Law of large Numbers?

Question

Can someone come up with another example than the one answer given in () below this question?

Using a coin toss and fictitious data, explain how the following scenario is possible: As the number of trials increases, the differences between the number of actual and expected successes tends to grow, but the difference between the percentage of actual and expected successes tends to decrease.

(something like For example, 3 heads in 10 tosses compared to 7 heads of 20 tosses would work. In the first trial there's a 2-head difference from expected (5-3) or 20% (50%-30%). In the second trial there's a 3 head difference (10-7) but just 15% (50%-35%). Do the same analysis with different data and explain why this occurs.)

Answer 1

Your question is explicitly asking for an example using "coin toss and fictitious data".

Do I am assuming you need another example like that is brackets but with different data.

Set the value for a head as 1 (or a “success” as it was phrased in the question), and for a tail as 0

So given samples of N1, N2, N3, ...., Nk values Xi (= 1 for head and 0 for tail)

The “differences between the number of actual and expected successes” is the same as the difference between the sum of Xi (which is the sum of heads of 1’s as the tails being 0 do not contribute to the sum) and the “expected no of successes” which is the P(head) value multiplied by the number of samples, and P(head) is 0.5. So:

Also looking at the sample “answer” it the figures shown are expected – actual. So I think the absolute value is what needs to be considered, so

δk = P(head)Nk - sum(Xk) = 0.5Nk - sum(Xk); summing from 1 to Nk samples

Let Σk = sum(Xr) for r = 1 to Nk, and Σk+1 = sum(Xr) for r=1 to Nk+1

And if this grows then

δk+1 > δk so

0.5Nk+1 – Σk+1 > 0.5Nk - Σk

Σk+1 - Σk < 0.5(Nk+1- Nk) - strictly speaking this should be ≤ but the = is not needed

And as Σk+1 = Σk + Σdiff - i.e. is just Σk plus the Ndiff extra values between Nk and Nk+1

Σdiff < 0.5Ndiff

Now the second condition that the “difference between the percentage of actual and expected successes”, which is the same as the difference between the expectation value and the mean. Again, the sample “answer” shows this reversed, so the absolute value must be what we consider to be decreasing.

P(head) - mean(Xk) = 0.5 - sum(Xk)/Nk; summing from 1 to Nk samples

And as this difference has to decreasing:

0.5 - Σk+1/Nk+1 < 0.5 - Σk/Nk

And then

-Σk+1/Nk+1 < -Σk/Nk => Σk+1/Nk+1 > Σk/Nk

And as above Σk+1 = Σk + Σdiff and Ndiff = Nk+1 - Nk

(Σk + Σdiff)/(Ndiff + Nk) > Σk/Nk

So

Σdiff > ((Ndiff + Nk)/Nk)Σk - Σk = (((Ndiff + Nk)/Nk)-1)Σk

And

Σdiff > ((Ndiff + Nk)/Nk)Σk - Σk = ((Ndiff/Nk + Nk/Nk)-1)Σk = (Ndiff/Nk) Σk

So we have two conditions to meet:

(i) Σdiff < 0.5Ndiff ; and
(ii) Σdiff > (Ndiff/Nk) Σk

Which can be reduced to:

0.3Ndiff < Σdiff < 0.5Ndiff

So consider the case of 3 heads in 10 tosses:

Σk = 3, Nk = 10, so the conditions reduce to:

(i) Σdiff < 0.5Ndiff ; and
(ii) Σdiff > (Ndiff/10) 3 => Σdiff > 3Ndiff/10 => Σdiff > 0.3Ndiff

So as long as we choose:

0.3Ndiff < Σdiff < 0.5Ndiff
Or
0.3 < (Σdiff/Ndiff) < 0.5

So if we choose Ndiff = 10

0.3 < 0.1Σdiff < 0.5
And
3 < Σdiff < 5

Which means we need 4 heads – as in the sample.

You can use this to construct as many examples as you like

For one more example, let’s take the last sample of 7 heads in 20 samples. So now:



Σk = 7, Nk = 20, so the conditions reduce to:

(i) Σdiff < 0.5Ndiff ; and
(ii) Σdiff > (Ndiff/20) 7 => Σdiff > 7Ndiff/20 => Σdiff > 0.35Ndiff

So as long as we choose:

0.35Ndiff < Σdiff < 0.5Ndiff
Or
0.35 < (Σdiff/Ndiff) < 0.5

So if we choose Ndiff = 10

3.5 < Σdiff < 5

So again only an extra 4 heads in the extra 10 samples, so the next sample is 11 heads in 30 samples.

Answer 2

When flipping a coin, you would expect 5 out of 10 heads, and 5 out of 10 tails.

3 heads in 10 tosses is 2 off, a small number but still 20%
7 heads in 20 tosses is 3 off, a larger number but "only" 15%

I'm not convinced the difference between actual and expected successes tends to grow, I think it is entirely possible to get 7 heads in the next 10 tosses, for a total of 14 heads in 30 tosses, which means only 1 off.

I think what they want you to understand here is this: only 1% wrong out of a thousand, is 10 wrong. "10" seems largish, but not when compared to "1000".

If you'd compare "10" against "2", you might think "10" is worse than "2" while in reality you should see that "10 out of 1000" is much better than "2 out of 10".

Answer 3

Perhaps your question relates to the theory of large numbers.

This concept describes the degree of accuracy of things such as the average lifespan of humans. If you observe that a person dies at ge 20, you might assume that that is the average life expectancy. If you observed the age at death of aids victims you might think the average lifespan is more like 30 or 40. If you toss a coin three times and it comes up heads twice and tails once, you might think that on the average it will come up tails one third of the time. If you do it 10,000 times you would find that it comes up tails about half of the times.

Insurance companies use theory of large numbers to estimate life spans so they can determine insurance costs, Casinos do too, to determine how to set the odds on gambling games.

Answer 4

I ask one person their age. Say 33. the next person is 48, next 23,14,69,etc. So the running average is 33, 40.5, 34.7, 29.5, 37.4
at each step lets count the number of people over the average:
0,1,1,2,2,... eventually, lets say I have everybody's age. How many people will be over that number (answer half or about 3 billion) what % will be over that number (answer 50%, obviously)
Which is bigger 50% or 3 billion?
This was a weak example because we were changing the expectation value (the average) as we went along. We had no a priori expectation value but lets say that you folowed me and got the same information but hadn't crunched the numbers until after I did. If I gave you the average (median), as you began to input your numbers, the law of large numbers would suggest that as more and more people answered, more and more (half if selected at random) would be over the median age but the proportion would settle down to be closer and closer to 50%.

Answer 5

Here is how understand it. Suppose you flip a coin one million times. The expected, or average, number of heads is 500,000, but it is EXTREMELY unlikely, that you would get exactly 500,000 heads.

However, the average number of heads is 50% of all of the flips, and it is EXTREMELY unlikely that you would have more than 51% (510,000) heads or less than 49% (490,000) heads.

Note that it you can be precise about the EXTREMELY's by using Sterlings formula in the first case, and a bit of stats in the second case (figure the Z-score).

Hope that helps.