Probability problem of a slot bonus

fruitman

New member
Joined
Jun 3, 2015
Messages
7
Hi All,

Im new so please be gentle :)

Im trying to work out the probability of a slot bonus.


There is 20 boxes containing randomly seeded and hidden.

Assuming it dont cheat and using the values below im trying to work out the probability of getting each colour.

3 x BLUE
4 x PURPLE
5 x ORANGE
8 x GREEN

The player picks until 3 the same are revealed and then wins that colour.

So far I have worked out that there is only 27 outcomes for each colour due to the order isnt strictly relevent.

eg

BBOOPGGG is same as if it had been BOGOPBGG = same result GREEN AWARDED

so 4 colours = 108 actual outcomes to work out.

Im also assuming that each one of the 108 can be worked out to get an overall break down.

But this is where im getting stuck

im using the combin function in excel like this :-

BBPOOGGG

=COMBIN(3,2)*COMBIN(4,1)*COMBIN(5,2)*COMBIN(8,3)/COMBIN(20,8)/COMBIN(8,1)

= 0.006668254

but after working out the other 107 it is higher than 1 which isnt right! as of course they should all add to 1

Anyone know what im missing / doing wrong ?

Regards,

Steve.
 
Hmmm....maximum is 9 drawn balls: BBPPOOGGB ; right?
And this happens only if 2 of each color are drawn by 8th draw; right?
Minimum is 3, of course.


Hi Denis yes your correct the maximum can be 9 draws and only would happen if you had 2 of all the colours at this point.

And also correct that the min is 3 as you could hit 3 Green, Orange,Purple or EXTREMELY lucky 3 blues on your first 3 picks!

Do you agree with me that there is only 27 ways to get each colour that I need to work out the probability for?
 
Hmmm....maximum is 9 drawn balls: BBPPOOGGB ; right?
And this happens only if 2 of each color are drawn by 8th draw; right?
Minimum is 3, of course.


Hi Denis, thanks for the reply, i did post a reply but seems it didnt post?


Anyway I said yes thats right, max 9 but only if you manage to pick 2 of each and you can get any colour by getting 3 straight off the bat although blue would be super hard to get from the initial 3 picks!

So in the senario i gave can you see where im going wrong? because i clearly am :(

Regards,

Steve.
 
Thanks for that Denis, that's great! very interesting, and does seem to fit in with roughly what I would expect.

However I'm still a bit frustrated about working it out as I used to be able to do all this stuff but since I hit 45 I'm really struggling :(


Hopefully one of the people that you know are good at this stuff on here will see this and take pity on me and help me actually work it out exactly because its doing my head in.


I still think the key is to work out each of the separate outcomes for each of the 4 awards but no matter what I seem to try, its wrong.


Someone please help an oldie, I dont expect it to be handed on a plate but throwing me a bone would be nice :) please,

Regards,

Steve.
 
Im trying to work out the probability of a slot bonus.
There is 20 boxes containing randomly seeded and hidden.
3 x BLUE
4 x PURPLE
5 x ORANGE
8 x GREEN
The player picks until 3 the same are revealed and then wins that colour.
Some of what has been said is correct but most is just off the wall.
I do trust Denis' approximations from his simulations.

The minimum number of draws is three and the maximum is nine for each colour.
Thus there twenty-eight calculations to completely solve the question.

In order to have: BBPPOOGG we calculate \(\displaystyle \dfrac{8!}{2^4}\left(\dfrac{3\cdot 2}{20\cdot 19}\right)\left(\dfrac{4\cdot 3}{18\cdot 17}\right)\left(\dfrac{5\cdot 4}{16\cdot 15}\right)\left(\dfrac{8\cdot 7}{14\cdot 13}\right)\) Study that until you understand it.
The first term allows any order. So to be done in nine draws that is the probability.
 
Last edited:
...

3 x BLUE
4 x PURPLE
5 x ORANGE
8 x GREEN

The player picks until 3 the same are revealed and then wins that colour....
As mentioned previously, the minimum number of draws is 3 and the maximum number of draws is 9. I'm not claiming the following is correct, but I would like to know what is wrong with the reasoning, if anything.

First of all, I am only talking about the the time it goes to the maximum number of draws. If we have the maximum number of draws the there have been 2 each of each color drawn leaving
1 x BLUE
2 x PURPLE
3 x ORANGE
6 x GREEN
for a total of 12 balls. Thus the probability of winning a
BLUE ball = 1/12 ~ 8.3%
PURPLE ball = 2/12 ~16.7%
ORANGE ball = 3/12 = 25%
GREEN ball = 6/12 = 50%.

Now I know you need some more numbers to combine all the number of draws from 3 to 9 but, again, I'm just talking about the maximum number of draws.
 
I don't quite follow that.
Result is: ~.001806
My simulator gives: ~39,800 as the number of times in a
million tries that a 9th draw is required.
What am I missing?
There were two typos now corrected.
Have a look at this. We are not far off.
 
Hmmmm, ok so im still confused, if not even more now lol


Im trying to understand how working out how many picks 3 - 9 is really relevant.

As that only tells you how many out of 3 - 9 picks are expect but not what those picks result in?

surely to work out the overall expected outcome for all 4 colours is working out the possible outcomes that end with 3 the same no matter how many picks it took.


Also someone mentioned that there was 28 end results ? I got it to 104 ( 27 x 4 ) so how and why am i getting this so wrong and not getting it :(

I really think I must be going senile,

Can someone put it a bit simpler for me, before I lose my sanity and give up totally !!
 
Probability that the 3 Blues "appear" before 3 of any others:
42607 / 1175720 = .0362390705269982235
or 36,239 out of a million tries...


Hi Denis, Thanks for that but as that equates to 1 in 27 tries on average to get blue it just feels like its wrong ( too easy ) , I'm not saying it is of course, but I have tried several different ways to work this out and my latest one came to

0.014615441 which is 1 in 68 ave and feels a better fit? Although another of my methods was 0.043846324 so its still confusing the **** out of me...... I WILL NOT BE BEAT !! I will keep going till im happy I got it right lol


Would it be possible to post your working out so that I can see how you got to that figure and so that I can work out the other 3 ( Purple, Orange and Green ) using your method ?

Regards,

Steve.
 
Would it be possible to post your working out so that I can see how you got to that figure and so that I can work out the other 3 ( Purple, Orange and Green ) using your method
You have asked a very complicated question. Look at the string below. That is from a generating function.
(Sum[x^(k),k=0,2])^4=x^8+4 x^7+10 x^6+16 x^5+19 x^4+16 x^3+10 x^2+4 x+1

From four categories ( Blue, Purple, Orange and Green ) we can select three even if we repeat. These are called multi-selections. You can study that page.

The term \(\displaystyle 16x^3\) tells us that there are sixteen three-selections that do not contain three of one kind.
Consider the selection \(\displaystyle [P,O,G]\) there is no order there, just content.
The probability of selecting that content in that order is \(\displaystyle \dfrac{4\cdot 5\cdot 8}{20\cdot 19\cdot 18}\)
Now there are six ways to order that string.
The selection G,P,G can be ordered in three ways each with probability \(\displaystyle \dfrac{8\cdot 4\cdot 7}{20\cdot 19\cdot 18}\)

The term \(\displaystyle 10x^6\) tells us there are ten six-selections in which there are not three of a kind.
Example: \(\displaystyle [P,O,P,G,B,G]\) that can be ordered in \(\displaystyle \dfrac{6!}{2!^2}\) ways with probability \(\displaystyle \dfrac{4\cdot 5\cdot 3\cdot 8\cdot 3\cdot 7}{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15}\)

Now we must do that for all terms 3 to 8. It ain't easy.I have taught courses in counting theory for many years in an upper level mathematics course. It would take me about six weeks to get good students to understand why we do the above. I understand your desire to master this. If I were you, I would find a course in discrete mathematics to audit.

Again it will not be easy to jump right in.
 
Last edited:
This is what's required to hit the 42607 / 1175720:
..........................................
Conceptually build a tree of all possible cases.

Start with a root with probability 1, and 20 cards, then deal the first card, it is either 1, 2, 3, 4, make four branches from the root, with probability 3/20, 4/20, 5/20, 8/20.

Repeat for each node at the next level, but now there are only 19 cards remaining, and multiply probability as you go down the tree (i.e. probability for the particular path), e.g. (3/20)*(2/19)*(1/18) for the 1,1,1 case.

At each node keep track of how many cards of each number have been played. As soon as a given branch reaches a count of 3 in any number, then that branch terminates. If the it was card number 1 that reached a count of 3, then include the resulting probability in the solution.
..........................................
This isn't my work (got it from someone at another site).

I was simply happy to see that the results of .0362390705269982235
pretty well matched my simulation escapade of +- 36,200 per million.

If this is for some sort of casino game, why are you trying to be so precise?

Thanks, It was actually a fairground game I saw recently and wanted to work out the expected payback % as well as the probability for all the prizes.

I have now come to terms that it is way above my mathematical skills and not something I could have ever worked out :(

Thanks to all the very clever people that have helped and given input on, im a bit deflated now but alls good :)

Regards,

Steve.
 
Top