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# what is the probability of... #3

Tue 30 Aug, 2011 07:54 am
Have an example:
123455
111555

the sample space is 0..9
row 1 can pick 6 numbers out of which 2 are repeated
row 2 can pick 6 numbers out of which 3 by 3 are repeated

I want to know what is the real probability that row 1 will match with 2 distinct numbers numbers row 2, and a repeated number (which is one of those first two distinct numbers if they match row 2). Warning, the repeated number (6th in the example above) is not obligated to match one of those two numbers.

The chance that row 1 matches row 2 with 2 numbers is:
C(5,2)C(10-5,0)/C(10,2)=10/45
C(2,2)C(10-2,3)/C(10,5)=56/252

The chance that row 1 repeated number matches row 2 numbers is:
C(5,1)C(10-5,1)/C(10,2)=25/45
C(2,1)C(10-2,4)/C(10,5)=140/252

how to combine these two probabilities to find out the real probability?

can I multiply them like this? 10/45 x 25/45 = 5/81? The result seems to low in comparison.
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1
Tue 30 Aug, 2011 08:29 am
But if I calculate like this I got a different answer:
1/10 x 1/9 x 1/8 x 1/7 x 1/6 x 1/10?
10!/5!(10-5)!10
126/5
final probabilty: 5/126

How to calculate?
0 Replies

markr

1
Tue 30 Aug, 2011 01:27 pm
Assuming this is the problem:
- Digits are chosen from 0-9
- R1 picks 5 distinct digits and a 6th digit that is the same as one of the other 5
- R2 picks 3 instances of one digit and 3 instances of a second, distinct digit
- What is the probability that R1 and R2 will have 3 digits in common (R1's pair and another distinct digit)?

Given R1's choices, R2 has:
- C(1,1) ways to match R1's pair
- C(4,1) ways to match one of R1's other 4 digits
- C(10,2) total choices

P = C(1,1) * C(4,1) / C(10,2) = 4/45

Given R2's choices, R1 has:
- C(2,1) ways to select the pair
- C(1,1) ways to match the other triplet
- C(8,3) ways to select the 3 non-matching digits
- C(10,5)*C(5,1) total choices

P = C(2,1) * C(1,1) * C(8,3) / (C(10,5) * C(5,1)) = 4/45

1
Wed 31 Aug, 2011 01:08 am
@markr,
1) What is the formula you used
2)I could understand that C(2,1) * C(1,1) is a product coming from the multivariate hypergeometric but what about the other factor:
C(8,3) / (C(10,5) * C(5,1))

1
Wed 31 Aug, 2011 02:08 am
I also found proof of the result above in a simpler way:
C(5,2)C(10-5,0)/C(10,2)=10/45 for the hipergeometric part, then, because the repeated unit can only be one of those first two that matched, it has 2/10 chance to match them. The total probability becomes:
(10/45)*(2/10)=(1/45)*2=2/45

Still, what formula did you use?

1
Wed 31 Aug, 2011 03:11 am
Actually it's proof for the thread here:

-1
Wed 31 Aug, 2011 06:34 am
We can't have both 4/45 and 2/45 as the answer. Which one is right and why, and please write in the abstract formula so I can check for myself.
0 Replies

markr

1
Wed 31 Aug, 2011 09:12 am
Well, if you've got proof, you must be right. However, take a minute (that's all it should take) to run through a quick exercise:

1) Select an arbitrary set for row 1.
2) Determine how many winning possibilities there are for row 2. There are only 45 total; so even if you enumerate all of them (which I think you'll find unnecessary), it won't take long.

Did you get 2, 4, or something else?

1
Wed 31 Aug, 2011 10:07 am
@markr,
I may have asked 4 times what formula it is. I can't solve other problems by guessing around all the time from examples.

For the first part it looks like the multidimensional hypergeometric distribution, but nothing in the theory of the formula indicates your application, especially where you add two combinations in the numerator.

So I don't have time to count them because I've been trying to get a formula to work.
I don't get your second part at all, so I'm making a smaller example. No offense but your application looks adjusted, it may be very well inexperienced me, but I simply do not understant.

For example copying what I understood from your example:

12344
144
14(4)
24(4)
34(4)
13-
12-
23-

3 ways to match
3 ways to not match

C(1,1)C(2,1)?C(8?,2?)?/C(10,4)C(4,1)=?/?

144
12344
14(4)
24(4)
34(4)
13-
12-
23-

3 ways to match
3 ways to not match

C(1,1)C(3,1)/C(10,2)=3/45!?

1
Wed 31 Aug, 2011 11:12 am
Told you I got no time, I'm getting extremely bored
I know I messed up in a hurry above, but one's thing is certain:
the correct ratio starts from something like this
144
12344
14(4)
13
12
24
23
34

"The player" row1 with ? distributed 6 numbers from 0 to 9 can only match in maximum 45 ways with "the lottery" row2
?/3/C(10,2)=45
and it all depends on the player. Since the hypergeometric of the part w/o repetition is 10/45, the answer has to be between 1/45 and 10/45, that is certain. I haven't a clue which method is correct but I'm obsessed with "the right way" )

I've posted on 5 forums and nobody can do a scientific demo or interpret data.

Even if this matches it doesn't mean it's the right probability:
C(5,2)C(10-5,0)/C(10,2)=10/45
C(2,2)C(10-2,3)/C(10,5)=56/252
markr

1
Wed 31 Aug, 2011 11:29 am
My formula(s) and the thought process that led to them were included in my first response to this post.

"So I don't have time to count them because I've been trying to get a formula to work."

If you don't have time to count them, you might as well give up. Computing probabilities is all about counting. By that, I don't mean testing every possible outcome and determining if it is desired. I mean using combinatorial techniques to count all of the desired outcomes. There isn't going to be a one-size-fits-all formula for every problem. Rather than focusing on how you can (mis)apply the hypergeometric distribution to each your problems, focus on the specifics of that problem. You're coming up with problems with very specific constraints. I may be wrong, but I doubt you'll find solutions with the HGD.

Your response doesn't surprise me. I've sensed a need to race toward a solution by rushing down paths with no real justification for taking those paths.

"I can't solve other problems by guessing around all the time from examples."

You're absolutely right, but what you appear to be doing is guessing around with variations of standard formulas instead of analyzing the problem you posed and inventing your own solution.

As I stated, it should take you no more than a minute to come up with all possible row 2 "winners" for an arbitrary row 1. For instance:

row 1: 1, 2, 3, 4, 5, 5

And in doing so, you'll gain insight into your problem that will lead you toward a correct counting method.
0 Replies

markr

0
Wed 31 Aug, 2011 11:31 am
If you're getting bored with your own problems, then stop inventing them...

1
Wed 31 Aug, 2011 11:45 am
@markr,

You can only approximate samples w/ repetitions when the trial number is infinite. This is a discrete case AFAIK. First you didn't involve the ways to fail in both calculations, the used as the total number the other experiment, there's too many whys to understand your answer. I only managed to waste time now and have to repost because this one is too poluted.

I can also come up with matching hipergeometric distributions and invent a match to the problem and pretend it's proof, when it's not:
15/210=18/252
I't isn't easy in my problems to come up with combinations that yield 18 though
markr

0
Wed 31 Aug, 2011 12:03 pm
You're tired of me, and I've run out of patience with you. It's time to go our separate ways...
0 Replies

1
Thu 1 Sep, 2011 07:42 am
Problem solved experimentally, but need to find an elegant combinatorial formula for approximating the result instead.
Also need verification of the solution below.

Problem data is:
sample space is numbers from 0 to 9
lottery picks 6 numbers, but only 2 are distinct (Example: 111444)
player picks 6 numbers, 5 distinct numbers and a bonus ball that repeats with one of the 5 distinct (Example: 012344)

What is the probability that the player will match the lottery with 3 numbers?
Note that this means that the player matches the lottery with the bonus ball, the number it repeats and a distinct number.

Match example:
111444
012344
Match:1, 4, 4

Solution:

There are C(10,5)=252 ways to select the 5 distinct numbers;
There are 252 x 5 = 1260 ways to select 5 distinct numbers with another number (say {B}onus ball) that repeats with one of them ;
There are 45 x 1260 = 56700 ways to select the lottery and player;
There are 11340 ways from 56700 to select only the ones where the lottery has a match with B;
Note that 56700/11340=5;
There are 5040 ways from 11340 to select only the ones where the lottery has a match with B and a distinct number of the player;

The chance that this player matches the lottery so that they match 3 numbers is 1/5040
The chance that the player doesn't match the lottery so that they match 3 numbers is 1/56700-5040=1/51660

What is the formula that approximates the 1/5040 probability?
0 Replies

1
Thu 1 Sep, 2011 09:15 am
I didn't calculate the probability right in the last part, it's 5040/56700, which is 0.0(8), or 4/45.

Explanation:

Where did I go wrong:
"
C(5,2)C(10-5,0)/C(10,2)=10/45 for the hipergeometric part, then, because the repeated unit can only be one of those first two that matched, it has 2/10 chance to match them. The total probability becomes:
(10/45)*(2/10)=(1/45)*2=2/45
"

In actuality, the repeated units can only repeat with the other 5 distinct numbers (in 5 ways - row 1, the player) and it has 2 chances to match the other row 2 distinct numbers (row 2, the lottery) . This means 2/5 probability. Since the hypergeometric distribution probability is an independent event from the repeated unit and it's probability, because the repeated unit doesn't reduce the set w/o repetition when it's picked it can be multiplied with 2/5, the repeated unit probability.
(10/45)*(2/5)=(2/45)*2=4/45

This matches this solution:
http://en.wikipedia.org/wiki/Lottery_mathematics
Powerballs And Bonus Balls
"
The general formula for B matching balls in a N choose K lottery with zero bonus ball from the N pool of balls is: {N-K-K+B\over N-K}{K\choose B}{N-K\choose K-B}\over {N\choose K}
"
With the twist that the there's 2 attempts when picking the repeated number, the two distinct numbers from row 2, the lottery.

P.S. markr's reciprocal solution is adjusted to support its counterpart, or invented (that's how he described it). That's why he can't provide a scientific explanation for it.

1
Fri 2 Sep, 2011 11:04 am
Well, that wiki bit was inspiring, even if it that formula I posted is wrong for this problem because it's for no replacement balls.

I should've posted this one instead perhaps:

The general formula for B matching balls in a N choose K lottery with one bonus ball from a separate pool of P balls is: {1\over P}{K\choose B}{N-K\choose K-B}\over {N\choose K}

1/5C(2,1)C(10-2,2-1)]/C(10,2)=1/5)*2*8]/45=(16/5)/45
which is almost 4/45, which can be if I put up 1/4 instead, but
the truth is that the ball that repeats, it does so with 5 distinct numbers not 4

1
Wed 7 Sep, 2011 09:07 am
This is a conditional probability problem.

ABCDE;E
AAA;EEE
was calculated as 4/45
experiment sais 0.0659658333
3/45 is very close: 0.0(6)

Anyone knows how to calculate this probability properly?
0 Replies

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