Javascript percentage of winning algorithm












-1















I have a slots machine game HTML5 + Javascript.
I have found the below algoritm on Stackoverflow, but i need another percentage



this.d = Math.random();

if (this.d < 0.3333) {
this.nextSymbols = [
['320', '320', 'jackpot'],
['jackpot', '320', '400'],
['320', '320', 'jackpot'],
];
} else if (this.d < 0.005) {
this.nextSymbols = [
['320', 'jackpot', '400'],
['320', 'jackpot', '400'],
['320', 'jackpot', '400'],
];
} else {
this.nextSymbols = [
[Symbol.random(), Symbol.random(), Symbol.random()],
[Symbol.random(), Symbol.random(), Symbol.random()],
[Symbol.random(), Symbol.random(), Symbol.random()],
];
}


I need to have:



 1 of 3 spins to be a winning prize 2 

1 of 200 spins wins big prize


Any math expert advice would be much appreciate.



Thanks










share|improve this question




















  • 1





    How is "I have 10 big prizes" relevant for the probability of a single spin?

    – Bergi
    Nov 19 '18 at 22:04






  • 2





    Hint: 1/3 = 0.3333…, 1/200 = 0.005 (compared to 50% = 0.5, 20% = 0.2 in the existing code)

    – Bergi
    Nov 19 '18 at 22:05













  • @Bergi thanks. I figured out

    – PixelArtie
    Nov 19 '18 at 22:21






  • 1





    < means less than. So if (this.d < 0.005) means that the value is less than 0.005. Any value where that is the case is also less than 0.3333. However the if-test for <0.005 is in the else clause of the <0.3333, so nobody will ever win the big prize.

    – Richardissimo
    Nov 19 '18 at 22:39
















-1















I have a slots machine game HTML5 + Javascript.
I have found the below algoritm on Stackoverflow, but i need another percentage



this.d = Math.random();

if (this.d < 0.3333) {
this.nextSymbols = [
['320', '320', 'jackpot'],
['jackpot', '320', '400'],
['320', '320', 'jackpot'],
];
} else if (this.d < 0.005) {
this.nextSymbols = [
['320', 'jackpot', '400'],
['320', 'jackpot', '400'],
['320', 'jackpot', '400'],
];
} else {
this.nextSymbols = [
[Symbol.random(), Symbol.random(), Symbol.random()],
[Symbol.random(), Symbol.random(), Symbol.random()],
[Symbol.random(), Symbol.random(), Symbol.random()],
];
}


I need to have:



 1 of 3 spins to be a winning prize 2 

1 of 200 spins wins big prize


Any math expert advice would be much appreciate.



Thanks










share|improve this question




















  • 1





    How is "I have 10 big prizes" relevant for the probability of a single spin?

    – Bergi
    Nov 19 '18 at 22:04






  • 2





    Hint: 1/3 = 0.3333…, 1/200 = 0.005 (compared to 50% = 0.5, 20% = 0.2 in the existing code)

    – Bergi
    Nov 19 '18 at 22:05













  • @Bergi thanks. I figured out

    – PixelArtie
    Nov 19 '18 at 22:21






  • 1





    < means less than. So if (this.d < 0.005) means that the value is less than 0.005. Any value where that is the case is also less than 0.3333. However the if-test for <0.005 is in the else clause of the <0.3333, so nobody will ever win the big prize.

    – Richardissimo
    Nov 19 '18 at 22:39














-1












-1








-1








I have a slots machine game HTML5 + Javascript.
I have found the below algoritm on Stackoverflow, but i need another percentage



this.d = Math.random();

if (this.d < 0.3333) {
this.nextSymbols = [
['320', '320', 'jackpot'],
['jackpot', '320', '400'],
['320', '320', 'jackpot'],
];
} else if (this.d < 0.005) {
this.nextSymbols = [
['320', 'jackpot', '400'],
['320', 'jackpot', '400'],
['320', 'jackpot', '400'],
];
} else {
this.nextSymbols = [
[Symbol.random(), Symbol.random(), Symbol.random()],
[Symbol.random(), Symbol.random(), Symbol.random()],
[Symbol.random(), Symbol.random(), Symbol.random()],
];
}


I need to have:



 1 of 3 spins to be a winning prize 2 

1 of 200 spins wins big prize


Any math expert advice would be much appreciate.



Thanks










share|improve this question
















I have a slots machine game HTML5 + Javascript.
I have found the below algoritm on Stackoverflow, but i need another percentage



this.d = Math.random();

if (this.d < 0.3333) {
this.nextSymbols = [
['320', '320', 'jackpot'],
['jackpot', '320', '400'],
['320', '320', 'jackpot'],
];
} else if (this.d < 0.005) {
this.nextSymbols = [
['320', 'jackpot', '400'],
['320', 'jackpot', '400'],
['320', 'jackpot', '400'],
];
} else {
this.nextSymbols = [
[Symbol.random(), Symbol.random(), Symbol.random()],
[Symbol.random(), Symbol.random(), Symbol.random()],
[Symbol.random(), Symbol.random(), Symbol.random()],
];
}


I need to have:



 1 of 3 spins to be a winning prize 2 

1 of 200 spins wins big prize


Any math expert advice would be much appreciate.



Thanks







javascript algorithm






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 19 '18 at 22:07







PixelArtie

















asked Nov 19 '18 at 21:59









PixelArtiePixelArtie

43




43








  • 1





    How is "I have 10 big prizes" relevant for the probability of a single spin?

    – Bergi
    Nov 19 '18 at 22:04






  • 2





    Hint: 1/3 = 0.3333…, 1/200 = 0.005 (compared to 50% = 0.5, 20% = 0.2 in the existing code)

    – Bergi
    Nov 19 '18 at 22:05













  • @Bergi thanks. I figured out

    – PixelArtie
    Nov 19 '18 at 22:21






  • 1





    < means less than. So if (this.d < 0.005) means that the value is less than 0.005. Any value where that is the case is also less than 0.3333. However the if-test for <0.005 is in the else clause of the <0.3333, so nobody will ever win the big prize.

    – Richardissimo
    Nov 19 '18 at 22:39














  • 1





    How is "I have 10 big prizes" relevant for the probability of a single spin?

    – Bergi
    Nov 19 '18 at 22:04






  • 2





    Hint: 1/3 = 0.3333…, 1/200 = 0.005 (compared to 50% = 0.5, 20% = 0.2 in the existing code)

    – Bergi
    Nov 19 '18 at 22:05













  • @Bergi thanks. I figured out

    – PixelArtie
    Nov 19 '18 at 22:21






  • 1





    < means less than. So if (this.d < 0.005) means that the value is less than 0.005. Any value where that is the case is also less than 0.3333. However the if-test for <0.005 is in the else clause of the <0.3333, so nobody will ever win the big prize.

    – Richardissimo
    Nov 19 '18 at 22:39








1




1





How is "I have 10 big prizes" relevant for the probability of a single spin?

– Bergi
Nov 19 '18 at 22:04





How is "I have 10 big prizes" relevant for the probability of a single spin?

– Bergi
Nov 19 '18 at 22:04




2




2





Hint: 1/3 = 0.3333…, 1/200 = 0.005 (compared to 50% = 0.5, 20% = 0.2 in the existing code)

– Bergi
Nov 19 '18 at 22:05







Hint: 1/3 = 0.3333…, 1/200 = 0.005 (compared to 50% = 0.5, 20% = 0.2 in the existing code)

– Bergi
Nov 19 '18 at 22:05















@Bergi thanks. I figured out

– PixelArtie
Nov 19 '18 at 22:21





@Bergi thanks. I figured out

– PixelArtie
Nov 19 '18 at 22:21




1




1





< means less than. So if (this.d < 0.005) means that the value is less than 0.005. Any value where that is the case is also less than 0.3333. However the if-test for <0.005 is in the else clause of the <0.3333, so nobody will ever win the big prize.

– Richardissimo
Nov 19 '18 at 22:39





< means less than. So if (this.d < 0.005) means that the value is less than 0.005. Any value where that is the case is also less than 0.3333. However the if-test for <0.005 is in the else clause of the <0.3333, so nobody will ever win the big prize.

– Richardissimo
Nov 19 '18 at 22:39












1 Answer
1






active

oldest

votes


















0














To summarize - as @Bergi and @Richardissimo written in comments.
1 / 3 is 0.(3), but you can just write thid.d < 1/3 interpreter will calculate it for you.
And if you are making else if the second case will never fire. You have to change the order of conditions.






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    To summarize - as @Bergi and @Richardissimo written in comments.
    1 / 3 is 0.(3), but you can just write thid.d < 1/3 interpreter will calculate it for you.
    And if you are making else if the second case will never fire. You have to change the order of conditions.






    share|improve this answer




























      0














      To summarize - as @Bergi and @Richardissimo written in comments.
      1 / 3 is 0.(3), but you can just write thid.d < 1/3 interpreter will calculate it for you.
      And if you are making else if the second case will never fire. You have to change the order of conditions.






      share|improve this answer


























        0












        0








        0







        To summarize - as @Bergi and @Richardissimo written in comments.
        1 / 3 is 0.(3), but you can just write thid.d < 1/3 interpreter will calculate it for you.
        And if you are making else if the second case will never fire. You have to change the order of conditions.






        share|improve this answer













        To summarize - as @Bergi and @Richardissimo written in comments.
        1 / 3 is 0.(3), but you can just write thid.d < 1/3 interpreter will calculate it for you.
        And if you are making else if the second case will never fire. You have to change the order of conditions.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 19 '18 at 23:25









        KonowyKonowy

        369420




        369420
































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