Probability Question
New to this forum so hope I have done this right?
Recently I was playing a slot machine game cant remember the name of it :( and it had a feature where there was a Gold, Silver and Bronze pot.
When the feature was triggered, there was 15 boxes to choose from and that you could continue to pick until you got 3 the same.
Out of the 15, providing it showed true was 3 Gold, 4 Silver and 8 Bronze
What is the true probability of getting the Gold, Silver and Bronze ? and how do you work it out?
Could anyone give me a pointer in to how this is worked out? As im guessing that its not a case of that the bronze is 2 times easier than silver is it?
Any help would be great, thanks.
If there are no additional informations about the boxes available, there is no particular favourable picking strategy.
So we can just pick our boxes in a specific order (say from upper left to lower right) until we reveal 3 equal featues.
Then we just need to calculate the probability of results for all combinations:
3G, 2S, 2B
3G, 2S, 1B
3G, 2S, 0B
3G, 1S, 2B
3G, 1S, 1B
3G, 1S, 0B
3G, 0S, 2B
3G, 0S, 1B
3G, 0S, 0B
2G, 3S, 2B
2G, 3S, 1B
...
0G, 3S, 0B
2G, 2S, 3B
...
0G, 0S, 3B
For example, take the line "2G, 3S, 1B":
What is the probability getting 2 golds, 3 silvers, and 1 bronce in 6 draws (in any arrangement, without replacement) ?
There are 6! / (2! * 3! * 1!) = 60 different arrangements.
Each specific arrangement has the drawing probability 3/15 * 2/14 * 4/13 * 3/12 * 2/11 * 8/10 = 8/25025.
So the line "2G, 3S, 1B" gets the probability 96/5005.
Repeat for every other line, and sum things up.
Quote: MangoJThis is my first guess, so I don't know if it is correct of if there is a simpler one.
If there are no additional informations about the boxes available, there is no particular favourable picking strategy.
So we can just pick our boxes in a specific order (say from upper left to lower right) until we reveal 3 equal featues.
Then we just need to calculate the probability of results for all combinations:
3G, 2S, 2B
3G, 2S, 1B
3G, 2S, 0B
3G, 1S, 2B
3G, 1S, 1B
3G, 1S, 0B
3G, 0S, 2B
3G, 0S, 1B
3G, 0S, 0B
2G, 3S, 2B
2G, 3S, 1B
...
0G, 3S, 0B
2G, 2S, 3B
...
0G, 0S, 3B
For example, take the line "2G, 3S, 1B":
What is the probability getting 2 golds, 3 silvers, and 1 bronce in 6 draws (in any arrangement, without replacement) ?
There are 6! / (2! * 3! * 1!) = 60 different arrangements.
Each specific arrangement has the drawing probability 3/15 * 2/14 * 4/13 * 3/12 * 2/11 * 8/10 = 8/25025.
So the line "2G, 3S, 1B" gets the probability 96/5005.
Repeat for every other line, and sum things up.
I think this is close. But, the order matters just a little bit. In the case of 2G, 3S, 1B: the set {G, G, S, S, B} can be chosen in any order, then the last S will be chosen. The odds of this happening are combin(3,2)*combin(4,2)*combin(8,1)*combin(4-2,1)/combin(15,6)/combin(6,1) = 48/5005.
I realize that I can simplify the combin(x,1) to x, but I think it's clear to write it the way I did to show that this is a choice.
When you run through all of the possibilities, you should find that the bronze has a chance of 0.808391608, silver 0.142457542, and gold 0.049150849.
Quote: CrystalMathbronze has a chance of 0.808391608, silver 0.142457542, and gold 0.049150849.
I agree with/can confirm these figures.
Quote: CrystalMathI think this is close. But, the order matters just a little bit. In the case of 2G, 3S, 1B: the set {G, G, S, S, B} can be chosen in any order, then the last S will be chosen.
Yes you are perfectly right, the last box opened must contain the winning feature.
Many thanks for the answers and some of the workings.
However im stuggling a little to fully understand how you came to those figures?
Im fascinated with these kind of features and are great probability puzzles, but rather than ask you guys how they work it out all the time im really keen to be able to myself.
Having added your answers up it comes to 0.999999999 but it should be 1.00000 in all or was that just down to rounding?
I also am unsure why the order slightly matters in this senario as theres always an end result of either gold, silver or bronze being awarded, so what was the reason for allowing for what order they were picked if we are only working out the probability of which the player would get on average?
So if you could maybe say break down in fool proof workings say just the chance of say winning the gold, I could have a stab at working out the silver and bronze myself.
The reason im confused too is the fact that I got it to only 27 outcomes,
G S B
3 2 2
3 2 1
3 2 0
3 1 2
3 1 1
3 1 0
3 0 2
3 0 1
3 0 0
2 3 2
2 3 1
2 3 0
1 3 2
1 3 1
1 3 0
0 3 2
0 3 1
0 3 0
2 2 3
2 1 3
2 0 3
1 2 3
1 1 3
1 0 3
0 2 3
0 1 3
0 0 3
So im way off ?? on how to do it?
Quote: Loveslots20Having added your answers up it comes to 0.999999999 but it should be 1.00000 in all or was that just down to rounding?
Of course this is due to rounding, as the answers itself are rounded numbers.
Quote:
I also am unsure why the order slightly matters in this senario as theres always an end result of either gold, silver or bronze being awarded, so what was the reason for allowing for what order they were picked if we are only working out the probability of which the player would get on average?
The last feature you reveal is always the price you are awarded. You cannot get the gold award if your last feature opened is the bronce. So the order does matter. However you are lucky that this order will only affect the last opened feature.
Quote:
So if you could maybe say break down in fool proof workings say just the chance of say winning the gold, I could have a stab at working out the silver and bronze myself.
Excuse me Sir, but this is not Home Schooling. The method has been stated. Write down a list of all feature combinations, calculate their revealing probability, and sum things up. You already have the list, the calculation of the probability has been stated. I don't think you need lessons on calculating sums.
Quote: MangoJ
Excuse me Sir, but this is not Home Schooling. The method has been stated. Write down a list of all feature combinations, calculate their revealing probability, and sum things up. You already have the list, the calculation of the probability has been stated. I don't think you need lessons on calculating sums.
Well at 43 im a bit old to be home schooled ;)
Im sorry if I have crossed a line here, im new on here, I just didn't fully understand the method that was used, and im unclear on how many feature combinations there are as on an earlier post it mentioned 60? Yet I get 27?
So if you can clarify just 2 points I would be very grateful.
there is only 27 sets to calculate?
the set {G, G, S, S, B, S} eg one of the Silver wins, as stated by CrystalMath the odds of this happening are combin(3,2)*combin(4,2)*combin(8,1)*combin(4-2,1)/combin(15,6)/combin(6,1) = 48/5005
So am I right that this is because (2 of the 3 gold were picked) * (2 of the 4 silver picked) * ( 1 of the 8 bronze picked ) * ( 1 of the remaining 2 silvers picked ) / ( combin (total choices, total picked)) / ( the last combin(6,1) ) is what I dont know why that is needed??
If MangoJ or anyone else could just help with that bit, thanks in advance.
Using the method above "The probability" of getting
B B B is the same as B B S B ?
and
S S G G B B S same as S S B B S?
Both come out at 0.123076923 and 0.22377622 respectively. How can the odds be the same? a mistake maybe?
Finally for fun I worked out what would happen if I changed one bronze with an extra Silver, and did not get the result I was expecting?
Gold win went up slightly to 0.53446553 from 0.049150849 even tho no extra golds added? how so ?
Sorry to be a pain but it really bugs me when I cant work out or account for why something dont seem right :(
Or am I reading to much into it?
