the sum of these probabilities gives the probability when Ann wins the game P (Ann wins) = Since we already know that the sum of the probabilities is equal to 1 so, we can calculate bobs probability by using the formula If we considered a situation in which both players can roll the die twice and all the assumptions made above still apply to the game. And As the highest score obtained from the two rolls is to be chosen as a result we can make an assumption that Bobs possible outcomes are same as Anns.we can say that Bobs probability of winning the game when the Die is rolled twice by each player is higher than Anns.

Now to investigate the probabilities of Ann and Bob to win the game if they roll the die a multiple times, we formulate a general form or a general formula assuming that each of the player takes only the highest score into account and all the assumptions discussed in the above tasks remain the same. For this first we construct a table showing the possible outcomes and their probabilities in multiple rolls. The table is as follows: Possible outcomes 1st roll 2nd roll 3rd roll 1 2 3 4 5 6.

So, we come up with a formula, which states the probability of a number/ event X to the highest in n rolls: We get as the numerator; we can show this through the use of an example illustrated below: To find the probability of getting a 3 in the 2nd roll, it can be found using the probabilities of the outcomes before 3 they are 1 and 2. So the numerators of the probabilities of 1 and 2 are 1 and 3 respectively So when we add 1 and 3 we get 4, which is a square of 2 we, can derive this by using the variable X where X is the outcome for which we need to find the probability. Here X is 3. So as 4 is 2 square we can derive it as where n=2.

So to get a 5 which is the numerator of the probability of getting a 3 can be found using the formula Where n= the number of the roll and X= the outcome for which the probability needs to be calculated. If we start looking at the denominator it is defined as the total number of possible outcomes. In the case of a die, it is always equal to This can be explained by referring back to the probability tree diagram, where in case that a die that is thrown, the denominator will always be multiplied by 6 when the die is thrown an additional time. Another way in which we can show this is by using a sequence for the numerator.

First we take an example for the 2nd roll for which we get an arithmetic sequence for the numerator of the probabilities using the table above as we can rearrange this formula to generalize it for n rolls as the proof can be illustrated as follows: For X=3 and n=2 This can be verified from the table we can find the formula for the Arithmetic sequence using this is as follows. Taking n=2 We can verify this for n=3 whose formula for the arithmetic sequence = This can be verified from the table we can find the formula for the Arithmetic sequence using this is as follows.

Now to generalize this formula for multiple number of rolls by Ann and Bob we use the general form we found above: Assuming that we have to find the probability that Ann wins in multiple rolls Assuming that Ann wins so her highest score in multiple rolls = X So Bobs highest score when Ann wins= Assuming that Ann has n throws and Bob has m throws and that n and m can be equal to infinite. So, the probability that Ann wins That is, We have used the sigma sign, which stands for sum for calculating Bobs probability because for Ann to win bob can have any score less than And can have any value of

To prove the above formula we need to expand it the proof is as follows: First we expand the following term: The expansion is as follows where X= 1 By simplifying the expansion we get the following: The expansion is as follows where X= 2 By simplifying the expansion we get the following: The expansion is as follows where X= 3 By simplifying the expansion we get the following: The expansion is as follows where X= 4 By simplifying the expansion we get the following: The expansion is as follows where X= 5 By simplifying the expansion we get the following:

The expansion is as follows where X= 6 By simplifying the expansion we get the following: We do the same for Bob using the equation : For all cases in which Ann wins: Possible Outcomes for Ann Possible Outcomes for Bob 2 1 3 2,1 4 3,2,1 5 4,3,2,1 6 5,4,,3,2,1 Now to find Anns winning probability in multiple rolls we need to find the sum of the Products of the expanded equations of bob and Ann where Ann is represented by A and bob is represented by B. For X=1 For X=2 For X=3 For X=4 For X=5 For X=6 This can be written as the following:

Where, = = = = = So the above-expanded equations can be written as, The situation given to us can now be applied to a dice game in a casino where we are assuming that the casino/ bank is using a fair die and the player playing the game. We can also make an assumption that the casino/ bank has the same probability of winning the game as Bob whose probability of winning was calculated in the earlier situations given to us. Thus, as a result we can say that the bank/ casino wins every 21 games out of a total of 36 games.

And We can also assume that as the casino has the same probability of winning as Bob, the player playing the game has the same probability of winning as Ann that is the player will win every 15 games out of a total of 36 games as we have calculated earlier. Taking the above probabilities into account we can consider an example, which will help us formulate a general term for the game between the casino and the player. We assume that to enter each game a player has to pay $50, which he/she will get back with an amount of a profit.

Taking the probabilities into account we assume that the total number of outcomes is 36 so, as a result the player has to pay a total of to the casino or the bank. As we already know that the if the player wins he /she gets back the entry fee of the game plus an amount of a profit. This winning amount can be calculated using the players wining probability that is So the players winning amount per game = $ 120 assuming that the player plays a total of 36 games. We can conclude that the player makes a profit of $70 per game he/she wins. This would mean that the casino would neither make a loss nor a profit.

For the casino to make a profit they should ensure that the amount of profit should be less that $70. This can be summarized as follows: For the player the profit should be For the casino to make a profit the players profit should be As the above model we have presented is based only upon probabilities it may not always be true, it is merely dependent upon the number of games the player plays. Looking at the game from the perspective of both players the game can be considered worthwhile for both the player and the casino this is because the casino will always make a profit as long as enough games are played that is ?

36 As they have to present the player with a reward every time he/she wins the case chosen the reward equals to $70. The game can be considered worthwhile for the player because he/she has reasonably high chance of winning and if the player wins he/she makes a profit of $70 from the casino, as this is higher than the entrance fee of a game the player will be willing to play another game and even if he/she loses this time he/she will still make a profit of $20.

Now using the example above we are going to create model for a game so that the casino makes a reasonable profit in the case where the player rolls the red die once and the bank rolls the white die once. For the creation of the model we need to define certain variables they are as follows: Entry fee for the game E The money the player gets when he wins (entry fee + profit) or the money the casino needs to pay to the player. T As shown in the above example where E= $50 the money that the player wins (T) if he/she wins is found using the probability of Ann winning when she rolls the Die once found in the earlier parts and used in the example above that is.

So to find T we use the probability and simple unitary method. First to enjoy the benefit of unitary method we find the total entry fee payed by the player assuming the total possible number of outcomes is 36 or the total game played is 36. Total entry fee As the player wins the game only 15 times using unitary method we divide the total entry fee by 15 to find T.

And to find the money earned by the casino/bank we simply divide the total money entry fee of the player by 21 since the probability that the casino/bank wins is Now to conclude our investigation we need to find the probability of that the player wins and that the casino wins when there can be multiple number of throws that is that the number of games the player plays is not fixed and that the number of times the casino rolls the dice can be different.

As we assumed in the above investigation that the probability that the player wins is equal to the probability that Ann wins. So we use the same formula as we used to calculate that Ann wins when the number of rolls can be multiple, it is as follows:

where m= number of games casino plays and n= number of games player plays. By investigating the case of Ann and Bob we came to a conclusion that Bob wins when his outcomes are more than or equal to that of Ann this same assumption applies in the case of the player and the casino where the player is Ann and the casino is Bob. Taking this assumption into account we formulate an equation which calculates the maximum money the player gets when he wins (entry fee + profit) or the money the casino needs to pay to the player.

The equation is as follows: That is, We make such a formula because determines how many games the player wins from a total of n games and we multiply the formula by E because to know the total money the player has payed to the casino/bank, that is the total money in the game. Dividing this by the number of games the player actually won we get the maximum money the player gets when he wins (entry fee + profit) or the money the casino needs to pay to the player.

For this formula to work we need to keep in mind certain assumptions they are as follows: Looking at the game from the perspective of both players the game can be considered worthwhile for both the player and the casino this is because the casino will always make a profit as long as enough games are played that is ? 36 As they have to present the player with a reward every time he/she wins the case chosen the reward equals to $70.

The game can be considered worthwhile for the player because he/she has reasonably high chance of winning and if the player wins he/she makes a profit of $70 from the casino, as this is higher than the entrance fee of a game the player will be willing to play another game and even if he/she loses this time he/she will still make a profit of $20. Summary/ conclusion This investigation has analyzed various dice games, starting from the simple case where two players Ann and Bob roll one die each, with Ann winning if their rolled number is larger than that of player Bob.

Player further expanded this game to include the case in which the greater of two rolls made Ann was counted against the roll made by player Bob, the case in which each player was able to roll the die twice, and the case where the greatest of three rolls made by player A was counted against one roll made by player Bob. A general formula for player Anns chances of winning were determined; not however that this formula is limited to the instance in which all dice are fair and six sided, and each face on the each dice is assigned a different number between one and six.

Using this information, possible payouts and player entry fees were modeled for some casino games built on the dice game template. This investigation has made use of a custom-made program on Microsoft Excel to calculate probabilities using the general formula. Another program was used to count the probability of player Ann winning in the variations of the dice game in which both players roll twice and in which player Ann rolls three times but player Bob rolls once.

Considering a game with two players, Ann and Bob. Ann has a red die and Bob a white die. They roll their dice and note the number on the upper face. Ann wins if her score is higher than Bobs and Bob wins if the scores are the same. This can be summarized as follows: Number of Ann is greater than the number of Bob = Ann wins Number of Ann is less than equal to number of Bob = Bob wins Now considering the probabilities if both players roll their dice once each. To find the probabilities of both plays we can use a probability tree diagram, which is shown below:

The above tree diagram shows the probabilities and the winner of each game where the color red represents Anns wins and bobs wins are represented by the color blue. For the following results to be obtained we make an assumption that both player are rolling the same, fair die each once. The above probability results can also be obtained using a space diagram that is shown below: Possible outcomes 1 2 3 4 5 6 1 Bob wins Bob wins Bob wins Bob wins Bob wins Bob wins 2 Ann wins Bob wins Bob wins Bob wins Bob wins Bob wins 3 Ann wins Ann wins Bob wins Bob wins Bob wins.

Bob wins 4 Ann wins Ann wins Ann wins Bob wins Bob wins Bob wins 5 Ann wins Ann wins Ann wins Ann wins Bob wins Bob wins 6 Ann wins Ann wins Ann wins Ann wins Ann wins Bob wins The above space diagram shows the probabilities and the winner of each game where the color red represents Anns wins and bobs wins are represented by the color blue. For the following results to be obtained we make an assumption that both player are rolling the same, fair die each once. The probabilities for the above events presented in the space as well as the tree diagram are calculated below.

As we already know that the general formula which is as follows: We use the above formula to calculate the probabilities for occurring events. The calculations are as follows: As we have made an assumption that the die being used is a fair die the total number of outcomes (nT) = 36 And the number of ways in which each event can occur (nW)= 6 So, Thus, P (1) P (2) P (3) P (4) P (5) P (6) Thus we can calculate the sum of the above probabilities as 1, which is a condition for a fair die to be rolled.

We can conclude that as the above probabilities are of the first throw by Ann and that as bob is allowed to throw the die once this means that as they throw the same fair die the probabilities for each number will remain same. If we reconsider the situation given to us where, Number of Ann is greater than the number of Bob = Ann wins Number of Ann is less than equal to number of Bob = Bob wins By looking at the space diagram we can conclude that the number of times Ann wins is 15 and as the total number of outcomes is 36 the probability that Ann wins is.

Since we already know that the sum of the probabilities is equal to 1 so, we can calculate bobs probability by using the formula If a game with two players, Ann and Bob. Ann has a red die and Bob a white die that are fair. They roll their dice and note the number on the upper face. Ann wins if her score is higher than Bobs then her probability of winning is and Bob wins if the scores are the same thus bobs probability of winning is Now we consider the same game but where Ann can roll her die a second time and will note the higher score of the two rolls but Bob rolls only once. All the assumptions made above still apply to the game.

To advance towards solving the problem presented to us above we first find out the probabilities of Ann of rolling the die twice as her highest score using a space diagram shown below: Anns possible number of outcomes 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6 2 6 3 6 4 6 5 6 6 The color green represents Anns first roll and the color orange shows Anns second roll the highest score chosen from Anns rolls is shown in the table below in the color red and highlighted by the color yellow. 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2

6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6 2 6 3 6 4 6 5 6 6 Using this space diagram above we can calculate Anns probabilities using the same formula we used in the first case as we assume that Ann is using a fair Die she can only obtain the outcomes of 1,2,3,4,5, or 6. To calculate the probability we use the data from the table and the formula, P (when Anns highest score = 1) this is because the total number of Outcomes = 36 and the highest score =1 is only obtained once as we can see in the table above. P (when Anns highest score = 2) this is because the total number of

Outcomes = 36 and the highest score =2 is only obtained thrice as we can see in the table above. P (when Anns highest score = 3) this is because the total number of Outcomes = 36 and the highest score =3 is only obtained five times as we can see in the table above. P (when Anns highest score = 4) this is because the total number of Outcomes = 36 and the highest score =4 is only obtained seven times as we can see in the table above. P (when Anns highest score = 5) this is because the total number of Outcomes = 36 and the highest score =5 is only obtained nine times as we can see in the table above.

P (when Anns highest score = 6) this is because the total number of Outcomes = 36 and the highest score =6 is only obtained eleven times as we can see in the table above. Using the probabilities above we can now construct a new tree diagram adapted from the first case and making an assumption that bob rolls a fair die and can only obtain outcomes of 1 or 2 or 3 or 4 or 5 or 6. Possible outcomes 1 2 3 4 5 6 1 Bob wins Bob wins Bob wins Bob wins Bob wins Bob wins 2 Ann wins Bob wins Bob wins Bob wins Bob wins Bob wins 3 Ann wins Ann wins Bob wins Bob wins Bob wins Bob wins 4 Ann wins.

Ann wins Ann wins Bob wins Bob wins Bob wins 5 Ann wins Ann wins Ann wins Ann wins Bob wins Bob wins 6 Ann wins Ann wins Ann wins Ann wins Ann wins Bob wins Using the tree diagram and the space diagram above we can calculate the probability of the events in which Ann wins, this can be presented in the table below: Anns highest outcomes from two rolls Bobs Outcomes Probability of the event (Ann wins) 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 Using the above found probabilities in the table the sum of these probabilities gives the probability when Ann wins the game.

P (Ann wins) = Since we already know that the sum of the probabilities is equal to 1 so, we can calculate bobs probability by using the formula If we consider the same game where Ann can roll her die a second time and will note the higher score of the two rolls but Bob rolls only once and all the assumptions made above still apply to the game then Anns probability of winning the game as compared to Bob increases as shown above. Now considering a situation in which both players can roll the die twice and all the assumptions made above still apply to the game.

As the highest score obtained from the two rolls is to be chosen as a result we can make an assumption that Bobs possible outcomes are same as Anns presented in the table below. 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6 2 6 3 6 4 6 5 6 6 Using the table above we create another table, which tells us the outcomes in which Ann wins the game in the situation provided to us. 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6.

2 6 3 6 4 6 5 6 6 The color red represents the outcomes Ann needs to get to win the game when Bob gets the outcomes represented by the color blue. As we have already calculated the probability of Ann getting Specific outcomes in the earlier situation provided to us, we can say that Bobs probability of getting the specific outcomes is same as Anns. So the probability of Ann winning the game in this situation can be shown with the help of the table below: Anns highest outcomes from two rolls Bobs Outcomes Probability of the event (Ann wins) 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 Using the above found probabilities in the table the sum of these probabilities gives the probability when Ann wins the game P (Ann wins) = Since we already know that the sum of the probabilities is equal to 1 so, we can calculate bobs probability by using the formula If we considered a situation in which both players can roll the die twice and all the assumptions made above still apply to the game. And As the highest score obtained from the two rolls is to be chosen as a result we can make an assumption that Bobs possible outcomes are same as Anns.we can say that Bobs probability of winning the game when the Die is rolled twice by each player is higher than Anns.

Now to investigate the probabilities of Ann and Bob to win the game if they roll the die a multiple times, we formulate a general form or a general formula assuming that each of the player takes only the highest score into account and all the assumptions discussed in the above tasks remain the same. For this first we construct a table showing the possible outcomes and their probabilities in multiple rolls. The table is as follows: Possible outcomes 1st roll 2nd roll 3rd roll 1 2 3 4 5 6.

So, we come up with a formula, which states the probability of a number/ event X to the highest in n rolls: We get as the numerator; we can show this through the use of an example illustrated below: To find the probability of getting a 3 in the 2nd roll, it can be found using the probabilities of the outcomes before 3 they are 1 and 2. So the numerators of the probabilities of 1 and 2 are 1 and 3 respectively So when we add 1 and 3 we get 4, which is a square of 2 we, can derive this by using the variable X where X is the outcome for which we need to find the probability. Here X is 3. So as 4 is 2 square we can derive it as where n=2.

So to get a 5 which is the numerator of the probability of getting a 3 can be found using the formula Where n= the number of the roll and X= the outcome for which the probability needs to be calculated. If we start looking at the denominator it is defined as the total number of possible outcomes. In the case of a die, it is always equal to This can be explained by referring back to the probability tree diagram, where in case that a die that is thrown, the denominator will always be multiplied by 6 when the die is thrown an additional time. Another way in which we can show this is by using a sequence for the numerator.

First we take an example for the 2nd roll for which we get an arithmetic sequence for the numerator of the probabilities using the table above as we can rearrange this formula to generalize it for n rolls as the proof can be illustrated as follows: For X=3 and n=2 This can be verified from the table we can find the formula for the Arithmetic sequence using this is as follows. Taking n=2 We can verify this for n=3 whose formula for the arithmetic sequence = This can be verified from the table we can find the formula for the Arithmetic sequence using this is as follows.

Now to generalize this formula for multiple number of rolls by Ann and Bob we use the general form we found above: Assuming that we have to find the probability that Ann wins in multiple rolls Assuming that Ann wins so her highest score in multiple rolls = X So Bobs highest score when Ann wins= Assuming that Ann has n throws and Bob has m throws and that n and m can be equal to infinite. So, the probability that Ann wins That is, We have used the sigma sign, which stands for sum for calculating Bobs probability because for Ann to win bob can have any score less than And can have any value of

To prove the above formula we need to expand it the proof is as follows: First we expand the following term: The expansion is as follows where X= 1 By simplifying the expansion we get the following: The expansion is as follows where X= 2 By simplifying the expansion we get the following: The expansion is as follows where X= 3 By simplifying the expansion we get the following: The expansion is as follows where X= 4 By simplifying the expansion we get the following: The expansion is as follows where X= 5 By simplifying the expansion we get the following:

The expansion is as follows where X= 6 By simplifying the expansion we get the following: We do the same for Bob using the equation : For all cases in which Ann wins: Possible Outcomes for Ann Possible Outcomes for Bob 2 1 3 2,1 4 3,2,1 5 4,3,2,1 6 5,4,,3,2,1 Now to find Anns winning probability in multiple rolls we need to find the sum of the Products of the expanded equations of bob and Ann where Ann is represented by A and bob is represented by B. For X=1 For X=2 For X=3 For X=4 For X=5 For X=6 This can be written as the following:

Where, = = = = = So the above-expanded equations can be written as, The situation given to us can now be applied to a dice game in a casino where we are assuming that the casino/ bank is using a fair die and the player playing the game. We can also make an assumption that the casino/ bank has the same probability of winning the game as Bob whose probability of winning was calculated in the earlier situations given to us. Thus, as a result we can say that the bank/ casino wins every 21 games out of a total of 36 games.

And We can also assume that as the casino has the same probability of winning as Bob, the player playing the game has the same probability of winning as Ann that is the player will win every 15 games out of a total of 36 games as we have calculated earlier. Taking the above probabilities into account we can consider an example, which will help us formulate a general term for the game between the casino and the player. We assume that to enter each game a player has to pay $50, which he/she will get back with an amount of a profit.

Taking the probabilities into account we assume that the total number of outcomes is 36 so, as a result the player has to pay a total of to the casino or the bank. As we already know that the if the player wins he /she gets back the entry fee of the game plus an amount of a profit. This winning amount can be calculated using the players wining probability that is So the players winning amount per game = $ 120 assuming that the player plays a total of 36 games. We can conclude that the player makes a profit of $70 per game he/she wins. This would mean that the casino would neither make a loss nor a profit.

For the casino to make a profit they should ensure that the amount of profit should be less that $70. This can be summarized as follows: For the player the profit should be For the casino to make a profit the players profit should be As the above model we have presented is based only upon probabilities it may not always be true, it is merely dependent upon the number of games the player plays. Looking at the game from the perspective of both players the game can be considered worthwhile for both the player and the casino this is because the casino will always make a profit as long as enough games are played that is ?

36 As they have to present the player with a reward every time he/she wins the case chosen the reward equals to $70. The game can be considered worthwhile for the player because he/she has reasonably high chance of winning and if the player wins he/she makes a profit of $70 from the casino, as this is higher than the entrance fee of a game the player will be willing to play another game and even if he/she loses this time he/she will still make a profit of $20.

Now using the example above we are going to create model for a game so that the casino makes a reasonable profit in the case where the player rolls the red die once and the bank rolls the white die once. For the creation of the model we need to define certain variables they are as follows: Entry fee for the game E The money the player gets when he wins (entry fee + profit) or the money the casino needs to pay to the player. T As shown in the above example where E= $50 the money that the player wins (T) if he/she wins is found using the probability of Ann winning when she rolls the Die once found in the earlier parts and used in the example above that is.

So to find T we use the probability and simple unitary method. First to enjoy the benefit of unitary method we find the total entry fee payed by the player assuming the total possible number of outcomes is 36 or the total game played is 36. Total entry fee As the player wins the game only 15 times using unitary method we divide the total entry fee by 15 to find T.

And to find the money earned by the casino/bank we simply divide the total money entry fee of the player by 21 since the probability that the casino/bank wins is Now to conclude our investigation we need to find the probability of that the player wins and that the casino wins when there can be multiple number of throws that is that the number of games the player plays is not fixed and that the number of times the casino rolls the dice can be different.

As we assumed in the above investigation that the probability that the player wins is equal to the probability that Ann wins. So we use the same formula as we used to calculate that Ann wins when the number of rolls can be multiple, it is as follows:

where m= number of games casino plays and n= number of games player plays. By investigating the case of Ann and Bob we came to a conclusion that Bob wins when his outcomes are more than or equal to that of Ann this same assumption applies in the case of the player and the casino where the player is Ann and the casino is Bob. Taking this assumption into account we formulate an equation which calculates the maximum money the player gets when he wins (entry fee + profit) or the money the casino needs to pay to the player.

The equation is as follows: That is, We make such a formula because determines how many games the player wins from a total of n games and we multiply the formula by E because to know the total money the player has payed to the casino/bank, that is the total money in the game. Dividing this by the number of games the player actually won we get the maximum money the player gets when he wins (entry fee + profit) or the money the casino needs to pay to the player.

For this formula to work we need to keep in mind certain assumptions they are as follows: Looking at the game from the perspective of both players the game can be considered worthwhile for both the player and the casino this is because the casino will always make a profit as long as enough games are played that is ? 36 As they have to present the player with a reward every time he/she wins the case chosen the reward equals to $70.

The game can be considered worthwhile for the player because he/she has reasonably high chance of winning and if the player wins he/she makes a profit of $70 from the casino, as this is higher than the entrance fee of a game the player will be willing to play another game and even if he/she loses this time he/she will still make a profit of $20. Summary/ conclusion This investigation has analyzed various dice games, starting from the simple case where two players Ann and Bob roll one die each, with Ann winning if their rolled number is larger than that of player Bob.

Player further expanded this game to include the case in which the greater of two rolls made Ann was counted against the roll made by player Bob, the case in which each player was able to roll the die twice, and the case where the greatest of three rolls made by player A was counted against one roll made by player Bob. A general formula for player Anns chances of winning were determined; not however that this formula is limited to the instance in which all dice are fair and six sided, and each face on the each dice is assigned a different number between one and six.

Using this information, possible payouts and player entry fees were modeled for some casino games built on the dice game template. This investigation has made use of a custom-made program on Microsoft Excel to calculate probabilities using the general formula. Another program was used to count the probability of player Ann winning in the variations of the dice game in which both players roll twice and in which player Ann rolls three times but player Bob rolls once.