Noob question about "combined" probabilities.

#1
Hello!

Sorry for my poor english, I´m from Argentina.
And sorry too if I´m in the wrong place.
I need some knowledged about stats and this forum is what seems the best I could found for my needs, but not sure.

To the point:

I have an experiment with two possible results: A and B
I did the experiment 100 times and gives me results A 52 times and B 48 times.

In the other hand, I have another experiment with results C and D.
I did this experiment 100,000 times and gives me 51,000 times C and 49,000 times D.

Now I need two options for a one chance bet: A vs B is one option and C vs D is the other option.
Of course in case of the first option I must bet for A and in the second option I must bet for C.... but my question is... wich option I must choice:
A vs B or C vs D?

I´m prety sure I must choice C vs D no matter here the probabilities are 51% vs 49% instead of 52% vs 48% as it is in option A vs B cause I think the previous results had less possibilities to be just cause luck in second experiment than in first experiment. .... but I´m not able to cuantificate why that choice.

Thanks for any help and for your time! :)
 
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#2
You should evaluate the confidence intervals for success in each case.

(Hint: Both choices have sufficient trials and are close enough to 50:50 to justify using a normal approximation to a binomial distribution.)
 
#3
Thanks Con-tester!
Now I need to read about what is "confidence intervals";"normal aproximation" and "binomial distribution" :yup:
But it seems exists a solution for my problem, I´m right?
 

hlsmith

Omega Contributor
#5
You definitely have the right logic in your first post. You could also test in both scenarios what the probability of the true proportion being 50% in both experiments based on a binomial test (success, failures, probability=0.5)
 
#6
Mmm.... a little different case:

There is one tennis player that made 1000 singles matches and won 20% of them (200).
There is another tennis player that made just 10 matches and won 90% of them (9).
They join together to play doubles and want to estimate* how strong theire formation is.
*: I said "estimate" cause there is a far from zero factor about how they work together.... but even not knowing it, how good they are in singles must have high weight in the result as a couple.
So I will ignore the part about how they work together (the real cause because I do not take it in account is that I completely ignore how to calculate it).

If I suppouse that probabilities of a win in a doubles match is p=F(1000,200)+F(10,9), is there any way to know that F?

I imagine something like p = G(1000,200) x 0.2 + G(10,9) x 0.9 with G(1000,100) > G(10,9) cause first data have more fiability.

I could generalize it to a football (soccer) team:

p = G(m1,w1) x p1 + G(m2,w2) x p2 + .... + G(m11,w11) x p11

Where mi are matchs player i did, w1 the wins player i had and pi = wi/mi. Here the result is surely even more random cause each player did his matches with differents teammates. And I need to include tie results, not just wons and looses.
 
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BGM

TS Contributor
#8
I have not read all the replies but here is my two cents based on the 1st post.

There are some functions called utility function which try to model one's utility, and assume that people making choices to maximize his own utility. In statistic, one try to make a decision based on minimizing the loss function.

So depending on how you value/weight the payoff of such decision. If both give you the same payoff when guessing correctly, then the game AB seemingly should give you a higher expected payoff. However as mentioned, this "expected payoff" is just based on a point estimation only from the data given, thus one cannot said the proportion of A is strictly larger than proportion of C almost surely. And say if you want to maximize the expected utility, then the result may be different depends on the utility chosen. One typical example is that people are usually risk averse, and thus one will require the risk premium to compensate a risky decision. And that is why even there is 1% of difference in the point estimation, if the downside of it is huge, then people may not choosing it over the alternatives.
 
#9
Mmmm.... still trying to understand this I read (read, not understood) about multinomial distribution, confidence intervals.
Now I try to ask about a new example I think it must be known:

There are 3 candidates (candidate1, candidate2 and candidate3) for some elections.
The real distribution of total votes will be c1, c2, c3.

I just made a poll with N people and obteined n1 for candidate1, n2 for candidate2 and n3 for candidate3. Of course n1+n2+n3=N

I guess that as bigger n1, n2 and n3 are, close will be n1/(n1+n2+n3) to c1/(c1+c2+c3); n2/(n1+n2+n3) to c2/(c1+c2+c3) and n3/(n1+n2+n3) to c3/(c1+c2+c3)

If I call:
p1=n1/(n1+n2+n3)
p2=n2/(n1+n2+n3)
p3=n3/(n1+n2+n3)
and
r1=c1/(c1+c2+c3)
r2=c2/(c1+c2+c3)
r3=c3/(c1+c2+c3)

I think p1/r1 = F(n1,n2,n3) --> 1 when n1,n2,n3 --> to c1,c2,c3 in other words, p1/r1 -->1 with bigger N=n1+n2+n3
The same for p2/r2 and p3/r3

My question is..... how is F?

Another way to say it:

Must happens that pi-Dpi < ri < pi+Dpi and Dpi --> 0 with N


Note: I understood correctly that (pi-Dpi,p+Dpi) is called "confidence interval" and the fact to have 3 possible results is called multinomial distribution with k=3?

Thanks again for your time! :)
 
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BGM

TS Contributor
#10
Ok it seems that you are interested in the convergence issue right now.

Since you are not talking about the joint distribution of the estimators of multinomial proportions (which is the sample proportion in our context), it suffice to consider a simple Bernoulli model

[math] X_i \sim \text{Bernoulli}(p) [/math]

where [math] p [/math] be the theoretical proportion.

Let

[math] \hat{p} = \frac {1} {n} \sum_{i=1}^n X_i [/math] be the sample proportion

Then [math] n \hat{p} \sim \text{Binomial}(n, p) [/math]

So the ratio [math] \frac {\hat{p}} {p} [/math] is also a random variable, following a scaled Binomial distribution with the support

[math] \left\{0, \frac {1} {np}, \frac {2} {np}, \ldots, \frac {n} {np}, \right\} [/math]

And it will have a mode close to 1 and will become more and more concentrated around it - the peak will be sharper and sharper if you plot the pmf. So this partly explain what you are looking for - the function [math] f [/math] you mentioned.


For the convergence part, it depends on which metric you are using the measure the "distance" between [math] \hat{p} [/math] and [math] p [/math]. Some commonly used choices will be "in distribution", "in probability", "almost surely", "in mean square" etc. You can read further details to understand the definition of these convergence.
 
#11
Thanks guys.
Little by little I´m progressing.

New problem:

I have n data and with them I obteined the mean (M) and some deviation (D) (could be standart deviation or any other measure that gives me an idea about how spread the results are)

If I have a new data (d), the new mean (nM) is easy to obtein without need to know the previous data:
nM=(M*n+d)/(n+1)

Is there any way to obtein the new deviation (nD) without need to know the previous data?

If I had millions of data, and have the mean and the deviation of them, when I´ll obtein new million of data, this way I´ll do not need to have old data storaged.