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what is the probability of... #2

Fri 26 Aug, 2011 11:07 am
This question I solved in accordance with this answered question (I need verification):
http://able2know.org/topic/175253-1

So I'm wondering whether I applied the math correctly.
The new problem is:

The sample space are the numbers 0 to 9.
The background story is a 6/49 lottery, calculating in a particular case the chance of player number units to match lottery number units.
The scenario conditions are:
- the player units are not sequencial amongst themselves, meaning there is no 1 and 2 or 3 and 4 in the player group of units.
- the lottery units have one pair of sequencial units, meaning a 1 and 2 or 3 and 4

Observations:
there can't be 6 units with no sequencial pair from 0 to 9:
0 2 4 6 8 or
1 3 5 7 9
*considering that 0 and 9 or 9 and 0 is a sequence
one unit out of six must repeat, so we're left with only 5 distinct units.

What is the probability, that a group of 5 distinct numbers (player's) will match another group of 5 numbers (the lottery's), given that the lottery has one sequenced pair, for instance:
lottery: 01 3 5 7
player: 1 3 5 7
we observe that the player can only match 4 units w/o having a sequence, though it uses 5 units to guess the lottery's five.

There is also a particularity, in one case the player is able to match all 5 units with the lottery units:
lottery: 12 4 6 8
player: 0 2 4 6 8
Because there's 11 ways to have a sequenced pair from 0 to 9, with 9 and 0 making a pair also, then in 9.(09)% cases player can match all 5 units, in the rest of 90.(90)% the player can match only 4.

How to calculate the probability of the match? I'm using the hypergeometric distribution formula:

90.(90)% of time: C(5,5)*C(10-5,1)/C(10,5)
9.(09)% if time: C(5,5)*C(10-5,0)/C(10,5)

Is this correct?

I assumed that in the largest case, you can use 5 units to guess the other five, but only 4 can match, so it's C(10-5,1), while in the particular case all numbers can match, so it's C(10-5,0).
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1
Fri 26 Aug, 2011 12:26 pm
Oops, friday evening and I got careless again

This particularity doesn't exist:
"
There is also a particularity, in one case the player is able to match all 5 units with the lottery units:
lottery: 12 4 6 8
player: 0 2 4 6 8
"

The player can only match 4 units with the lottery, so only this is relevant:
C(5,5)*C(10-5,1)/C(10,5)

Still, is it correct?
0 Replies

markr

1
Sat 27 Aug, 2011 01:04 pm
Does this summarize what you're trying to solve?

- The sample space is the numbers 0 to 9.
- Five unique numbers are selected by the player and the lottery.
- The player's numbers contain no adjacent pairs (0 and 9 are adjacent).
- The lottery's numbers contain exactly one adjacent pair.
- What is the probability that the player matches exactly four of the lottery's numbers?

1
Sat 27 Aug, 2011 04:44 pm
@markr,
The lottery draws 6 numbers and the player picks 6 too, the particular case here is that we assume the lottery draws only one sequenced unit in those 6 numbers ( 1 and 2 for example in drawing: <1> <12> 14 36 38 44 )

Also we restrict the player in this case to having a pick where he has no sequenced units in his numbers, which means he's got to have a doubled unit, like in the example below.

That way he can at most match with the above example only 5 numbers:
lottery units: 12 4 6 8 4
player units: 1 4 6 8 4

The player's sixth unit would have to be a double, in this case 1, 4, 6 or 8

From here I observe the problem:
Because we have a repeated unit we count it as a single distinct number, but C(4,5) can't be, and if we do it from the lottery towards the player C(5,4), I don't see how we can have combinatorial proof.

Perhaps C(5,5) is wrong, though it has the same result as C(4,4) as it should have been perhaps. In this case I can pick a single losing number so it's C(10-4,1)? I don't know...

The player can't hit more than 4 distinct units with these constraints, but he can actually win 2nd category (guess 5 numbers, meaning guessing 4 distinct numbers plus a repeated unit).

The player can't win 1st category with these constraints because of the third rule (no adjacent pairs in sequence 1 and 2 or 5 and 6 etc.)

markr

1
Sat 27 Aug, 2011 05:00 pm
@markr,
Assuming what I posted above, I don't know of a formulaic method to solve this problem in a general way.

However, it turns out that the player can select numbers in only two ways:
0, 2, 4, 6, 8
1, 3, 5, 7, 9

It also turns out that the lottery can select numbers in only 40 ways (four ways for each of the ten adjacent pairs).

So, given that the player has made his choice, how many of the 40 lottery options result in winning combinations. That turns out to be easy to compute. Since four of the numbers must match, and there can be only one adjacent pair, all of the winning lottery combinations consist of replacing one of the numbers in the player's selection with one of the two numbers adjacent to it. For instance, if the 2 is to be replaced in 0, 2, 4, 6, 8:
0, 2, 4, 6, 8 => 0, 3, 4, 6, 8
or
0, 2, 4, 6, 8 => 0, 1, 4, 6, 8

There are 10 ways to do this:
5 numbers that can be replaced * 2 potential replacement numbers.

Therefore, the probability of winning is 10/40.
0 Replies

markr

1
Sat 27 Aug, 2011 05:10 pm
You're confusing me with examples that contain numbers greater than 9. Also, is <12> the number 12 or a 1 and a 2?

Can you publish the problem as it was originally given to you? It's hard for me to determine if you've properly reduced the problem to a simpler problem if I don't know what the original problem was.

1
Sat 27 Aug, 2011 06:15 pm
@markr,
To avoid confusion, I used the word "numbers" for numbers from 1 to 49 in this 6/49 lottery, and the word "units", for the units of those numbers, 0 to 9, meaning the least significant digit, e.g. in the number 49, the unit is 9.

Where I gave the example:
"The lottery draws 6 numbers and the player picks 6 too, the particular case here is that we assume the lottery draws only one sequenced unit in those 6 numbers ( 1 and 2 for example in drawing: <1> <12> 14 36 38 44 )"
This is what it means:
drawing (all numbers): <1> <12> 14 36 38 44
units (all least significant digits of numbers above): 1-2 4 6 8 4

With the "-" I mean the sequence, and with < and > I signal the numbers in which the units form a sequence.

The problem wasn't given to me, it simply is a problem I got to solve to draw some conclusions, that I will give a taste of after I figure out how to calculate this particular probability. This problem isn't much different perhaps from the one you solved that I gave the link in the first message.

I don't understand how the lottery can have 40 way to have a sequence of units from 0 to 9, 9 and 0 included as a sequence and variations up to 6 units if possible:
0-1, 1-2, ..., 9-0
There are 10 types of single sequences and the rest of the units don't matter as long as they are distinct units and don't form another sequence. If you consider 9-0 sequence, there's less ways we can arrange the rest of the distinct, non-sequenced units. For instance:
Units: 1-2 4 6 8
can be only arranged as 12 5 7 9, which means that for each sequence we can have only one other arrangement of the other distinct non-sequenced units, which is a total of 20 (which I don't believe it matters towards this problem). The probability 1/4 looks too big to be real, the player can miss way more than 3/4 on the above match:
not)1 3 5 7
not)1 4 6 8
not)1 5 7 9
not)1 4 7 9
etc.
For all we know he can get all zeros, it's allowed: 0 0 0 0 0 0
We're interested in the probability the player will guess the most units of the lottery, which is maximum 4, but he can miss all of them in a lot of ways.

Also I don't understand the bit where you replace units.
I'm interested in the probability the player can guess the lottery outcome with the above restrictions, so this case in particular looks like this:
lottery units: 1-2 4 6 8 4
*possible player unit matches:
player units: 1 4 6 8 4
player units: 2 4 6 8 4
*possible player units not matching:
player units: 0 0 0 0 0 0
player units: 1 1 1 1 1 1
player units: 0 1 0 1 0 1
etc.

The fifth and sixth unit of the player or lottery can only be a repeat of one of the units of the player or lottery respectively, which is inconsequential to the first problem, because we're only interested in distinct units probability. Though I would be interested in the total probability of repeated units too. The fifth and sixth unit has to be a repeat, and each can vary in 5 ways so it's an extra 1/5 x 1/5 chance, multiplied with the first probability for the whole probability? Added or multiplied with the first problem's probability? They seem to be independent events so I would say multiply.
markr

1
Sat 27 Aug, 2011 06:24 pm
I was solving a different problem. I assumed that all numbers were in the range 0-9. I'll digest your latest post and see if I can understand what you're trying to solve.

1
Sat 27 Aug, 2011 06:28 pm
@markr,
Please notice I edited the end of my last message. The end of this problem has a tail of an extra probability that has to be combined with the main probability for a better result, if it's an independent event how I figured and is as easy as multiplying probabilities.
0 Replies

markr

1
Sat 27 Aug, 2011 09:54 pm
" For instance:
Units: 1-2 4 6 8
can be only arranged as 12 5 7 9, which means that for each sequence we can have only one other arrangement of the other distinct non-sequenced units, which is a total of 20"

Not true. There are four arrangements for each sequence.
1-2 4 6 8
1-2 4 6 9
1-2 4 7 9
1-2 5 7 9

"For all we know he can get all zeros, it's allowed: 0 0 0 0 0 0"

Actually, the most you can get of any particular unit is five (e.g. 1, 11, 21, 31, 41), and there are only four zeros (10, 20, 30, 40).

Since duplicates are allowed, the player should be able to match five of the lottery's digits. For example:
Player: 1, 3, 5, 7, 9, 11
Lottery: 1, 3, 5, 8, 9, 11

Does this sum up your problem?

- There are 49 balls numbered 0-9.
- Four balls contain 0, five balls contain 1, five balls contain 2, ..., five balls contain 9.
- Player selects six balls.
- Lottery selects six balls.
- Player's balls may not contain any adjacent digits (0 and 9 are adjacent).
- Lottery's balls contain exactly one pair of adjacent digits.
- What is the probability that five of player's balls match exactly five of lottery's balls?

1
Sun 28 Aug, 2011 05:34 am
@markr,
Right, they are 40. Good spirit of observation
The problem definition seems to be a bigger problem than the math

"
- What is the probability that five of player's balls match exactly five of lottery's balls?
"
Yes, but I've noticed I'm interested to obtain the separate probability of the distinct units first, then the distinct probability of the repeating units, then perhaps multiply them to form the final probability.

So that becomes:

- a) What is the probability that 4 distinctly numbered player's balls match 4 of the 5 distinctly numbered lottery's balls? E.g.:
lottery units: 1-2 4 6 8
player's units: 1 4 6 8

- b) What is the probability that after a), the player's other 2 repeated units will match the lottery's repeated unit:
- b^i) player's repeated units can be equal in value in which case, all player's 6 balls can match the lottery's 6 balls
E.g.:
lottery units: 1-2 4 6 8 8
player's units: 1 4 6 8 8 8
- b^ii) player's repeated units can't be equal in value in which case, only 5 of player's balls can match only 5 out of 6 of the lottery's balls
lottery units: 1-2 4 6 8 8
player's units: 1 4 6 8 8 1

- c^i) What is the total probability for a) and b^i),
- c^ii) What is the total probability for a) and b^ii)

P.S.
If it matters (it should help at guessing what the probabilities should be),
these are the relevant frequencies for repeated units in draws of 6, regardless of adjacent units.
(2)Statistical frequency of 2 repeated units: ~50%
(22)Statistical frequency of 3 repeated units. E.g. 1-1-1, 2-2-2 etc.: ~19%
(3)Statistical frequency of 3 repeated units. E.g. 1-1 and 2-2 etc.: ~7%

and these are the relevant frequencies for adjacent units in draws of 6, regardless of repeated units.
(0)Statistical frequency of 0 adjacent units: ~7.5%
(2)Statistical frequency of 2 adjacent units: ~27%
markr

1
Sun 28 Aug, 2011 12:18 pm
This gets crazier and crazier. It's not at all clear to me that multiplying any of the results will yield anything meaningful. I think it won't.

I'm surprised that you consider this to be a 6-ball match:
lottery units: 1-2 4 6 8 8
player's units: 1 4 6 8 8 8

But, given that, I'm more surprised that you consider this to be only a 5-ball match:
lottery units: 1-2 4 6 8 8
player's units: 1 4 6 8 8 1

There seems to be some dependence on the 6th ball matching the 5th ball.

I suspect you've got a final result in mind (c^ii?), and that you've attempted to break the problem into (simpler?) sub-problems that can be combined (multiplied?) to produce the final result. However, I don't think the approach is valid.

I don't know how to proceed from here without a concise and complete statement of the problem(s) - preferably in the style that I've used. That statement would need to make clear why the first example above is a valid 6-ball match and why the second example is not. For example:

- A player's ball is considered to match a lottery's ball if any lottery's ball contains the same number as the player's ball.
- A single lottery ball may be considered to match multiple player's balls.

Intuition would have told me that:
lottery units: 1-2 4 6 8 8
player units: 1 1 1 1 1 1
is a 1-ball match. However, I'm not sure, based on your examples, if this is a 1-, 5-, or 6-ball match.

Also, does this work in the other direction? For example:
lottery units: 1-2 4 4 4 4
player units: 0 2 4 6 8 8
Is this a 2-, 4-, or 5-ball match?

"The problem definition seems to be a bigger problem than the math"

I think you have an idea in your mind of what you want, but I don't think that has been completely or correctly conveyed for someone else to understand. I also suspect you've incorrectly reduced your original problem into sub-problems with the expectation that they can be combined to provide a solution to the original problem.

"Yes, but I've noticed I'm interested to obtain the separate probability of the distinct units first, then the distinct probability of the repeating units, then perhaps multiply them to form the final probability."

Why?

1
Sun 28 Aug, 2011 01:56 pm
@markr,
"
But, given that, I'm more surprised that you consider this to be only a 5-ball match:
lottery units: 1-2 4 6 8 8
player's units: 1 4 6 8 8 1
"
You didn't notice player's units don't contain the unit 2, while the lottery does, that's a mismatch. It means the player can only guess 5 numbers out of the lottery's 6. Try using another unit and you'll notice you'ld be breaking the problem rules about the player having no ajancent or sequenced units. This is b^ii) case, the player can match all 6 according to b^iii) case.

"Why?"
This is only going to confuse you more perhaps
I don't think it's much different from your answer here though, and I explained after the solution what I did with it:
http://able2know.org/topic/175253-1

for the units drawn w/o replacement we use the hypergeometric distribution formula, and for the repeated units plain chances math, and because the repeated drawings are with replacement, we can multiply the two probabilities because they're just as independent as the factors in calculating the probability of a lottery draw (49x48x47x46x45x44)/6!(49-6)!, then divide it by the number of possible matches of repeated units with replacement multiplied with sequenced units w/o replacement. I think the formula would resemble the hypergeometric

a) probability ratios between all the particular cases or possible unit sequences outcomes tells the player what combination is better to play to win for 1st or 3rd category win, that is a 6 out of 6 match or 4 out of 6 match
b) probability of repeated units combined with probability of sequenced units tells the probability to be used in c) to ascertain both repeated and sequenced choices when choosing to play and win a certain category
markr

1
Sun 28 Aug, 2011 04:07 pm
In your b^i example, the player also didn't have a 1, but you called it a 6-ball match.

"and because the repeated drawings are with replacement"

How do you figure? There are four zeros. You can't draw five or six zeros. You're not replacing; you just happen to have multiples of each digit.

"I don't think it's much different from your answer here though, and I explained after the solution what I did with it"

I couldn't make sense of your follow-up post (what you did with it). I don't know if that was valid or not.

If you can give a concise, clear, unambiguous definition of a problem, I'll give it a try. I suggest you state the problem in terms of what you're really after and not in terms of sub-problems that may or may not be combined to produce a valid result.

1
Mon 29 Aug, 2011 05:23 am
@markr,
I think I solved it for the most of it, but i need help with b), the numbers don't add up to a equivalent verification. Look after the explanations for your last post below.

"
In your b^i example, the player also didn't have a 1, but you called it a 6-ball match.
"
You're right, the player can't ever match all 6, I ignored the 2, my bad

lottery units: 1-2 4 6 8 8
player's units: 1 4 6 8 8 8

"
How do you figure? There are four zeros. You can't draw five or six zeros. You're not replacing; you just happen to have multiples of each digit.
"
Right again, there can only be 4 repeated zeroes and 5 of other units from 0 to 9, that was a quick example though. This means we must proportion the probability of b) extracting 1/5 from 1/10 of the whole, because 0 (zero) lacks a fifth appearance in the sample space.

---------------------------------------------------------------------------------------
Correct me if I'm wrong please; I've yanked standard approach by the hair a bit here

the solution for a) should be:
a) because we're having sampling w/o replacement for the distinct units part, e.g.
lottery units: 1-2 4 6 8
player units: 1 4 6 8
we can apply the hypergeometric distribution:

Player) because it picks with 5 distinct units out of which 4 match 4 out of 5 distinct units of the lottery, we have C(5,4) way to pick success, C(10-5, 1) way to fail (that's the unit that always fails actually), and total sample space is C(10,5), thus the probability is:
C(5,4)*C(10-5,1)/C(10,5)=25/252

Lottery) because it picks with 5 distinct units out of which 4 match 4 out of 5 distinct units of the player, we have C(5,4) way to pick success, C(10-5, 1) way to fail (that's the unit that always fails actually), and total sample space is C(10,5), thus the probability is:
C(5,4)*C(10-5,1)/C(10,5)=25/252

----------------------------------------------------------------------------------
Now for b^i and b^ii, they are one and the same case (separating them is rather irrelevant to the problem I think now), that is from the remaining two units out of the 6, only one is able to match with the repeated unit of the lottery, or both.
*Because we're sampling with replacement in this case, we can use normal probability math, and corroborate with the binomial distribution afterwards.

Now the problem is that the player has one more unit, that was used to help the probability of the sampling w/o replacement at a). Should this unit contribute to the probability of the sampling w/ replacement b)? Since it fails to be a distinct unit to be more exact in a) I think yes, it should contribute to the probability of b) because it can only be a repeated unit in accord with the rule that the player set of units contain no adjacent or sequenced units.

The question is then, what is the probability that 2 random units will hit 1 unit picked out of 4 random units from 0 to 9. Probability is (4/10) + (4/10)=8/10=4/5?

4/10 is from one of the two player units that will be repeated with a unit from the player set of units that matches the lottery units (they can't really be repeated with anything else but that set in actuality), and has 1/10 chance of being any of the units from 0 to 9, times 4 units that match between player and lottery (it can match in 4 ways a matching unit between the lottery and the player), which is in total: 4/10 chance times 2 tries = 4/5.

Viceversa, the probability that the repeated unit in the lottery set hits only the matching units (and also repeated units of the player) between the lottery and the player is 4/10 for one of the player's repeated units, 4/5 for both, and 1/10 chance that it will match the unit that doesn't match with the player. This means 4/5 of the time the player can match 5 units (4 distinct 1 repeated) with the lottery, and 1/5 of the time only 4 distinct units.
The respective probabilities are 4/5 probability *4/5 of the time = 16/25 and 1/5 of the time * 1/10 probability = 1/50, where the total probability of success is (80*16/25 + 20*1/50)/100=0.516

Another approximation is the binomial distribution:
C(10,5)[80*(4/5)^5] *[20*(5/5-4/5)^5]=42.27858432/100=0.423... (rather far apart from the mean approximation). A little help here with verification?

The first attempt before verifying it with the player side, and I thought the probability was 4/10 not 4/5, the average was 0.34 and the binomial was 3.2105299968/10?=0.321... like this:
C(10,5)[80*(4/10)^5] *[20*(10/10-4/10)^5]
Seems to me that the lottery's timing with whether it repeats the 6th unit with the adjacent unit that doesn't match with the player is an absolute matter in the account.

Verification from player's point of view:
4/5 chance times 4/5 of the time success = 16/25 chance for success
1/10 times 1/5 of the time success = 1/50 chance for failure
(80*16/25 + 20*1/50)/100=
=(16*16/5 + 2/5)/100=0.516

I could adjust the above probability with minus 2% here, for zero not making a fifth appearance, which is 10th part (1/10) times a fifth (1/5) = 1/50, and because we're talking continuous distribution here on the long run, that would mean minus the fiftieth part of the whole, which is 2% I think.

For now, I'm interested in matching 5 units between player and lottery, which is optimal. Combining the probability w/o replacements with the one w/ replacements is irrelevant because the repeated units can't be avoided for once, and the ratio I need can be extracted from the probabilities w/o replacement for sequenced units alone, which was the first problem. Not to mention it would be hard to prove the combination of probabilities, even if correct

Total probability would be:
(25/252) *0.516=0.05112... (1 sequenced units and one repeated units, matching)

and the verification would be:
(3)(22)and(2)repeated units occur:
96% of the time with an ~0.2326... probability=~0.223296...
times
?% of the time sequenced units with a ? probability
(haven't done sequenced units calculations yet)

~0.223296 * x = 0.05112
x=0.05112/~0.223296 =0.2289... probability of 2 sequenced units ?% of the time (perhaps about 96% too)
which looks good to me at this point, though not verified yet thoroughly and scientifically.

0 Replies

1
Mon 29 Aug, 2011 05:48 am
@markr,
The explanation for C(5,4)*C(10-5,1)/C(10,5) is in my opinion:

since the player can't ever match one of the units of the sequence in the lottery set, I register that as an automatic fail C(10-5,1), but we can't ignore the fact it contributes to the probability of matching, hence it's C(5,4) for the player too, then the ways to choose the set is C(10,5)

Otherwise we could put out the unit in the lottery sequence of units that will always fail, and it will look like this:

C(4,4)*C(10-4,0)/C(10,4) = 1/210

Which do you think it's closer to the truth?
markr

1
Mon 29 Aug, 2011 01:48 pm
Your problem (a) looks a lot like the problem I already solved above:

- The sample space is the numbers 0 to 9.
- Five unique numbers are selected by the player and the lottery.
- The player's numbers contain no adjacent pairs (0 and 9 are adjacent).
- The lottery's numbers contain exactly one adjacent pair.
- What is the probability that the player matches exactly four of the lottery's numbers?

I've already shown that answer to be 1/4. If you don't believe me, you can easily enumerate all cases (two options for the player, 40 options for the lottery).

Here's an example where the lottery uses sequence 0-1. The result applies to all other nine sequences the lottery can use.
A: 0-1, 3, 5, 7
B: 0-1, 3, 5, 8
C: 0-1, 3, 6, 8
D: 0-1, 4, 6, 8

Recall that the player only has two options:
X: 0, 2, 4, 6, 8
Y: 1, 3, 5, 7, 9

AY and DX are the only winners.

1
Tue 30 Aug, 2011 01:01 am
@markr,
No-no, I didn't restrict the player, remember I said the player can repeat all units: 999999 and alternate 898989 etc. This doesn't change the number of times the player and lottery match, but the ratio of the match number to the fail number, which means the probability is lower than 1/4.

Someone said use the multivariate hypergeometric, but I got no proof is the right probability: C(2,1)C(3,3)/C(10,5) = 2/252

Here's all the sequences for example:
hypergeometric: C(5,5)*C(10-5,0)/C(10,5)=1/252
no sequence - 0
0 2 4 6 8
0 2 4 6 8

hypergeometric: C(5,4)*C(10-5,1)/C(10,5)=25/252?
multivariate hypergeometric: C(2,1)C(3,3)/C(10,5)=2/252?
1 sequence(2) - 1
12 4 6 8
1 4 6 8

2 sequences(22) - 2
12 45 7 9
1 4 7 9

2 sequences(3) - 1
012 4 6 8
0 2 4 6 8

3 sequences(32) - 2
012 45 7
0 2 4 7

3 sequences(222) - 3
01 34 67
0 3 6

4 sequences(33) - 1
012 456
0 2 5 6

4 sequences(4) - 2
0123 5 7
0 2 5 7

5 sequences(5) - 2
01234 6
0 2 4 6

6 sequences(6) - 3
012345
0 2 4

1
Tue 30 Aug, 2011 02:04 am
@markr,
Found a relation with proof that matches the problem:

C(4,4)(10-4,1)/C(10,5)=6/252
C(5,4)(10-5,0)/C(10,4)=5/210
6/252=5/210

Lottery: 5 distinct
Player: 4 distinct

If we fix the lottery (5) and calculate the player:
There are C(4,4) ways to select the 4
There are C(10-4,1) ways to select a losing variant
There are C(10,5) ways to select the set of 5
C(4,4)(10-4,1)/C(10,5)=6/252

If we fix the player(4) and calculate the lottery:
There are C(5,4) ways to select the 4
There are C(10-5,0) ways to select a losing variant
There are C(10,4) ways to select the set of 4
C(5,4)(10-5,0)/C(10,4)=5/210

P.S.
The multivariate hypergeometric is = C(2,1)C(3,3)/C(10,5)=2/252?
The logic is that after we pick an unit out of the sequenced ones, the other sequenced unit

becomes invalid for picking, however which probability is closer to the truth, the multivariate

or the normal hypergeometric distribution?

1
Tue 30 Aug, 2011 02:59 am
What's interesting is that if we combine the multivariate and simple hipergeometric distribution we obtain proof for the simple hipergeometric distribution, like this: C(2,1)C(5-2,1)C(3,3)/C(10,5)=6/252

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