How to test the bias of a coin? How to choose the right test? I came across this question on cross-validated around bias of a coin. My initial instinct was to go for a chi-square test. 
The other answers provided were also correct with binomial probability calculation and normal approximation.
There is a slight variation in the p-values because of continuity corrections. 
Briefly, the question asked is "out of 900 trials, we have 490 heads - is the coin biased?"
p-value: binomial: 0.008468
p-value: normal approximation: 0.0078
p-value: chi-square: 0.00766

All these values are close by and would have 'rejected the null hypothesis of the coin being unbiased'
Let's assume we are testing for a significance level of 1% and the coin is moved towards being unbiased. 
At some point, the binomial test will say the coin is unbiased, but normal approximation and chi-square will say biased. Because a coin toss is a binomial process, we should trust the binomial test more.
In general, what is the intuition behind picking the right test, when there are two or more hypothesis tests that could work?
 A: The binomial test is the exact one, so if you're near the 5% level--and have a reason to reject only if the P-value $\le 0.05,$ then use the binomial test to be sure.
Suppose you are testing $H_0: p = 1/2$ vs. $H_a: p > 1/2$ based on $n = 900$ tosses. Then, under $H_0,$ the distribution of the number $X$ of Heads observed
is $X \sim\mathsf{Binom}(n=500, p=1/2).$ If you observe $X = 490$ Heads,
then the P-value of the test is
$$
P(X \ge 490) = 1 - P(X \le 489) = 0.0042$$
1 - pbinom(489, 900, .5) 
[1] 0.00420954

If you are doing a two-sided test $H_0: p = 1/2$ vs. $H_a: p \ne 1/2,$ then
you have to include the probability of being just as far below 450 as 490 is above 450. Thus the two-sided P-value is 0.00842.
1 - pbinom(489, 900, .5) + pbinom(410, 900, 1/2)
[1] 0.00841908

Note: Because the binomial distribution is discrete, the actual significance
level of this two-sided test cannot be exactly 5%. If we reject when
$X \le 420$ or $X \ge 480,$ then the exact significance level is $0.04916.$
1-sum(dbinom(421:479, 900,.5))
[1] 0.04916137

For any practical purpose this is close enough to 5%. (You would have to do a million tests to notice the difference between 0.050 and 0.049.) If you use a normal or chi-squared approximation, the discreteness of the
binomial distribution is not 'on display', so people don't usually know or worry about slight inconsistencies.   
