Toss coin probabilities (Bino vs Gauss approx)

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    Gauss Probabilities
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Discussion Overview

The discussion revolves around the comparison of probabilities calculated using the binomial distribution and the Gaussian approximation for a scenario involving coin tosses. Participants explore the implications of using different methods to calculate the probability of obtaining between 3 and 6 heads in 10 trials.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant calculates the binomial probability for getting between 3 and 6 heads in 10 tosses as approximately 0.7734375 and the Gaussian approximation as approximately 0.633505, questioning the reason for the discrepancy.
  • Another participant suggests that the discrepancy arises because the parameters np and n(1-p) are both equal to 5, which may not be large enough for the Gaussian approximation to be accurate.
  • Further, it is noted that while the Gaussian approximation can be off by about 10%, other rules of thumb may provide better estimates but are more complex to calculate.
  • A later reply introduces the concept of continuity correction, proposing that adjusting the limits of integration to 2.5 and 6.5 yields a Gaussian approximation of approximately 0.7717.

Areas of Agreement / Disagreement

Participants express differing views on the accuracy of the Gaussian approximation in this context, with some suggesting that the sample size is too small for reliable results, while others introduce corrections that may improve the approximation. No consensus is reached regarding the best approach or the reasons for the differences in calculated probabilities.

Contextual Notes

Participants mention various rules of thumb and the importance of continuity correction in the Gaussian approximation, indicating that the discussion is nuanced and dependent on specific conditions and assumptions.

ChrisVer
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Hi, I calculated the probability to this scenario:
getting between 3 and 6 Heads after tossing a coin for n=10 trials...

The binomial probability for this is:
[itex]P(3 \le k \le 6) = \sum_{k=3}^6 Bi(k;p,n)= \sum_{k=3}^6 \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} =\sum_{k=3}^6 \frac{10!}{k!(10-k)!} (0.5)^{10} = 0.7734375[/itex]

Whereas for the Gaussian approximation to the binomial:
[itex]G(x) = \frac{1}{\sqrt{2 \pi n p (1-p) }} \exp\Big( - \frac{ (x-np)^2}{2np(1-p)} \Big)[/itex]

I get:
[itex]P(3 \le x \le 6) = \int_3^6 dx~ G(x) = ... = 0.633505[/itex]

My question is why are the two probabilities so off from each other? Is it because [itex]n[/itex] cannot be considered as "large" ? I think when extracting the Gauss approximation, the additional higher orders go as [itex]\frac{1}{n}= 0.1[/itex] but that's for the gaussian distribution ##G## and not the integral of it.
Any feedback?
 
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Simon Bridge said:
n=10 p=0.5 gives [np, n(1-p)] = [5,5] and you really want them both bigger than 5.

So the difference is mainly due to the fact that [itex]np= n(1-p)=5[/itex] is not large enoguh? So Gauss can go a bit off in its predictions (here ~10%)...

The strange thing is that this is the only rule the approximation break, whereas the other two rules give:
[itex]\Big| \frac{1}{\sqrt{n}} \Big( \sqrt{\frac{q}{p}} -\sqrt{\frac{p}{q}} \Big) \Big| = 0 <0.3[/itex]
and
[itex]np \pm 3 \sqrt{npq} \in [0.26 , 9.74 ][/itex] which is in [0,n=10]...

Simon Bridge said:
There are online calculators that will help you check your answers.
i.e. http://stattrek.com/online-calculator/normal.aspx

For the calculations I try to build my own source codes just for a practice :smile: (or check with the formulas in wolframalpha), and so far my numerical methods work fine.
 
You are forgetting the continuity correction: $$ P(3 \le x \le 6) \approx \int_{2.5}^{6.5} G(x)~dx \approx 0.7717 $$
 

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