Modelling a dice game in a casino

Hi! So here's the problem I hope you can help me:

Consider the game in a casino where the player has a red die and the bank has a white die. The casino is granted the tie.
Find a model for a game so that the casino makes a reasonable profit in case where the player rolls the red die once and the bank rolls the white die once.
It must be considered how much a player must pay to play the game and how much the bank will pay if the player wins. Consider this from the perspective of both the player and the casino so that it is worthwhile for the player and the casino.

Now consider other models for the game so that a) there are multiple players involved in the game and b) the player rolls the red die multiple times

P.s. How can it be calculated a general formula when the number of dice thrown by the casino increases but the player rolls only once (considering the casino is granted the tie)?

Thanks! ;)

Quote: MathsMarie

Hi! So here's the problem I hope you can help me:

Consider the game in a casino where the player has a red die and the bank has a white die. The casino is granted the tie.
Find a model for a game so that the casino makes a reasonable profit in case where the player rolls the red die once and the bank rolls the white die once.
It must be considered how much a player must pay to play the game and how much the bank will pay if the player wins. Consider this from the perspective of both the player and the casino so that it is worthwhile for the player and the casino.

Now consider other models for the game so that a) there are multiple players involved in the game and b) the player rolls the red die multiple times

P.s. How can it be calculated a general formula when the number of dice thrown by the casino increases but the player rolls only once (considering the casino is granted the tie)?

Thanks! ;)

Somebody asked this same question earlier this year. Must be a new semester! :-)

Quote: MathsMarie

Hi! So here's the problem I hope you can help me:

Consider the game in a casino where the player has a red die and the bank has a white die. The casino is granted the tie.
Find a model for a game so that the casino makes a reasonable profit in case where the player rolls the red die once and the bank rolls the white die once.
It must be considered how much a player must pay to play the game and how much the bank will pay if the player wins. Consider this from the perspective of both the player and the casino so that it is worthwhile for the player and the casino.

Now consider other models for the game so that a) there are multiple players involved in the game and b) the player rolls the red die multiple times

P.s. How can it be calculated a general formula when the number of dice thrown by the casino increases but the player rolls only once (considering the casino is granted the tie)?

Thanks! ;)

This is little different than Casino War, except it is with dice.

The casino profit is really simple on this one. The casino wins in the event of a tie, and that is really all they need.

A tie will occur: (1/6 * 1/6) or = .1666666667 * .1666666667 = .02777777783% of the time.

For every $10 bet, the casino can be expected to win $10 * .2777777783 or $0.28 (rounding up)

The casino's edge is 2.8% on the game. This is a lesser edge than the casino experiences with Roulette, but the value for the casino is strong, in my opinion, because the game would go really fast.

You would have to avoid dice setting by having similar rules to craps in the manner in which the dice must be rolled.

A.) If there are multiple players, you would simply have more than two dice colors and a chip for each player matching the color of that player's die. You could have White, Blue, Red, Pink, Green and Yellow, for example, thereby enabling six players to play the game at once.

The casinos die would be compared to all of the other dice rolled in terms of value, payouts would be made accordingly. In the event of a Dealer, "Six," all bets could be cleared automatically as the casino wins regardless.

B.) What do you mean rolls the red die multiple times? The more rolls or the more dice, the less likely the event of a tie thereby diminishing the casino's edge.

P.S. Why would the casino roll more than once?

Quote: Mission146

A tie will occur: (1/6 * 1/6) or = .1666666667 * .1666666667 = .02777777783% of the time.

You thinketh too much. A tie will occur 1/6 of the time. You needed a formula to mess up this easy one?

Quote: MathsMarie

Now consider other models for the game so that a) there are multiple players involved in the game and b) the player rolls the red die multiple times

P.s. How can it be calculated a general formula when the number of dice thrown by the casino increases but the player rolls only once (considering the casino is granted the tie)?

I'm not sure if there is an easy "general" formula for this, but if one side (i.e. player or dealer) throws N dice, and P(X) is the probability that the highest number rolled is X, then:
P(1) = (1/6)^N
P(2) = (1/3)^N - (1/6)^N = (2^N - 1) / 6^N
P(3) = (1/2)^N - (1/3)^N = (3^N - 2^N) / 6^N
P(4) = (2/3)^N - (1/2)^N = (4^N - 3^N) / 6^N
P(5) = (5/6)^N - (2/3)^N = (5^N - 4^N) / 6^N
P(6) = 1 - (5/6)^N = (6^N - 5^N) / 6^N

The probability of the player winning is P(6) * (P(1) + P(2) + P(3) + P(4) * P(5)) + P(5) * (P(1) + P(2) + P(3) + P(4)) + P(4) * (P(1) + P(2) + P(3)) + P(3) * (P(1) + P(2)) + P(2) * P(1); the probability of the dealer winning = 1 - the player's winning probability.

Note that K^N - (K-1)^N = K^N-1 + K^N-2(K-1) + K^N-3(K-1)² + K^N-4(K-1)³ + ... + K (K-1)^N-2 + (K-1)^N-1, but I am not sure if there is a more general solution for that sum other than K^N - (K-1)^N.

To make the house edge more reasonable you would have to have some additional rule to help the player, like a a player 5 is an automatic winner. Or how about a player bets 6 to win 7, casino wins ties? That would be like a place 6 in craps.

Thank you very much, that is really helpful!

A) Just wondering, how does probability of both the players and the bank get affected as the number of players increases? I mean, it is obvious that the casino will havee less chances to win and so will the players but I still don't get why. What confuses me a lot is the fact that casinos normally have large tables of players and still make a profit. So even if their chances are lower they end up winning. I must be because all player pay but only one wins, right? How could I calculate the changing probabilities as the number of players increeases? Is it still 1/36 for each?

B) Yes, I know it doeasn't make sense that the casino rolls more thant once. Normally in casinos the bank and the player roll the same number of times at the same time isn't it? I was trying to ask that if the player would have a higher winning probability if the player rolled more than once. However, the player has to pay for every game right? So, I'm guessing that even if the player won, he/she would have still given more money to the bank than what the player might receive right? Basically, does the probability change as the dice are thrown more times?

p.s. You're really smart! ;) Thanks for your time!

How about putting the tie bet in prison, only to be collected by the house if the next bet is a tie ?

Thank you so much for your help!!!
These is absolutely awesome! ;)
I would have never thought about it hahaha

A.) If there are multiple players, you would simply have more than two dice colors and a chip for each player matching the color of that player's die. You could have White, Blue, Red, Pink, Green and Yellow, for example, thereby enabling six players to play the game at once.

The casinos die would be compared to all of the other dice rolled in terms of value, payouts would be made accordingly. In the event of a Dealer, "Six," all bets could be cleared automatically as the casino wins regardless.

B.) What do you mean rolls the red die multiple times? The more rolls or the more dice, the less likely the event of a tie thereby diminishing the casino's edge.

P.S. Why would the casino roll more than once?

Thank you very much, that is really helpful!

A) Just wondering, how does probability of both the players and the bank get affected as the number of players increases? I mean, it is obvious that the casino will havee less chances to win and so will the players but I still don't get why. What confuses me a lot is the fact that casinos normally have large tables of players and still make a profit. So even if their chances are lower they end up winning. I must be because all player pay but only one wins, right? How could I calculate the changing probabilities as the number of players increeases? Is it still 1/36 for each?

B) Yes, I know it doeasn't make sense that the casino rolls more thant once. Normally in casinos the bank and the player roll the same number of times at the same time isn't it? I was trying to ask that if the player would have a higher winning probability if the player rolled more than once. However, the player has to pay for every game right? So, I'm guessing that even if the player won, he/she would have still given more money to the bank than what the player might receive right? Basically, does the probability change as the dice are thrown more times?

p.s. You're really smart! ;) Thanks for your time!

Cool!!!! So that the player gets tricked that the house is not always winning but the truth is the bank has higher probability and gets granted the tie so will win a lot more? hahaha Any suggesctions abouth how much the player should pay to play the game? So that the player is attracted to play but the casino makes a profit?
Thank you!

Ohh that's interesting... though I'm not completely sure how to do it. :/ I mean that varies probabilities?
Also, would it be reasonable to play the player by the difference of scores it has with the casino? I.e. if the player wins by 5 points pay $x, if the player wins by 4 points pay $y (less) and so on?
Thank you!

Thanks! I meant a formula if you increase the number of dice... ;)

Quote: SOOPOO

You thinketh too much. A tie will occur 1/6 of the time. You needed a formula to mess up this easy one?

That was hilarious!

I calculated a tie for a specific number, for whatever reason!

EDIT: In all fairness, I'd been doing complicated BJ math for the better part of two hours just before that. I can't switch from complicated to simple on a time. I'd have probably whiffed on 2 + 2 at that moment!

Quote: MathsMarie

Thank you very much, that is really helpful!

A) Just wondering, how does probability of both the players and the bank get affected as the number of players increases? I mean, it is obvious that the casino will havee less chances to win and so will the players but I still don't get why. What confuses me a lot is the fact that casinos normally have large tables of players and still make a profit. So even if their chances are lower they end up winning. I must be because all player pay but only one wins, right? How could I calculate the changing probabilities as the number of players increeases? Is it still 1/36 for each?

B) Yes, I know it doeasn't make sense that the casino rolls more thant once. Normally in casinos the bank and the player roll the same number of times at the same time isn't it? I was trying to ask that if the player would have a higher winning probability if the player rolled more than once. However, the player has to pay for every game right? So, I'm guessing that even if the player won, he/she would have still given more money to the bank than what the player might receive right? Basically, does the probability change as the dice are thrown more times?

p.s. You're really smart! ;) Thanks for your time!

A.) The probability for the players and the bank would not change regardless of the number of players. The game is simply the casino's die vs. each individual player's die, regardless of the number of dice thrown.

-What do you mean by only one player wins? Why would only one win? You wouldn't change the Rules of a Table Game based on the number of players, it is simply the casino's die vs each individual player's die. Every bet is treated separately.

Even if you wanted to make a game where the players played against one another and the House (dividing ties evenly, except a dealer, "Six") then everyone would have to bet the same amount for it to work. I don't think that would be as attractive, some people like to bet more than others.

B.) No, you would be making a bet on each individual die. I suppose you could hedge by betting $5 on each die and rolling two dice, but there's statistically no difference between that and just betting $10.00.

IMPROVED PLAYER ODDS

You can use the patent pending MISSION146 dumb mistake rule. Highest number wins, a tie is a Push, except a tie on a, "Six," which is a house win. You couldf make it a tie on a, "One," any number you want, really.

Actually, make it a tie on a one wins for the House. The player might feel ripped off losing on a six.

Here are the numbers I get for various numbers of player and house dice, limited to values where the house edge > 0:

Player Dice	House Dice	House Edge %
1	1	16.67
2	2	22.07
3	2	5.68
4	3	19.45
5	3	12.05
6	3	6.47
7	3	2.17

If the house gets 4 dice, the probability of the house throwing at least one 6 is 671/1296 = about 51.775%, so even if you assume the player always rolls a 6, the house edge is 3.55%.

The best chance for the player without it being a positive expectation game is 3 dice against the dealer's 7.

There are ways of making the house edge lower, which have the added benefit of making the game more exciting for the player; for example, increasing the odds if all of the player's dice are the same number.

Thank you MISSION146!!! I'm seriously grateful for all yor help!!!

There are ways of making the house edge lower, which have the added benefit of making the game more exciting for the player; for example, increasing the odds if all of the player's dice are the same number.

Thanks!!!! But I'm not completely sure how would you do this? What else could you do? Could you pleasse explain it a bit more? ;)

Quote: MathsMarie

Thanks!!!! But I'm not completely sure how would you do this? What else could you do? Could you pleasse explain it a bit more?

Here's an example:

If you are playing the version where you roll three dice and the house rolls 2, normally the house edge is 5.68%.

However, if you have a special rule where you win ties instead of losing if all three of your dice are the same, the house edge is now only 4.76%.

You could also, for example, have some sort of "super bad beat" where if all five dice (your 3 and the house's 2) are the same, it pays off something like 20-1. This happens only one time in 1296, so it lowers the house edge by 1.54%.

Wow that's seriously awesome plus it would attract a lot of players!!!!!
I' had no idea you could do that! Though I guess it's like boxing right, when one guy has small probabilities of winning so if you bet on him and he wins then you'll get a crazy amount of money! Hahaha Thank you ThatDonGuy! ;)

Very similar idea is currently on the floor at Turning Stone Casino...

https://wizardofodds.com/games/beat-the-dealer/

Thanks winmonkeyspit3! I'll check it out!