Alethiar
Alethiar
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October 12th, 2011 at 2:43:58 PM permalink
Hi, I joined this forum seeing that people have solid math knowledge here, and hoping I could have some help with a game or two I've found.

In this first game I have to choose 11 numbers out of 80, those being 1-80. 20 Winning numbers are then chosen, and if you had picked at least 6 of the numbers, you keep count of how many you had and start a new game with the same rules.

Every time you get a total of 11 or a multiple, you receive a prize; there are 96 games per day - each day the counter is reset -, so getting at least 11 numbers shouldn't be too hard.

What I'd like to know is the chance I have each day to get 1 prize, 2 prizes, etc. and the math I need to calculate it.

Anyone can help?
MathExtremist
MathExtremist
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October 12th, 2011 at 3:04:43 PM permalink
You're describing a Markov chain on a Keno game. The Wizard has a keno page which includes a calculator. What you need is the probability of catching 0 through 11 numbers, and then you use that distribution in a Markov process with 96 steps. The state-space range will be between 0 and 96*11, and you need to find the probabilities of landing on any of states n*11 at any of the 96 steps, followed up by the aggregate probabilities of landing on 0..96 of the n*11 states.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
miplet
miplet
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October 12th, 2011 at 3:06:10 PM permalink
Go here and choose 11 picks. That will give you the odds of matching 0 to 11 numbers. You'll get to add to your count about 1 in 41 draws. You'll average .15 per game or 14.4 per day.
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CrystalMath
CrystalMath
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October 12th, 2011 at 9:56:19 PM permalink
I calculate that you will get 0.12 prizes, on average. This average includes getting multiple awards, but I don't have the answer broken down this way which would be significantly more difficult. If you are only interested in calculating your return, you would multiply 0.12 by the expected average value of a prize. I think I'll follow up with a simulation tomorrow to verify.
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CrystalMath
CrystalMath
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October 14th, 2011 at 9:14:35 AM permalink
I messed up and found an error in my math. The number of prizes, on average, is 0.008018835. The probability of getting more than one prize is virtually nill.
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MathExtremist
MathExtremist
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October 14th, 2011 at 10:31:20 AM permalink
Is it that every time you catch 6 or more, you add one to your counter and you get paid when the counter reaches 11, 22, 33, etc.?
Or is it that every time you catch 6 or more, you add the number you caught to your counter (6 or more) and you get paid when the counter reaches 11, 22, 33, etc.?
If it's the latter, the chance of winning a prize is almost zero.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
CrystalMath
CrystalMath
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October 14th, 2011 at 11:18:00 AM permalink
Quote: MathExtremist

Is it that every time you catch 6 or more, you add one to your counter and you get paid when the counter reaches 11, 22, 33, etc.?
Or is it that every time you catch 6 or more, you add the number you caught to your counter (6 or more) and you get paid when the counter reaches 11, 22, 33, etc.?
If it's the latter, the chance of winning a prize is almost zero.



My calculation assumes that they kept a cumulative counter of the total hits and increment it only when 6 or more have been hit in a game, so you would get paid when the counter reaches 11, 22, 33, ...

I also did not award anything if the counter skips one of those numbers. For instance, if the player hit 6 on one game then 6 on a later game, the total would be 12 and the player would not win anything.

In re-reading the question, it seems that hitting or exceeding 11 might get paid, and that the player would get a prize when the total hits or surpasses 11, 22, 33, ... In that case, the player could win a lot more often.

Without additional input from the original poster, I think we're at a stand still. Oh well.
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Alethiar
Alethiar
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October 15th, 2011 at 11:01:07 PM permalink
To answer some of the questions, the counter does indeed work on the numbers you catch, so a 6-hit ticket will get you 6 added to the counter.

Also, it is not needed to get to exactly 11,22,33... to get the prize: getting a 6-hit and a 7-hit tickets would bring the counter to 13, and that would mean you'd both get a prize for reaching 11, and be nearer to the next winning total of 22. So, the 11,22,33... are simply goals that can be reached and surpassed.

I can see why the question can be taken as getting those winning totals exactly, but that's not the case here. So, I guess there is a different set of calculations to be applied?
CrystalMath
CrystalMath
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October 16th, 2011 at 7:29:52 PM permalink
Given the new information, this is what I calculate:

# of WinsProbability
0 0.320699587396175
1 0.473935841278694
2 0.171714275344839
3 0.0287861050507089
4 0.00438475327129188
5 0.000455538856391035
6 2.30170184701225E-05
7 8.48688643064339E-07
8 3.2132512694606E-08
9 9.43287480004254E-10
10 1.86792959974164E-11
Average Number of Prizes 0.923683723158769
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Alethiar
Alethiar
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October 17th, 2011 at 9:56:50 PM permalink
Could you give me the calculations you have done to get those results, CrystalMath?

I'd like to understand what to use in cases like this.
CrystalMath
CrystalMath
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October 18th, 2011 at 8:45:21 AM permalink
PM me with your email address and I'll send it to you.
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