Math Problem
I have a null 3x3 matrix (9 squares). The probability for a symbol x to be shown on each square is 0.2 and it is the same for all squares.The game plays one round at a time. If a symbol x appears on a square stays on it and the game runs one more round.
Is there a formula to calculate the probability of symbol x at round n (1<=n<=9)?
1. In each round, does each square without an x have a 0.2 probability of showing an x in that round?
2. Does the game end if there is a round where no new x's are created?
Quote: Alexander20Hi all.
I have a null 3x3 matrix (9 squares). The probability for a symbol x to be shown on each square is 0.2 and it is the same for all squares.The game plays one round at a time. If a symbol x appears on a square stays on it and the game runs one more round.
Is there a formula to calculate the probability of symbol x at round n (1<=n<=9)?
Omg, 😱 you're trying to solve the probabilities for Royal Slots :/
P.S. The payouts on the demo in no way reflect the probabilities of the payouts! I purposefully choose figures so far from the actual probabilities just to through of the mathematicians here, so I could see I you confused as the payout values I chose ;)
Quote: USpapergamesOmg, 😱 you're trying to solve the probabilities for Royal Slots :/
P.S. The payouts on the demo in no way reflect the probabilities of the payouts! I purposefully choose figures so far from the actual probabilities just to through of the mathematicians here, so I could see I you confused as the payout values I chose ;)
Trolling much?
Quote: CrystalMathTrolling much?
How is a skin a legitimate question trolling? If he's talking about my game I want to know about it! Honestly Crystal Math, I'm starting to think you're trolling me!
At round 1 probability is 0.2
At round 2 probability is (1-0.8^8) * (0.8 * 0.2) = 0.8322 * 0.16 = 0.1331
I can' t find a formula for the rest rounds.
Quote: ThatDonGuyQuestions:
1. In each round, does each square without an x have a 0.2 probability of showing an x in that round?
2. Does the game end if there is a round where no new x's are created?
Yes
Quote: USpapergamesOmg, 😱 you're trying to solve the probabilities for Royal Slots :/
P.S. The payouts on the demo in no way reflect the probabilities of the payouts! I purposefully choose figures so far from the actual probabilities just to through of the mathematicians here, so I could see I you confused as the payout values I chose ;)
I do not know your Slots. This mini game could be considered as a simplified version of Hold n Spin feature which is well known for several years in gaming industry. I am trying to figure out the math behind this feature and it is all about personal amusement and not commercial use.
Quote: Alexander20I do not know your Slots. This mini game could be considered as a simplified version of Hold n Spin feature which is well known for several years in gaming industry. I am trying to figure out the math behind this feature and it is all about personal amusement and not commercial use.
I have figured out the math behind the hold and spin feature. It's extremely complicated. Not sure I'm willing to just post openly about my results.
You don't specify, but if the intent is that you get three spins to hit an x before the game is over (and if you do hit an x, you reset to three spins), then the probability under your scenario of filling the screen is 0.3519481466.
Quote: rsactuaryI have figured out the math behind the hold and spin feature. It's extremely complicated. Not sure I'm willing to just post openly about my results.
You don't specify, but if the intent is that you get three spins to hit an x before the game is over (and if you do hit an x, you reset to three spins), then the probability under your scenario of filling the screen is 0.3519481466.
Thanks for your response. I am trying to figure out this feature for quite long time. I can understand why you are not willing to share your knowledge openly. If you would like to give some hints/help via private message it would be great.
I still can’t figure out what you’re asking. How can the probability of at least one X be LESS after 2 rounds (0.1331)than after 1 round (0.2) ?Quote: Alexander20I am looking for the probability of a symbol X to be shown in any square at n round.
At round 1 probability is 0.2
At round 2 probability is (1-0.8^8) * (0.8 * 0.2) = 0.8322 * 0.16 = 0.1331
I can' t find a formula for the rest rounds.
Quote: Ace2I still can’t figure out what you’re asking. How can the probability of at least one X be LESS after 2 rounds (0.1331)than after 1 round (0.2) ?
If symbol X appears in a square at the first round then and only then I proceed to the second round. At the second round, I already have one symbol X and I am looking for another symbol X to run the third round. This probability of the second symbol X is 0.1331 according to my calculations. At the third round, I have two symbols X and I want to calculate the probability to win another symbol X in order to proceed to the fourth round, etc.
Quote: Alexander20If symbol X appears in a square at the first round then and only then I proceed to the second round. At the second round, I already have one symbol X and I am looking for another symbol X to run the third round. This probability of the second symbol X is 0.1331 according to my calculations. At the third round, I have two symbols X and I want to calculate the probability to win another symbol X in order to proceed to the fourth round, etc.
If I understand the rules, the game terminates on the first round where you do not get an X in an empty square. Therefore, the game could end on round 1 by filling all squares, but must end by round 9. Is that right?
The binomial distribution function of any spreadsheet will solve the problem. I don't want to do the work for you. I think you will learn more by attacking it yourself.
https://wiki.openoffice.org/wiki/Documentation/How_Tos/Calc:_BINOMDIST_function
Hint: If you get two Xs on the first round, you have to solve the same original problem with seven squares.
