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Let's assume, we threw a coin $110$ times and in $85$ tosses it was head. What is the probability that the coin is biased towards head?

We can use chi squared test to test, whether the coin is biased, but using this test we only find out, that the coin is biased towards heads or tails and there seems to be no one tailed chi squared test.

The same problem seems to appear when using z-test approach.

What is the correct way to solve this problem?

tach
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2 Answers2

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Let's assume that you have a fair coin $p=.5$. You can approximate a binomial distribution with a normal distribution. In this case we'd use a normal distribution with mean $110p=55$ and standard deviation $\sqrt{110p(1-p)}\approx5.244$. So getting 85 heads is a $(85-55)/5.244\approx5.72$ standard deviation event. And looking this value up on a table (if your table goes out that far, lol) you can see that the probability of getting 85 heads or more is $5.3\times10^{-9}$. An extremely unlikely event. That is approximately the probability you have a fair coin.

timidpueo
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  • Yes, computing the probability of a fair coin is not a problem for me. However, alternative to this hypothesis is, that the coin is biased towards either heads or tails. I would like to know the probability that the coin is biased towards heads. I know, that the difference will be small in this example, but I still need it. – tach Mar 24 '13 at 16:45
  • If your coin was actually biased towards TAILS, the probability that you'd get 85 heads is even lower. So the probability that your coin is biased towards heads is AT MOST $5.3\times10^{-9}$. – timidpueo Mar 24 '13 at 16:49
  • To answer your question exactly it is actually a little trickier. I don't think this is what the original question is getting at but you need an additional a priori assumption about the coin's probability. What I mean by this is where did you get this coin? From the bank? Then your a priori assumption could be that the coin is very likely to be fair. What if you got it from your prankster friend? Then maybe the probability that it's a biased coin is much higher now. It is outside the scope of this comment but look up maximum likelihood estimators. – timidpueo Mar 24 '13 at 16:53
  • *I meant probability that your coin is biased towards heads is at LEAST $1-5.3\times10^{-9}$ – timidpueo Mar 24 '13 at 16:57
  • The question can be restated. For example. I have a large population of people. I randomly pick 110 people and 85 of them are females. What is the probability that there are more than 50% females in the whole population? It really doesn't matter if it is coin problem. – tach Mar 24 '13 at 16:58
  • My answer would be the same for the restated question...? – timidpueo Mar 24 '13 at 17:02
  • My point is (as I understand it) is that the question as it is stated is impossible to answer without some additional assumptions. A question we CAN answer is what is the probability that the coin is fair. I would guess that that is what the question was intended to get at... – timidpueo Mar 24 '13 at 17:04
  • I don't think, we need additional assumptions. Lets assume similar problem. We have population of N people, and we picked 110 of them, 85 females. What is the probability, that there are N/2 or N/2+1 or N/2+2 ... N females in the population? This is actually exactly our problem if the N in infinite. – tach Mar 24 '13 at 17:24
  • let us continue this discussion in chat – timidpueo Mar 24 '13 at 17:32
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I actually think I found a solution, but it is very ugly one. Let's have similar problem:

We have population of $1 000 000$ people. We take $110$ random samples - $85$ females, $25$ males. What is the probability, that there are more females than males?

This means that we need to know, what is the probability, there are $500 001$ females in the population or more. The solution is following: $$\frac {\sum_{i=1}^{500000} \binom {25} {500000-i} \binom {85} {500000+i}} {\binom {110} {1000000} }$$

So for the general solution, it would be $$\lim_x^\inf \dfrac {\sum_{i=1}^{x/2} \binom {25} {x-i} \binom {85} {x+i} } {\binom {110} {x} }$$

This is pretty ugly solution and I hope there is better solution to this problem.

tach
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