Hi @Kevin_Shey!

Thanks for your post and your question.

Firstly, I would note that in this case we are dealing with the actual probability and not the likelihood. The likelihood is usually the value without the denominator in the Bayes Theorem, because for maximizing/minimizing or, more generally, sorting purposes, it does not matter a constant positive term in the denominator.

In this case, it is still more likely for theta to be 0.5 because there is one chance in four of picking the coin with theta = 0.8. To make yourself more convinced, I have made a small python code to run this experiment.

```
import numpy as np
from scipy.stats import bernoulli
res = np.array([1,1,1,1,1,1,1,0,0,0], dtype = np.int64) ## Resulting array to compare. 7 1's and 3 0's
def throw_coins(theta = 0.5, sample_size = 10):
return bernoulli.rvs(p = theta,size=sample_size)
p_05 = .5
p_08 = .8
match_05 = 0
match_08 = 0
for i in range(100000):
coin_picked = bernoulli.rvs(p = 0.75, size = 1)[0] ## Picks the coin. 1 if 0.5 is picked, 0 if 0.8 is picked
if coin_picked == 1: # If coin with theta = 0.5 is picked, throw it 10 times and check if it gives 7 1's and 3 0's
res_05 = throw_coins(p_05)
if np.allclose(res,res_05):
match_05 +=1
else: # If coin with theta = 0.8 is picked, throw it 10 times and do the same checl
res_08 = throw_coins(p_08)
if np.allclose(res,res_08):
match_08 += 1
print(f"Number of times the coin with theta = 0.5 outputs 7 1's and 3 0's: {match_05}")
print(f"Number of times the coin with theta = 0.8 outputs 7 1's and 3 0's: {match_08}")
```

```
Number of times the coin with theta = 0.5 outputs 7 1's and 3 0's: 76
Number of times the coin with theta = 0.8 outputs 7 1's and 3 0's: 50
```

And from every entry with 7 1’s and 3 0’s (total of `76 + 50 = 126`

)

Proportion of times that the outcome came from the coin with theta = 0.5: `76/126 = 0.603`

Proportion of times that the outcome came from the coin with theta = 0.8: `50/126 = 0.397`

I’ve run the code above with 500000 iterations (it took some minutes) and the result was:

```
Number of times the coin with theta = 0.5 outputs 7 1's and 3 0's: 363
Number of times the coin with theta = 0.8 outputs 7 1's and 3 0's: 208
```

So the proportion is:

- theta = 0.5:
`363/(363+208) = 0.636`

- theta = 0.8:
`208/(363+208) = 0.364`

Which is precisely the value we’ve encountered theoretically.

I hope that helps.

Cheers,

Lucas