Assume the following dataset:
| X1 | X2 | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 2 | 1 | 0 |
| 3 | 2 | 0 |
| 4 | 2 | 0 |
| 4 | 4 | 1 |
| 4 | 5 | 1 |
| 5 | 5 | 1 |
| 6 | 6 | 1 |
| 7 | 6 | 1 |
I now want to compute $P(Y=1 | X1 = 6 \wedge X2 = 5)$. I make the assumption that the two variables are independent of each other (in the context of the Naive Bayes classifier). With this and the Bayes' theorem I now have:
$$ \begin{aligned} P(Y=1 | X1 = 6 \wedge X2 = 5) &= P(Y=1 | X1 = 6) \cdot P(Y=1 | X2 = 5) \\\\&= \frac{P(Y=1) \cdot P(X1 = 6 | Y=1)}{P(X1 = 6)} \cdot \frac{P(Y=1) \cdot P(X2 = 5 | Y=1)}{P(X2 = 5)} \\\\&= \frac{0.5 \cdot 0.2}{0.1} \cdot \frac{0.5 \cdot 0.4}{0.2} \\\\&= 1 \end{aligned} $$
All good so far. However, if I decide to apply Bayes' theorem first and then split up the conditional variable I get a wrong result:
$$ \begin{aligned} P(Y=1 | X1 = 6 \wedge X2 = 5) &= \frac{P(Y=1) \cdot P(X1 = 6 \wedge X2 = 5 | Y=1)}{P(X1 = 6 \wedge X2 = 5)} \\\\&= \frac{P(Y=1) \cdot P(X1 = 6 | Y=1) \cdot P(X2 = 5 | Y=1)}{P(X1 = 6 \wedge X2 = 5)} \\\\&= \frac{0.5 \cdot 0.2 \cdot 0.4}{0.1 \cdot 0.2} \\\\&= 2 \end{aligned} $$
I can clearly see that I am missing a 0.5 (or even a P(Y=1)) somewhere but I can not see where/why. To me, both equations make perfect sense. What am I doing wrong in the second one?
Bonus question: In this video: https://youtu.be/lFJbZ6LVxN8?t=557 the person tries to do something very similar. He explains that he will leave out the denominator for comparing probabilities because it is the same in both cases. But if I were to add the denominator back in, the probabilities would be >1 in that case too? Is he doing a mistake as well or is my logic flawed somewhere? I.e. the probabilities would be this instead:
$$ \begin{aligned} P(Y=1 | X = (0,2)) &= \frac{P(Y=1) \cdot P(X = (0,2) | Y=1)}{P(X = (0,2))} \\\\&= \frac{P(Y=1) \cdot P(X1 = 0 | Y=1) \cdot P(X2 = 2 | Y=1)}{P(X1 = 0) \cdot P(X2 = 2)} \\\\&= \frac{\frac{4}{10} \cdot \frac{3}{4} \cdot \frac{2}{4}}{\frac{4}{10} \cdot \frac{3}{10}} \\\\&= 1.25 \end{aligned} $$
This is again >1.