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How to calculate probability

I have Keno game with modified rules and i need to calculate probability to win jackpot. (not the game itself).
Where is 75 numbers all together, five last numbers (71..75) are jackpot numbers.
For each game 10 to 15 random numbers are selected (from 1 to 75 without duplicates).
For the range from 1..70 ALWAYS exactly 10 numbers are selected.
If generated numbers are from 71 to 75 when overall count of selected numbers will be more than 10.
Once 10 numbers from 1 to 70 are selected game will stop.

Game examples:
selected numbers: 1,2,3,4,5,6,7,8,9,10
or : 71,1,2,3,4,5,6,7,8,9,10
or : 1,2,71,72,73,74,75,3,4,5,6,7,8,9,10 - in such case player wins jackpot.

Cost of each game is - 1$
Jackpot payout - 4000$
I need to know what is probability to win jackpot?
If probability is more than 1/4000 when game will go into negative payouts and it's a bad thing...

I don't know how to solve this correctly because it is not like 6/49 probability calculations since total count of selected numbers is not fixed.

Thank you!
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DRY_GIN
Asked:
DRY_GIN
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1 Solution
 
InteractiveMindCommented:
I'm struggling to figure this too, but my simulations consistently give 0.0001 < 1/4000.
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d-glitchCommented:
This is equivalent to the game you are playing:

There are 70 white balls numbered 1-70.
There are 5 gold balls numbered 71-75.
We don't care about any of the numbers.

You will always pick out 15 balls.
If all 5 gold balls are picked, somebody wins the jackpot.

There are C(75,15) = 2.28 x10^15  ways to pick 15 balls.

There are C(70,10) = 3.96 x10^11  ways to pick 10 white balls.
There are C(5,5) = 1 way to pick 5 gold balls.

The jackpot odds are 1.74 x10^-4     or   1 in 5747

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DRY_GINAuthor Commented:
2 d-glitch:
Thanks for posting d-glitch!
This probability you've calculated resulting in 5747.382617, i also got it using 6/49 formula
But because you assume where is exactly 15 pics out of 75 this is actually a "top measurement" meaning the real probability is somewhat less than this because you not picking up exactly 15 numbers most of the time, so the real probability will be less than that.
or maybe i'm mistaken , please correct me if is so.
again, because you not allowed to pick any more numbers if 10 numbers from 1..70 are already selected - the probability will be lower.
anyway, my main concern was to be sure it is less than 4000  and this is true.
2 InteractiveMind::
I also run simulations of that game for 50 million runs, and i got similar results to yours
approximately 1 in 9400
so i'm safe ;)
2 d-glitch: as you may see from simulations the value is almost twice as match as 5747 to 9400
so where is a big error in your approach
anyway, if nobody will provide correct formula I will happily split points to both of you.
here is visual basic code for Microsoft Excel I used to simulate the game results


Private Sub CommandButton1_Click()
Dim nums(75) As Boolean, wins As Long, i As Long, snum As Integer
Randomize
For i = 1 To 50000000
 For m = 1 To 75:  nums(m) = False: Next
lbl:
 aa = Int(1 + 76 * Rnd(5))
 If (aa > 75) Then GoTo lbl
 If nums(aa) = True Then GoTo lbl
 nums(aa) = True
 
 snum = 0
 For m = 1 To 70
   If nums(m) = True Then snum = snum + 1
 Next
 If snum < 10 Then GoTo lbl
 
 For m = 71 To 75
  If nums(m) = False Then Exit For
 Next
 If m = 76 Then wins = wins + 1
 If Int(i / 1000) * 1000 = i Then 'report once in 1000 runs
  Cells(8, 1) = i
  Cells(9, 1) = wins
  'Cells(10,1) is Cells(8, 1)/Cells(9, 1) - the odds of winning jackpot
  DoEvents
 End If
Next
End Sub

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DRY_GINAuthor Commented:
actually this is more optimized
Private Sub CommandButton1_Click()
Dim nums(75) As Boolean, wins As Long, i As Long, snum As Integer, jpnum As Integer
Randomize
For i = 1 To 50000000
 For m = 1 To 75:  nums(m) = False: Next
 jpnum = 0: snum = 0
lbl:
 aa = Int(1 + 76 * Rnd(5))
 If (aa > 75) Then GoTo lbl
 If nums(aa) = True Then GoTo lbl
 nums(aa) = True
 If aa < 71 Then snum = snum + 1 Else jpnum = jpnum + 1
 If snum < 10 Then GoTo lbl
 If jpnum = 5 Then wins = wins + 1
 If Int(i / 10000) * 10000 = i Then:   Cells(8, 1) = i:    Cells(9, 1) = wins:    DoEvents: 'report once in 10000 runs
Next
End Sub

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ozoCommented:
It was not explicitly stated in the question
but I infer from the discussion that "to win jackpot"
occurs when all the  (71..75) numbers are selected


> You will always pick out 15 balls.
> If all 5 gold balls are picked, somebody wins the jackpot.

Not exactly,  If a gold ball is the 15th ball, then the game would have ended before it was picked
so you must pick 5 gold balls out of the first 14 picks.
C(70,9)*C(5,5)/C(75,14)
or 1 in about 8621
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ozoCommented:
which is equivalent to coloring 14 balls blue, and 61 balls red,
then picking 5 of them, and requiring then to be all blue
C(14,5)/C(75,5) is a little easier to compute and is also about 1 in 8621
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DRY_GINAuthor Commented:
ozo - you right!
but you somehow should include if   a gold ball is the 14th ball, and 13 numbers already contain 10 blue balls it is also not going be picked (and so on)
C(14, 5) / (C(75, 5) + C(75, 4) + C(75, 3) + C(75, 2) + C(75, 1)) ???
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ozoCommented:
If 13 numbers already contain 10 blue balls, then 14 numbers will also contain 10 blue balls, so that is accounted for.
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DRY_GINAuthor Commented:
2 ozo
you've calculated 5 out of 14 selected out of 75
that is not the case.
in reality is 5 out of 14 out of 75 munus all 13 out of 75 and minus 12 out of 75 and - c(11,75) - c(10,75)

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ozoCommented:
Do you have an example of 5 gold balls out of 14 picks that does not win jackpot?
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DRY_GINAuthor Commented:
5 gold out of 14 - no,
but where is some combinations that will no lead for this to happen
this formula C(14,5)/C(75,5)  assumes you pick "any" 5 out of "14" if im not mistaken
but it's not any 5 out of 14 its special "5" that means if 4 out of 13 do not contain at least 4 out of 5 - you will never pick the fifth one.
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DRY_GINAuthor Commented:
sorry.
maybe i lost, where should be a total number of combination's possible
and because it is depending on what numbers are already selected it should affect the final formula.
in your case it is not, it does no include cases when first 10 numbers was less than 71 and so the rest of combination's should be omitted.
again I'm lost (whats why i post this question in the first place)
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d-glitchCommented:
>>  Not exactly,  If a gold ball is the 15th ball, then the game would have ended
       before it was picked.  So you must pick 5 gold balls out of the first 14 picks.

      C(70,9)*C(5,5)/C(75,14) => 1 in about 8621

ozo is certainly correct.

I realized my mistake on the train about 15 minutes after logging off yesterday.
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jtm111Commented:
I think I see dry_gin's hesitation.

Think of the draws as a binary tree structure. First node, pick = white or gold, within white, second pick = white or gold, and within gold, second pick = white or gold, etc. Each branch is a path.

You will have multiple paths through this tree, but all have different lengths -- e.g., one path is gold, gold, gold, gold, gold (length = 5), another path is gold, white, gold, gold, gold, gold (length = 6), etc.

But each path has a different length, and a winner is when you get 5 gold out of a series of sequential draws, up until 14 draws at which point = loser.

If you follow the way I describe each path, you can see that each path follows a hypergeometric probability distribution. Think it over. In hypergeometric distribution paradigm you have 5 marbles where 2 are gold and 3 are white, what's the probability of getting both gold in 2 draws, in 3 draws, in 4 draws, etc.

http://en.wikipedia.org/wiki/Hypergeometric_distribution

Since others have already programmed simulations, I'm not going to duplicate their effort. But my way of framing the problem might help. Perhaps?
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d-glitchCommented:
You need to count the number of Jackpot combinations.

You will pick balls out of your group of 75 until you have 10 white ones.

All of the Jackpots must have 5 gold balls in the first 14.
If there are only 4 gold balls in the first 14, then there are 10 white and you stop.

I agree with ozo's analysis completely.  It gives you a way to count the number of
Jackpots out of C(75,14) possibilities.

In ozo's analysis, order doesn't matter, and in your game it does.  But the Jackpot calculations are still correct.

Think of it this way:
You will always pick 15 balls and keep them in order.
After you've picked 14 balls, you can tell if it's a Jackpot.  
Jackpot odds are 1 in 8621 per ozo.

After you've picked 15 balls this way, you do it your way.
Pick the same balls again, in the same order, stopping after
you get 10 white.
There is no loss of generality.

I think this is much clearer than hypergeometric distributions and binary trees.





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ozoCommented:
> when first 10 numbers was less than 71 , the first 14 numbers will not contain 5 gold balls,
winning the  jackpot would occur when and only when the first 14 numbers contain  5 gold balls,
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DRY_GINAuthor Commented:
I had to agree with ozo
i tried few different approaches, but the bottom line is that the last 15-th ball should be "white" in all such cases user wins jackpot
and C(70,9)/C(75,14) is the correct formula giving 1 in 8621
also d-glitch provided C(75,15) / C(70,10)  giving 1 in 5747 what is not considering last ball should be white, and it becomes  8621 if you divide it by 1.5 (that is probability that 10/15 should be the last ball)
so im closing the questions
many thanks to everyone!
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