# Bayesian approach to inference. Why do we pick specific values as our hypothesis?

So say I'm dealing with M&M. I have 5 M&Ms and one of them is yellow. I want to know if the true proportion of M&Ms in the entire bag is 10% or 20%.

In the frequentist approach, my hypothesis would be: H0: P = 10% H1: P > 10%

However, in the Bayesian approach, I would say:
H0: P = 10%
H1: P = 20%

Philosophically, why is this?

Updated question

I'm going to attempt to solve this using the Bayesian approach. What am I missing?

Say our priors are:
P(H1) = 0.5
P(H2) = 0.5

$P(k = 1 | H1) = 0.33$ (I used binomial distribution)
$P(k = 1 | H2) = 0.41$ (I used binomial distribution)

So,

P(H1 | k = 1) = $\frac{0.5 * 33}{0.5 * 33 + 0.5 * 0.41} = 0.45$
P(H2 | k = 1) = $0.55$

So we choose H2.

• Just an observation: you can test $H_0 : p=10\%$ against $H_1 : p=20\%$ using classical/frequentist methods. In fact, when both the null and the alternative are simple, there is a most powerful test available (which would probably have great appeal to a classical/frequentist statistician): en.wikipedia.org/wiki/Neyman%E2%80%93Pearson_lemma
– Zen
Commented May 26, 2017 at 0:20
• This is not correct: a Bayesian analysis can evaluate any (point or compound) hypothesis against any (point or compound) hypothesis. Commented May 29, 2017 at 7:21

For instance if your posterior is absolutely continous, making statements suck as $$H_0 = k$$ does not make any sense. This is because to make bayesian hypothesis tests you simply have to compute the probability of the $$H_0$$ statement from the posterior and atoms under continous distributions have 0 probability.
From your question I can see $$P(H_0) = .45$$ this is $$H _0: p = .1$$, it is clear that $$P(H _1) = .65$$ where $$H_1: p > .1$$ (this a kolmogorov axiom of probability), then you reject $$H _0$$