I am a newbie Game Designer. I have a project that needs to recreate the game's math model.
I have read lots of your Article and Q&A post in " for me to have an idea on how to do my task.
Can you help me please by giving me a sample on how to compute the RTP for a HI-LO game?
Thank you.
RULES
NUMBERS
1-48
NEW GAME
Press PLAY to start a new game.
A number will be automatically generated on the first position.
PLAYING THE GAME
You must guess if the new number that will be generated is going to be higher or lower than the previous one. For this, the buttons HIGHER or LOWER can be pressed.
HIGHER - by clicking this button, you will submit your guess that the next generated number will be higher than the previous number
LOWER - by clicking this button, you will submit your guess that the next number will be lower than the previous number
BURN
If you submit that the new number will be higher than the previous one, but the number was lower or equal, you will get burned.
If you submit that the new number will be lower than the previous one, but the number was higher or equal, you will get burned.
The new number will always be different than the current number.
If you get burned once, the multiplier of the win that you would receive at the end of the row, will be reduced to x1.
If you get burned twice, you will lose and the game will end.
COLLECT WIN
To be able to collect or risk more your win, you must reach the end of a line without being burned twice.
If you will complete 1 line without getting burned at all, the next win multiplier will multiply your current win by x2.
If you will complete an additional 1 line (that means 2 lines since the game was started) without getting burned at all, the next win multiplier will multiply your current win by x4. Then by x5 and finally by x6.
ADVANCING TO NEXT LINE
If you will reach the end of a line, you can decide to TAKE WIN or PLAY NEXT ROW.
TAKE WIN - will add the amount listed at the end of the previous row to your balance and will end the game.
PLAY NEXT ROW - will not add any amount to your balance. This will allow you to risk your previous win in order for chances to win even more.
thank you!
also i created a game simulator of this game and gives me 87-88% RTP.
but the game site claims that the RTP is 97.59%
that is the reason why I am having second thought now.
again thank you.
paulo
Quote: GWAEThere are people on here that will do it for you but not for free. Are you looking to hire someone?
Nope. I want to do it myself. That is why I was asking for a sample at first because I just want to get some ideas.
Thank you for asking.
I hope there are some here that can give me some samples or ideas.
The bigger problem is making sure you're analyzing the right game. If your game sim doesn't use exactly the same rules as the actual game, the results will be off.
Quote: oluapseyer
BURN
If you submit that the new number will be higher than the previous one, but the number was lower or equal, you will get burned.
If you submit that the new number will be lower than the previous one, but the number was higher or equal, you will get burned.
The new number will always be different than the current number.
Perhaps this conflict in the rules is messing with your calculation?
if ever do you think you can check my excel? i could send it to you if thats okay with you.
thank you.
paulo
possible balls(for HI and for LO) divided by 47 since the new number should always be different with the previous number.
thank you again
There's really not enough context here -- the rules you posted aren't sufficient. But are you saying that previously used numbers are recycled back into the deck one turn after they were used? Because if that's not how it works -- that is, if the number is only going to be used once during the whole game -- then the odds change.Quote: oluapseyernope. i am sure i got this rule. the probability table is computed as
possible balls(for HI and for LO) divided by 47 since the new number should always be different with the previous number.
thank you again
And you're also assuming that the numbers are distributed fairly and that the game actually behaves the way it looks like it behaves. That definitely isn't a reasonable assumption for an unregulated Internet casino, if that's where this game is.
Is the maximum win 240?
What have you calculated as the probability to complete a row with 0 burns? 1 burn?
Can a number ever repeat in the same row?
Quote: CrystalMathHow many decisions do you make per row?
Is the maximum win 240?
What have you calculated as the probability to complete a row with 0 burns? 1 burn?
Can a number ever repeat in the same row?
1. The number ranges from 1-48
2. 7 numbers per row. 4 rows in a game
3. The numbers repeats in a row but cannot be the same as the previous draw.. so that the choice is HI or LO only no TIE result
RULES
NEW GAME
Press PLAY to start a new game.
A number will be automatically generated on the first position.
PLAYING THE GAME
You must guess if the new number that will be generated is going to be higher or lower than the previous one. For this, the buttons HIGHER or LOWER can be pressed.
HIGHER - by clicking this button, you will submit your guess that the next generated number will be higher than the previous number
LOWER - by clicking this button, you will submit your guess that the next number will be lower than the previous number
BURN
If you submit that the new number will be higher than the previous one, but the number was lower or equal, you will get burned.
If you submit that the new number will be lower than the previous one, but the number was higher or equal, you will get burned.
The new number will always be different than the current number.
If you get burned once, the multiplier of the win that you would receive at the end of the row, will be reduced to x1.
If you get burned twice, you will lose and the game will end.
COLLECT WIN
To be able to collect or risk more your win, you must reach the end of a line without being burned twice.
If you will complete 1 line without getting burned at all, the next win multiplier will multiply your current win by x2.
If you will complete an additional 1 line (that means 2 lines since the game was started) without getting burned at all, the next win multiplier will multiply your current win by x4. Then by x5 and finally by x6.
ADVANCING TO NEXT LINE
If you will reach the end of a line, you can decide to TAKE WIN or PLAY NEXT ROW.
TAKE WIN - will add the amount listed at the end of the previous row to your balance and will end the game.
PLAY NEXT ROW - will not add any amount to your balance. This will allow you to risk your previous win in order for chances to win even more.
a) What is the "amount listed at the end of the previous row"?
b) Does each new row start with a newly revealed number just like the first row, or does the player make a high/low pick based on the last number of the previous row? In other words, is the player making 6 picks per row regardless, or does the player make 7 picks for all rows after the first?
c) Are burns cumulative or do they reset after each row? If I get burned on row 1 but make it, and I get burned on row 2, am I done or do I keep playing?
d) Are the multipliers cumulative? If I complete two rows without getting burned at all and then want to take the win, do I have a 4x multiplier or an 8x multipler (2x4)?
e) If I got burned on the first row but then complete rows 2,3,4 with no burns, what is the multiplier?
I'd bet that somewhere in your spreadsheet the multiplier rules aren't being handled right. At least that's where I'd look first.
Quote: MathExtremistStill not enough detail:
a) What is the "amount listed at the end of the previous row"?
the amount won. or bet x multipier... x1(5 correct guess) , x2,x4,x5,x6(6correct guesses) or 0 if no won.
Quote: MathExtremistb) Does each new row start with a newly revealed number just like the first row, or does the player make a high/low pick based on the last number of the previous row? In other words, is the player making 6 picks per row regardless, or does the player make 7 picks for all rows after the first?
all rows starts with newly revealed number. 6 picks per row.
Quote: MathExtremistc) Are burns cumulative or do they reset after each row? If I get burned on row 1 but make it, and I get burned on row 2, am I done or do I keep playing?
burns reset in each row
Quote: MathExtremistd) Are the multipliers cumulative? If I complete two rows without getting burned at all and then want to take the win, do I have a 4x multiplier or an 8x multipler (2x4)?
cumulative... the won amount per row will be the bet for the next row.
Quote: MathExtremiste) If I got burned on the first row but then complete rows 2,3,4 with no burns, what is the multiplier?
if burned once the multiplier is x1 for that row. if burned twice the game ends.
thank you
You say there are 7 numbers per row, but I assume the first one is the given number, so you must make 6 decisions.
I calculate the following probabilities for a single row:
p(lose) 0.742696404
p(win) 0.152509584 (0 burns)
p(push) 0.104794011 (1 burn)
What does the player win in the following situation:
Finishes the first row with zero burns.
Finishes the second row with one burn.
Finishes the third row with one burn.
Finishes the fourth row with zero burns.
Based on my understanding, they would win 8x their bet.
1 1.000000 0.000000 HI 1.000000
2 0.978723 0.021277 HI 0.978723
3 0.957447 0.042553 HI 0.957447
4 0.936170 0.063830 HI 0.936170
5 0.914894 0.085106 HI 0.914894
6 0.893617 0.106383 HI 0.893617
7 0.872340 0.127660 HI 0.872340
8 0.851064 0.148936 HI 0.851064
9 0.829787 0.170213 HI 0.829787
10 0.808511 0.191489 HI 0.808511
11 0.787234 0.212766 HI 0.787234
12 0.765957 0.234043 HI 0.765957
13 0.744681 0.255319 HI 0.744681
14 0.723404 0.276596 HI 0.723404
15 0.702128 0.297872 HI 0.702128
16 0.680851 0.319149 HI 0.680851
17 0.659574 0.340426 HI 0.659574
18 0.638298 0.361702 HI 0.638298
19 0.617021 0.382979 HI 0.617021
20 0.595745 0.404255 HI 0.595745
21 0.574468 0.425532 HI 0.574468
22 0.553191 0.446809 HI 0.553191
23 0.531915 0.468085 HI 0.531915
24 0.510638 0.489362 HI 0.510638
25 0.489362 0.510638 LO 0.510638
26 0.468085 0.531915 LO 0.531915
27 0.446809 0.553191 LO 0.553191
28 0.425532 0.574468 LO 0.574468
29 0.404255 0.595745 LO 0.595745
30 0.382979 0.617021 LO 0.617021
31 0.361702 0.638298 LO 0.638298
32 0.340426 0.659574 LO 0.659574
33 0.319149 0.680851 LO 0.680851
34 0.297872 0.702128 LO 0.702128
35 0.276596 0.723404 LO 0.723404
36 0.255319 0.744681 LO 0.744681
37 0.234043 0.765957 LO 0.765957
38 0.212766 0.787234 LO 0.787234
39 0.191489 0.808511 LO 0.808511
40 0.170213 0.829787 LO 0.829787
41 0.148936 0.851064 LO 0.851064
42 0.127660 0.872340 LO 0.872340
43 0.106383 0.893617 LO 0.893617
44 0.085106 0.914894 LO 0.914894
45 0.063830 0.936170 LO 0.936170
46 0.042553 0.957447 LO 0.957447
47 0.021277 0.978723 LO 0.978723
48 0.000000 1.000000 LO 1.000000
Per Draw Average Winning % = 75.53% < - Average Greater % winning
6 Winning Streaks % = 18.57% < - 75.53% ^6 or 75.53% *75.53% *75.53% *75.53% * 75.53%*75.53%
Atleast 5 Wins in a round = 70.77% <- Excel Formula -> 1-BINOMDIST(4 , 6 , .7553, FALSE)
DO you think this make sense?
Maybe I'm misunderstanding the rules, but if the chance of an individual win is 75.53%, I agree that the chances of 6 wins in a row is 18.57% but wouldn't 5 wins out of 6 be 36.09%? 6C5 * 75.53%^5 * (1-75.53%). I think you need a 5 in your BINOMDIST function, not a 4 (only 4 wins means you lose) and you're not supposed to subtract it from 1. Subtracting from 1 gives you the odds of not having 5 wins, but that's not what you want.Quote: oluapseyerPer Draw Average Winning % = 75.53% < - Average Greater % winning
6 Winning Streaks % = 18.57% < - 75.53% ^6 or 75.53% *75.53% *75.53% *75.53% * 75.53%*75.53%
Atleast 5 Wins in a round = 70.77% <- Excel Formula -> 1-BINOMDIST(4 , 6 , .7553, FALSE)
DO you think this make sense?
Quote: MathExtremistMaybe I'm misunderstanding the rules, but if the chance of an individual win is 75.53%, I agree that the chances of 6 wins in a row is 18.57% but wouldn't 5 wins out of 6 be 36.09%? 6C5 * 75.53%^5 * (1-75.53%). I think you need a 5 in your BINOMDIST function, not a 4 (only 4 wins means you lose) and you're not supposed to subtract it from 1. Subtracting from 1 gives you the odds of not having 5 wins, but that's not what you want.
What do you mean by 6C5?
Thank you really.
Sorry, that was shorthand for the combinations function. In Excel it's COMBIN(6,5).Quote: oluapseyerQuote: MathExtremistMaybe I'm misunderstanding the rules, but if the chance of an individual win is 75.53%, I agree that the chances of 6 wins in a row is 18.57% but wouldn't 5 wins out of 6 be 36.09%? 6C5 * 75.53%^5 * (1-75.53%). I think you need a 5 in your BINOMDIST function, not a 4 (only 4 wins means you lose) and you're not supposed to subtract it from 1. Subtracting from 1 gives you the odds of not having 5 wins, but that's not what you want.
What do you mean by 6C5?
Thank you really.
Quote:Sorry, that was shorthand for the combinations function. In Excel it's COMBIN(6,5).
Thank you.
Quote: CrystalMath
p(lose) 0.742696404
p(win) 0.152509584 (0 burns)
p(push) 0.104794011 (1 burn)
I clearly have a problem with the 1 burn calculation. I'm confident in the 0 burns calculation. I'll look at it again tomorrow.
p(lose) 0.467505362
p(win) 0.152509584
p(push) 0.379985054
The reason that the probability of winning 6 decisions in a row isn't simply (0.7553^6) is because the distribution of numbers on the second decision is not uniform assuming that you won. The distribution tends to favor numbers in the middle, which are worse for the player.
Likewise, if you lose a decision, you are more likely to win the next decision, so the probability of getting 1 burn is slightly higher than 0.7553^5*(1-0.7553)*combin(6,5).
The return of the game, in my opinion, is 76.93%. This is the ratio of the final win to the initial bet. In domestic jurisdictions, this is how the return would be calculated.
The OP considered each level to be an independent wager. Calculating using this method results in a return of 87.0951%. This also results in a small strategy change: in my game, I will not play the top row if I have had 1 win and two pushes.
If you advance on the first burn, does that mean getting to the second to the last card with no burns is an automatic win for the row?
Quote: CrystalMathI've corrected my calculations:
p(lose) 0.467505362
p(win) 0.152509584
p(push) 0.37998505
The OP considered each level to be an independent wager. Calculating using this method results in a return of 87.0951%. This also results in a small strategy change: in my game, I will not play the top row if I have had 1 win and two pushes.
what do mean you will not play the top row if you had one win and two pushes?
push means 5 correct guesses or x1 of the bet, right?
thank you for the time and effort.
Quote: AyecarumbaWhen you get burned, is the figure that "burned" you replaced, (so that you have to guess at the position again), or do you simply "advance" to the burn card whilst taking a hit to your burn counter.
If you advance on the first burn, does that mean getting to the second to the last card with no burns is an automatic win for the row?
when burned for the first time, simply advance to the next guess and take hit on your burn counter.
Quote: CrystalMathI've corrected my calculations:
p(lose) 0.467505362
p(win) 0.152509584
p(push) 0.379985054
The reason that the probability of winning 6 decisions in a row isn't simply (0.7553^6) is because the distribution of numbers on the second decision is not uniform assuming that you won. The distribution tends to favor numbers in the middle, which are worse for the player.
Likewise, if you lose a decision, you are more likely to win the next decision, so the probability of getting 1 burn is slightly higher than 0.7553^5*(1-0.7553)*combin(6,5).
The return of the game, in my opinion, is 76.93%. This is the ratio of the final win to the initial bet. In domestic jurisdictions, this is how the return would be calculated.
The OP considered each level to be an independent wager. Calculating using this method results in a return of 87.0951%. This also results in a small strategy change: in my game, I will not play the top row if I have had 1 win and two pushes.
My numbers are very similar.
I even came to the same conclusion that the optimal strategy would be to take the win if you have 1 win and 2 pushes.
I was coming here to post just that.
(Also optimal...is to take the win if you push the first line...I think that should be kind of apparent)
Quote: oluapseyerwhat do mean you will not play the top row if you had one win and two pushes?
push means 5 correct guesses or x1 of the bet, right?
thank you for the time and effort.
It means you should take the win if you won the first line then push the second and third lines. Therefore, you do not play the top line in this case.