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

javascript algorithm

share|improve this question

edited Nov 19 '18 at 22:07

PixelArtie

asked Nov 19 '18 at 21:59

PixelArtiePixelArtie

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
  

  
  
  
  

add a comment | 

-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

javascript algorithm

share|improve this question

edited Nov 19 '18 at 22:07

PixelArtie

asked Nov 19 '18 at 21:59

PixelArtiePixelArtie

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
  

  
  
  
  

add a comment | 

-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

javascript algorithm

share|improve this question

edited Nov 19 '18 at 22:07

PixelArtie

asked Nov 19 '18 at 21:59

PixelArtiePixelArtie

43

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

javascript algorithm

share|improve this question

edited Nov 19 '18 at 22:07

PixelArtie

asked Nov 19 '18 at 21:59

PixelArtiePixelArtie

43

share|improve this question

edited Nov 19 '18 at 22:07

PixelArtie

asked Nov 19 '18 at 21:59

PixelArtiePixelArtie

43

share|improve this question

share|improve this question

edited Nov 19 '18 at 22:07

PixelArtie

edited Nov 19 '18 at 22:07

PixelArtie

edited Nov 19 '18 at 22:07

PixelArtie

asked Nov 19 '18 at 21:59

PixelArtiePixelArtie

43

asked Nov 19 '18 at 21:59

PixelArtiePixelArtie

43

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  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

add a comment | 

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.

share|improve this answer

answered Nov 19 '18 at 23:25

KonowyKonowy

339420

add a comment | 

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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

answered Nov 19 '18 at 23:25

KonowyKonowy

339420

add a comment | 

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

answered Nov 19 '18 at 23:25

KonowyKonowy

339420

add a comment | 

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

answered Nov 19 '18 at 23:25

KonowyKonowy

339420

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

answered Nov 19 '18 at 23:25

KonowyKonowy

339420

share|improve this answer

share|improve this answer

answered Nov 19 '18 at 23:25

KonowyKonowy

339420

answered Nov 19 '18 at 23:25

KonowyKonowy

339420

answered Nov 19 '18 at 23:25

KonowyKonowy

339420

339420

add a comment | 

add a comment | 

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