June 23rd, 2013 at 1:24:23 PM
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Hello guys,
Please help me understand how to calculate the probability of the following situation.
Let’s say there are 80 cards on a table. Every card has a number from 1 to 80 and each card has a different number. I don't see the numbers.
1. What is the chance that I will pick the card with the number 3 (only 3) if I have 20 trials? Each card I picked goes away from the table after I pick it up.
2. What is the chance that I will pick the card with the number 5, 15, 25, 35, 45, 55, 65, 75 if I have 20 trials? Each card I picked goes away from the table after I pick it up.
Many thanks in advance :-)
Stats Beginner
Please help me understand how to calculate the probability of the following situation.
Let’s say there are 80 cards on a table. Every card has a number from 1 to 80 and each card has a different number. I don't see the numbers.
1. What is the chance that I will pick the card with the number 3 (only 3) if I have 20 trials? Each card I picked goes away from the table after I pick it up.
2. What is the chance that I will pick the card with the number 5, 15, 25, 35, 45, 55, 65, 75 if I have 20 trials? Each card I picked goes away from the table after I pick it up.
Many thanks in advance :-)
Stats Beginner
June 23rd, 2013 at 2:20:46 PM
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The probability of catching a certain number in Keno is described by the hypergeometric distribution.
Let:
N = number of spots you mark
K = number of spots you catch
R = range of numbers drawn from
D = size of draw
Then the probability p of catching K out of N, when the game draws D from a field of R, is:
p(N,K,R,D) = C(N, K) * C(R-N, D-K) / C(R, D), where C(n, r) is the combinations function: how many unordered ways to select r elements from a set of n without replacement.
For question 1:
C(1, 1) * C(79, 19) / C(80, 20) = 0.25. In other words, you have a 1 in 4 chance of winning a one-spot keno ticket.
I'll let you figure out the rest.
Let:
N = number of spots you mark
K = number of spots you catch
R = range of numbers drawn from
D = size of draw
Then the probability p of catching K out of N, when the game draws D from a field of R, is:
p(N,K,R,D) = C(N, K) * C(R-N, D-K) / C(R, D), where C(n, r) is the combinations function: how many unordered ways to select r elements from a set of n without replacement.
For question 1:
C(1, 1) * C(79, 19) / C(80, 20) = 0.25. In other words, you have a 1 in 4 chance of winning a one-spot keno ticket.
I'll let you figure out the rest.
"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
June 24th, 2013 at 12:37:15 AM
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Hi MathExtremist,
I feel it's not so hard for you ;-)
Many thanks for your answer, I understood how this works and even succeeded to build this function in the Excel. But in order to answer my second question I don't understand what value I need to change? I thought that now K should be 8, but the Excel says me that there is an error with this number in the formula I build. Can you help please?
I feel it's not so hard for you ;-)
Many thanks for your answer, I understood how this works and even succeeded to build this function in the Excel. But in order to answer my second question I don't understand what value I need to change? I thought that now K should be 8, but the Excel says me that there is an error with this number in the formula I build. Can you help please?
June 24th, 2013 at 2:58:26 AM
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Both N and K need to be changed to 8.
In your first question, you wanted to know the probability of matching K=1 of N=1 spots.
In your second question you want to know the probability of matching K=8 of N=8 spots.
The following Excel formula answers your first question: =HYPGEOMDIST(1,1,20,80)
The following Excel formula answers your second question: =HYPGEOMDIST(8,8,20,80)
The following Excel formula calculates the probability of matching X of Y picks in Keno:
Just substitute X and Y accordingly. For example, to calculate the probability of matching 4 numbers when 9 were marked:
In your first question, you wanted to know the probability of matching K=1 of N=1 spots.
In your second question you want to know the probability of matching K=8 of N=8 spots.
The following Excel formula answers your first question: =HYPGEOMDIST(1,1,20,80)
The following Excel formula answers your second question: =HYPGEOMDIST(8,8,20,80)
The following Excel formula calculates the probability of matching X of Y picks in Keno:
=HYPGEOMDIST(X,Y,20,80)
Just substitute X and Y accordingly. For example, to calculate the probability of matching 4 numbers when 9 were marked:
=HYPGEOMDIST(4,9,20,80)
June 24th, 2013 at 7:21:20 AM
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Hi JB,
Many thanks,
But I think that something is wrong with this stuff. I have used K=8 and N=8 and the result was totally surprising, p=0.0000043....
Before my calculations I felt that in the second case the probability should me much higher cause instead of catching only one number I can catch 8 numbers.
Can you please check it and tell me what I did wrong.
I have use the combin funcion in Excel as MathExtremist proposed.
p(N,K,R,D) = COMBIN(N, K) * COMBIN(R-N, D-K) / COMBIN(R, D)
I was confused using the HYPGEOMDIST function cause I didn't know exactly what is my Sample_s, Number_sample, Population_s and Number_pop
I am a real beginner, sorry :-)
Thanks for your help
Many thanks,
But I think that something is wrong with this stuff. I have used K=8 and N=8 and the result was totally surprising, p=0.0000043....
Before my calculations I felt that in the second case the probability should me much higher cause instead of catching only one number I can catch 8 numbers.
Can you please check it and tell me what I did wrong.
I have use the combin funcion in Excel as MathExtremist proposed.
p(N,K,R,D) = COMBIN(N, K) * COMBIN(R-N, D-K) / COMBIN(R, D)
I was confused using the HYPGEOMDIST function cause I didn't know exactly what is my Sample_s, Number_sample, Population_s and Number_pop
I am a real beginner, sorry :-)
Thanks for your help
June 24th, 2013 at 7:57:52 AM
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0.0000043 is correct. It is far less likely to match all 8 numbers (provided that you picked 8 numbers, in other words, a "bullseye") than to only match one of the 8 numbers. In fact, it is about 61317 times more likely to match 1 of 8 than it is to match 8 of 8.
June 24th, 2013 at 8:34:34 AM
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By clicking on the name of the function in Excel it (the click) will open up the Help section for you.Quote: StatsbeginnerI was confused using the HYPGEOMDIST function cause I didn't know exactly what is my Sample_s, Number_sample, Population_s and Number_pop
This is what my version of Excel spit out and more
"Syntax
HYPGEOMDIST(sample_s,number_sample,population_s,number_population)
Sample_s is the number of successes in the sample.
Number_sample is the size of the sample.
Population_s is the number of successes in the population.
Number_population is the population size.
Remarks..."
Teach yourself statistics
http://stattrek.com/probability-distributions/hypergeometric.aspx?Tutorial=Stat
winsome johnny (not Win some johnny)
June 24th, 2013 at 10:33:37 AM
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JB, you are totally right, matching 8 numbers is much harder than matching 1 number. And I really appreciate your help and the knowledge you gave me :-)
Simply I didn't explain my self properly.
So, there are 80 cards on a table. Every card has a number from 1 to 80 and each card has a different number. I don't see the numbers and I have 20 trials.
In order to win I need to pick one of the following cards: 5, 15, 25, 35, 45, 55, 65, 75. Each card I picked goes away from the table after I pick it up.
What is the chance?
Exactly regarding this case, I thought that K should be 8 but the Excel says me that there is an error with this number in the formula I built.
Maybe for this case the probability is higher than 1? And the Excel goes crazy because of this? :-)
Simply I didn't explain my self properly.
So, there are 80 cards on a table. Every card has a number from 1 to 80 and each card has a different number. I don't see the numbers and I have 20 trials.
In order to win I need to pick one of the following cards: 5, 15, 25, 35, 45, 55, 65, 75. Each card I picked goes away from the table after I pick it up.
What is the chance?
Exactly regarding this case, I thought that K should be 8 but the Excel says me that there is an error with this number in the formula I built.
Maybe for this case the probability is higher than 1? And the Excel goes crazy because of this? :-)
June 24th, 2013 at 10:37:01 AM
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Hi 7craps,
Many thanks for this link and excellent explanation, I will study this.
Many thanks for this link and excellent explanation, I will study this.
June 24th, 2013 at 11:12:43 AM
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Did you get confused by JB's post? Wrong colors??Quote: StatsbeginnerSo, there are 80 cards on a table. Every card has a number from 1 to 80 and each card has a different number. I don't see the numbers and I have 20 trials.
In order to win I need to pick one of the following cards: 5, 15, 25, 35, 45, 55, 65, 75. Each card I picked goes away from the table after I pick it up.
What is the chance?
Read and learn from his post.
=HYPGEOMDIST(8,8,20,80) 0.000004345660666
=HYPGEOMDIST(1,1,20,80) 0.25
=HYPGEOMDIST(4,9,20,80) 0.114105182095472
=HYPGEOMDIST(X,Y,20,80) X of Y
added:(after a swim and lunch)
OK.
You are not drawing 20 cards like Keno draws always 20 balls.
You draw one card until you win.
So the # of draws is a variable between 1 and 20 inclusive
so the prob of NOT drawing 1 of your 8#s on the 1st draw = 72/80
2nd draw: 71/79
3rd draw: 70/78
and so on to the 20th. Multiply them together
in Excel I get
P = 0.088266238 of NOT drawing 1 of your 8#s in 20 draws so
1-P should be your answer. Do we match??
1-P should be your answer. Do we match??
winsome johnny (not Win some johnny)
June 24th, 2013 at 11:31:26 AM
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Do you need to pick exactly one of the cards? If so, then use =hypgeomdist(1,8,20,80).
Can you pick more than one of the cards and still win? If so, then use =1-hypgeomdist(0,8,20,80).
Can you pick more than one of the cards and still win? If so, then use =1-hypgeomdist(0,8,20,80).
I heart Crystal Math.
June 26th, 2013 at 12:28:54 AM
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7craps,
Many thanks, your solution is perfect! I got exactly the same result! Respect!
CrystalMath,
Thanks, I understood, further I will use this formulas.
Have a nice day, guys!
Many thanks, your solution is perfect! I got exactly the same result! Respect!
CrystalMath,
Thanks, I understood, further I will use this formulas.
Have a nice day, guys!