# Thread: Whats the chances of...

1. ## Whats the chances of...

Hi,
My very first post so please be gentle lol

Im programming a simple slot machine that has 5 reels and a has 1 "star" on each.

What i want to know is if the Star can be on, above or below the pay line what is the probabilty of getting ANY 3 stars , Any 4 and any 5 etc

As for working out the 5 im thinking its a simple as 3 positions on each reel divide by 24 on each =

3 x 3 x 3 x 3 x 3 = 243
24 24 24 24 24 = 7962624

32768 / 243 = 1 in 32768 ??

but how do i work out the chances of getting ANY 3 or ANY 4 and not just say the first 3 ?

i did work out 3 as 1 in 512 but i think it must be different to that as you have 5 chances ( reels ) to get ANY 3 etc

Hope this makes sence to someone

2. It is quite possible that several of us can help you. But we must be able to understand the problem. Please give a simple description of each ‘reel’ in question.
How many positions are there on each reel?
There is exactly one star on each reel: CORRECT?
The reels operate independently: correct?
There are only five reels?

3. Originally Posted by banditman
Hi,
My very first post so please be gentle lol

Im programming a simple slot machine that has 5 reels and a has 1 "star" on each.

What i want to know is if the Star can be on, above or below the pay line what is the probabilty of getting ANY 3 stars , Any 4 and any 5 etc

As for working out the 5 im thinking its a simple as 3 positions on each reel divide by 24 on each =

3 x 3 x 3 x 3 x 3 = 243
24 24 24 24 24 = 7962624

32768 / 243 = 1 in 32768 ??

but how do i work out the chances of getting ANY 3 or ANY 4 and not just say the first 3 ?

i did work out 3 as 1 in 512 but i think it must be different to that as you have 5 chances ( reels ) to get ANY 3 etc

Hope this makes sence to someone
This is a binomial distribution problem.

On any reel the prob that the star is in a favourable position is $p=3/24=1/8$.

Then the probability of n from N reels show a star in a favourable position is
a binomial random variable $\sim B(N,p)$ so:

$
p(n;N,p) =\frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}
$

RonL

4. Originally Posted by Plato
It is quite possible that several of us can help you. But we must be able to understand the problem. Please give a simple description of each ‘reel’ in question.
How many positions are there on each reel?
There is exactly one star on each reel: CORRECT?
The reels operate independently: correct?
There are only five reels?

24 positions

1 star on each reel but can be in any of 3 positions so i suppose it is like haveing 3 ?

reels are independant and every spin has equal chance as previous spin

yes only 5 reels

Hope this is clearer

5. Originally Posted by CaptainBlack
This is a binomial distribution problem.

On any reel the prob that the star is in a favourable position is $p=3/24=1/8$.

Then the probability of n from N reels show a star in a favourable position is
a binomial random variable $\sim B(N,p)$ so:

$
p(n;N,p) =\frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}
$

RonL

Ok is there any chance i can have the last bit in simpler terms please eg what is the ! all about ??

Thanks

6. Hello, Banditman!

You're working on a Probability problem
. . and don't know about Combinations and Factorials
. . or anything about Binomial Probabilities?
I suggest a Google search . . .

I'm programming a simple slot machine that has 5 reels.
Each wheel has 24 positions and has one "star".

What I want to know is: if the Star can be on, above or below the pay line,
what is the probabilty of getting: 5 stars, 4 stars, 3 stars, etc.

The probability of a "win" on a wheel is: . $P(\star) \,=\,\frac{3}{24} \,=\,\frac{1}{8}$

The probability of "not win" on a wheel is: . $P(\star') \,=\,\frac{7}{8}$

Here are the probabilities:

$P(5\,\star) \;=\;C(5,5)\left(\frac{1}{8}\right)^5\left(\frac{7 }{8}\right)^0 \;=\;\frac{1}{32,768}$

$P(4\,\star) \;=\;C(5,4)\left(\frac{1}{8}\right)^4\left(\frac{7 }{8}\right)^1 \;=\;\frac{35}{32,768}$

$P(3\,\star) \;=\;C(5,3)\left(\frac{1}{8}\right)^3\left(\frac{7 }{8}\right)^2 \;=\;\frac{490}{32,768}$

$P(2\,\star) \;=\;C(5,2)\left(\frac{1}{8}\right)^2\left(\frac{7 }{8}\right)^3 \;=\;\frac{3,430}{32,768}$

$P(1\,\star) \;=\;C(5,1)\left(\frac{1}{8}\right)^1\left(\frac{7 }{8}\right)^4 \;=\;\frac{12,005}{32,768}$

$P(0\,\star) \;=\;C(5,0)\left(\frac{1}{8}\right)^0\left(\frac{7 }{8}\right)^5 \;=\;\frac{16,807}{32,768}$

7. I do know the basics and i do use spreadsheets to do the work for me it was just the first time i used this kinda win method

The machine is only going to give you a win for any combination of 3 stars where as all the other wins have to be on the payline and left to right

Many thanks for taking the time to help me out