Different Methods of Calculating Probability?

[Numnum]

If I were to calculate the probability of getting 6 out of 6 numbers correct in a lottery where you choose six numbers out of 49 (the lottery is called 6/49), the probability would be
(6C6)(43C0)/(49C6), right? But what are different methods of getting this same answer without using combinations?

 

[Simon Bridge]

The same way you work out any probability ... and strategy that let's you count how many ways there are to get a desired result and how many possibilities there are in total will work.

The permutations and combinations are just fancy words for what you would do if you didn't know about factorials.
http://betterexplained.com/articles/easy-permutations-and-combinations/

 

[HallsofIvy]

I assume you mean "exactly 6 out 6 numbers chosen from 49 numbers right without regard to the order". The probability that you get the first number right is 1/49. Assuming that the 6 numbers cannot repeat, there are now 48 numbers left so your probability of getting the second number right is 1/48. Similarly, the probability of the third number right is 1/47, the probability of the fourth being right is 1/46, fifth, 1/45, and 6, 1/44.

The probability of all 6 being right- in the right order- is (1/49)(1/48)(1/47)(1/46)(1/45)(1/44). We can write that in terms of factorial by multiplying both numerator and denominator by the numbers missing from the denominator: 43(42)...(3)(2)(1)= 43! and get 43!/49!

To find the probability of getting the same 6 numbers in any order, multiply by the number of different orders of 6 things, 6!. That gives (43!)(6!)/49! which is the same as $1/\begin{pmatrix}49 \\ 6\end{pmatrix}$.

As for your expression, it should be clear that both $\begin{pmatrix}6 \\ 6\end{pmatrix}$ and $\begin{pmatrix}43 \\ 0 \end{pmatrix}$ are equal to 1.

 

[chiro]

You can also use Stirlings approximations for factorials especially if they are large:

http://en.wikipedia.org/wiki/Stirling's_approximation

 

[magravat]

I derived and published my own probability algorithm and distribution in ( Agravat SP03Pharmasug 2011). This new algorithm distribution assumes independence. The methods of probability are different answers based on the distribution used. The uses can be vast because there are infinite numbers the conditions can allow at any discrete level. Plost are incredible as well.

The new (Agravat’s) probability algorithm and new (Agravat’s) distribution is included for discussion because it allows you to deal with the probability of events as in the case of the data taken 1 at a time for three variables for example for the probability of outcome.


A sample calcuation is below:

P(n|i)=(n-i)*(n-i)/(n-i)**n
P(n=3|i=0)=1/3




For the last condition,


1) For n=3 and i=0:
The derivation below explains the checking of the answers for the above algorithm that addresses a general question that if there are three possibilities to start, and there exist i different possibilities, what is the probability of outcome? In the case of P (3|2) inspections leads one to conclude the answer is 0, but the answer is supported by the derivation below. The probability of P (3|1) is given below as well. This algorithm may work to calculate essential or important probabilities more than demonstrated.

Ln(P(n|i=0))=
Ln(n-i)+ Ln(n-i)- nLn(n-i)-ln1/3+
Ln(n-i)+ Ln(n-i)- nLn(n-i)-ln1/2+
Ln(n-i)+ Ln(n-i)- nLn(n-i)-ln1+
Ln(n-i)+ Ln(n-i)- nLn(n-i)-ln1/6+
=8ln (n-i)-4n*ln (n-i)-ln2
8/4n*2=1/n>>n=3 so P(n|i=0) =1/3




The distribution is similar to the binomial or exponential distributon with mean and variance like the gamma distribution due to proportions.


magravat

 

[Simon Bridge]

Welcome to PF. I looked up your paper with interest:
http://www.pharmasug.org/proceedings/2011/SP/PharmaSUG-2011-SP03.pdf
Excuse me... it reads like the output of one of those paper-generator websites.

PharmaSUG is the Pharmaceutical Industry SAS Users Group.
The purpose of their conferences is to network SAS users... so it does not look like a scientific or mathematical conference.

P(n|i)=(n-i)*(n-i)/(n-i)**n
... you are using a double-star (**) to indicate a power?
surely the RHS simplifies to 1/(n-i)**(n-2) ?

P(3|0) then indeed comes to 1/3 1/3**1 = 1/3 simplifying the calculation?
But we are still missing the context... what is this the probability of?

 

[magravat]

Mr. Simon Bridge,
My Pharmasug 2011 paper involves Statistics and Pharmaceutcics for the division I was in. You are correct that (n-i)**n is (n-i) to the power of n. To the other reply about it simplifying by the derviatives using limits, it does simply to something like that that I have derived before. I have done that simplification for another method that I had submited in Europe for a paper(submission and review) for a comparison of methods with and wihtout limits. But correct, the simplification is correct for anwer I checked! Statistics is important to me:

According to Paul Dirac “Some time before the discovery of quantum mechanics people realized that the connection between light waves and photons must be of a statistical character” (Dirac, 1958).

P(3|0) =1/3 means for the probability of 0 events out of three happening is 1/3 I believe. There is proof of independence too.

Thank you,
magravat

 

[magravat]

Numnum,
For your question above, for P(49|6) using my new algorithm is (from Agravat Pharmasug 2011):
1.68X10-77.

magravat

 

[viraltux]

Numnum,
For your question above, for P(49|6) using my new algorithm is (from Agravat Pharmasug 2011):
1.68X10-77.

magravat

 

[D H]

Numnum,
For your question above, for P(49|6) using my new algorithm is (from Agravat Pharmasug 2011):
1.68X10-77.

magravat

That's just wrong. You might want to think of going back to the drawing board.

The correct probability is that shown by HallsofIvy in post #3, 1/(49 choose 6) = 1/13,983,816.

 

[magravat]

Sir,

I agree ,trying to point out a tangent. The textbooks do use combinations correct. I am trying an application of what if different probability than Combinations:

1) since 6|6 will be 1 that is:P(49|6)...P(49|0) is approximately 1
2) 1-sum of (P(49|49...7))= a different answer though. Some try in combinations that if
probability of the remainder may hold a clue from another probability.
P(49|47)=7.105Xe-15

Only trying to look another way, sorry. I know text I used (Wackerly, Mendenhall, and Scheaffer) shows:n!/(r!*(n-r)!) for right answer of Hallsof Ivy.
sum of P(49|49)..P(49|7)=
The highest value is from P(49|47) =7.105e-15 hence reciprocal is close but different from combinations method of:1/1.407X10+14 so sum will approach it. It is different value from point of view of conditional probability.
magravat
so reciprocal is

 

[Simon Bridge]

P(3|0) =1/3 means for the probability of 0 events out of three happening is 1/3 I believe. There is proof of independence too.

Yeah I had a feeling you were using a non-standard notation: P(x|y) usually denotes a conditional probability.

Note: usually, if three things (numbered 1..3) can happen, then the probability that one will happen in one trial is 1/3 but the probability that none of them will happen is 0. If "none of the above" is to be included as one of the possible events, then the probability that it happens is 1/4... since there are now 4 possible events. (Assuming independence and all events equally likely.)

So either your method is incorrect or the description of what it is intended to say is incorrect. Can you provide a practical example to illustrate your method?

To the other reply about it simplifying by the derviatives using limits, it does simply to something like that that I have derived before.

I did not use any limits or derivatives to perform the simplification - just normal algebra. Perhaps it is just unfamiliarity with English but the way you use these words suggests you don't know what they are normally taken to mean in mathematics. Practical illustration is probably the best way to deal with this.

I notice that your solution:

7.105e-15 hence reciprocal is close but different from combinations method of:1/1.407X10+14

is nothing like the actual one of 7.1511e-08 ... you are 6-7 orders of magnitude too small.
The sums do not add up enough to make up the difference. But it is not clear that you are working OPs problem.

So here's the suggestion: work through your method, step-by-step, for the problem of getting 6 numbers correct out of six trials selecting without replacement out of 49 possibilities and regardless of the order. We'll see if the result comes out the same as the combinations method, ad also gain an understanding of what your method actually entails. Would you do that for us?

 

[magravat]

Mr. Simon Bridge,
Originally, I used my algorithm to view points in space to observe in my distribution graphically. My description was incorrect.
I derived in my Pharmasug 2011 paper on page 38-39. I derived the new distribution in the paper. I used SAS software to plot the probability from P(n|i) as ‘Pr’ and ‘omp’ as 1-Pr. The plot shows me more about generated slopes and intercepts based on the ‘discrete random variables’ probabilities. The plot may have meaning or hold importance for viewing points in space in the new distribution I derived.
In the next paper, my most recent in Global SAS Forum 2012, I derived the new use that is Z score
using the new Probability Mass Function from it. One can observe it below.
You are right, that according to conditional probability which my algorithm is different from.
My algorithm is actually meant for ‘discrete levels of probability’.
The answer as you said P(3|0)=0 . Originally, I had hoped to derive based on only deriving probability.
If a (problem 3.5 Wackerly Mendenhall and Scheaffer pg.87) takes is to match pictures to words and
the three words at random to three pictures, find the probability distribution Y for correct matches:
A B C No. matches Pr P(i) Pr=1/(No.)**power No. Power(n=3) No.=n-i
A B C 3 P(0)=1/3 P(0) 1/3**1 3 n-2=1 3-0
A C B 1 P(1)=1/2 P(1) 1/2**1 2 n-2=1 3-1
B A C 1 P(2)=0 P(2) 0 1 n-2=1 3-2
B C A 0 P(3)=1/6 P(3) 1/1(actually(1/n!)) 0 n-2=1 3-3
C A B 0
C B A 1


The algorithm and distribution work very well to produce a new Z score for probability mass function that
Works very well. For the last condition, P(n|i) the answer is 1/n! The My algorithm shows the answers
except for next to last which gives 0 from inspection. P(3|2). My derived answer using natural log also gives 0.
The algorithm gives answers that can be used for probability mass function that do produce good Z scores from paper calculations rather than tables, also if you want to see and Z scores.







P(i) Pr=1/(n-i)**n-2 No. Power No.=n-i New Factor**2
P(0) 1/1296(.000771) 6**4 n-2 6-0 6 6*sqrt(6)
P(1) 1/625(.0016) 5**4 n-2 6-1 5 5*sqrt(5)
P(2) 1/256(.00390) 4**4 n-2 6-2 4 4*sqrt(4)
P(3) 1/81(.0123) 3**4 n-2 6-3 3 3*sqrt(3)
P(4) 1/16(.0625) 2**4 n-2 6-4 2 2*sqrt(2)
P(5) 1(0) 1**4 n-2 6-5 1
P(6) 1/6!=1/720 0**4(actually1/6!) n-2 6-6 0

Based on this simplification, for P(49|6):Expect:1/(49-6)**49-2=1/(43)**47 but:
This method is a permutation like procedure for last step. However, for P(n=49|i=6). The last step is not
Asked for. Only the 43rd to the 47 th power. The math is like Pr(49|6)=: n-2=47
n=49
n-i=43 :

That is different from the answer of 1/49C6=:1/13,983,816==7.15X10-8
To develop further the differencs is (i-2) so the New Method is from my algorithm simplification plus (Agravat’s algorithm):

Pr(49|6)=1/(i-2)*(n-i)*((n-i)*sqrt(n-i))*2=1/13,675,204=7.31X10-8
Pr(n|i)= 1/(i-2)*(n-i)*((n-i)*sqrt(n-i))*2=

Pr(49|7)= 1/ (i-2)*(n-i)*((n-i)*sqrt(n-i))*2=7.07X10-8 about 1/(49C7)=1/85900584=1.16X10-8

Thank you,

magravat

 

[Simon Bridge]

I'm sorry - you don't seem to have supplied the probability asked for.
Not only that but your step-by step includes statements like "However, for P(n=49|i=6)." which, although it ends at a period, is not a complete sentence.

Am I to take to that your final answer $1.16 \times 10^{-8}$ is the asked for probability computed by the method described in your paper? Because it is still way off - about 1/6th of the correct answer by counting. You did say it was "approximate" but this is a bit too far off for SAS purposes so I don't think this is what you mean.

Do I understand correctly that you have submitted the paper for peer review to a European journal of some kind? If so, which journal may I ask?

On evidence so far I cannot possibly recommend this approach.

 

[viraltux]

That is different from the answer of 1/49C6=:1/13,983,816==7.15X10-8

To develop further the differencs is (i-2) so the New Method is from my algorithm simplification plus (Agravat’s algorithm):

Pr(49|6)=1/(i-2)*(n-i)*((n-i)*sqrt(n-i))*2=1/13,675,204=7.31X10-8

Hello Mr. Magravat,

Your algorithm seems similar to Stirling's formula approximation for big factorials, using Stirling's formula for the big factorials in the OP problem we get 7.1494271324217539E-8 which is a better approximation to the real value 7.151123842018516E-8 than your algorithm with 7.3125051735974104E-8.

Is there any advantage using your algorithm over using Stirling's formula?

Thank you.

 

[magravat]

At this time, I feel it is only a possibility. I had intended it for Z scores with the use of probability mass function that is derived from it for which it works very well. I am hoping to develop it further. For P(49|6), the answer at this time posted was:
7.31x10-8, until I develop and explore more possibilities .

Thank you,
magravat

 

[Simon Bridge]

OK - so it does not work to compute probabilities then. No worries.
Back to the original question - methods of calculating probabilities as per example post #1 without using combinations explicitly...
... you can reason through the problem, counting the different ways.
post #3 demonstrated this by multiplying probabilities.

The fancy notations are shortcuts. Not using combinations is a bit like wanting to find the area of a field without using this fancy multiplication thing. However, I am sympathetic - I recall being terribly confused at High School stats because the factorial and combination and permutation notation was used by definition without explaining where it comes from. I was taught to identify the distribution involved and then apply the formula for that distribution by rote. Knowing how to do it the long way helps understanding... and it's not always longer.

And we have seen that for really big numbers it can be more efficient to use an approximation.

 

[magravat]

Mr. Simon Bridge,

For calculating probabilities in case of P(y|x) a random discrete probability when x is even, this formula does come close:
Pr(n|i)= 1/ (i*i)(i-2)*(i-1)*(n-i)*((n-i)*sqrt(n-i))*2
The approximation I gave is somewhat correct for even ‘i’ values and the new algorithm is modified:
Pr(n|i)= (i-2)*(n-i)**(i-2)
For 1/Pr(49|6) and other even ‘i’ discrete levels yields a probability of:
7.31250517359741e-08 for Pr(49|6). If n is small, the method is an approximation of probability and is simpler to understand and calculate.


magravat

 

[Simon Bridge]

So it is not a method of "computing" probabilities, but for approximating them? Can you also figure out the likely accuracy of the approximation? What is this method's advantage over Stirling's method?

In fact, on my reading this is the sort of information that is missing from your paper - you need to motivate the algorithm (what problem does it solve?) and compare it, directly, with other approximation methods... preferably against a metric justified by the motivation.

If you did supply this information, you need to put it more prominently and in simpler language right at the beginning - and, very briefly, in the abstract.