First off, I have to correct an earlier mistake: a single die roll is fully independent from other rolls, but a sequence is not. If the ask is about the probability of rolling a sequence and you've already rolled some, then you can't ignore what's been rolled so far. It doesn't affect the next roll, but it does affect the probability of the sequence overall.

As for your nonsensical question, you could still apply Bayes' to independent events, but P(B|A) is just P(B), which cancels out with the denominator, leaving just P(A|B) = P(A) (which is to be expected, by definition).

Going back to the dice sequence: if dice sequences were fully independent, then P(3 6s given 6 6s) wouldn't be 1, but be (1/6)^3 , by definition. But that's silly, because we know the given here is that we already rolled them.

But we're now talking in circles. You insist on updating the expected scenario at each step (first, you ask about 6 6s, then about 5 6s, then 4 6s...) while I keep the information along the way (first, 6 6s, then 6 6s given 1 6, then 6 6s given 2 6s...). The answers were get are the same, but my notation preserves the original ask and the history of events.

The preservation of information between rounds is important, because it will impact the probablilty of arriving at that original end goal. The Monty Hall problem is a classic example of this: if you treat the second decision as independent of the first, then the odds for each door from your perspective becomes (1/2); but if you include the given information about prior events, then the odds of your first door being right becomes (1/3) and the other door becomes (2/3).

>That's still just the odds of rolling 3 in a row.

Technically yes, but also technically no. It's similar to saying (2*2) is the same as 4. They are equivalent, but also technically different. P(3 6s) is equivalent to P(6 6s given 3 6s), but the latter has information about prior events that the former doesn't; it also is describing a different set of events.

You can only ignore the earlier rolls because each roll is independent of the rest. If the event in question is not fully independent, then it will alter the probability in a way that doesn't let you simply eliminate the prior events. This is what I meant by the dice example being trivial; because they're independent, calculating the future probability is trivial, regardless of the given priors.

This is literally Bayes' theorem. P(A|B) = (P(B|A)*P(A))/P(B). In words, the probability of A given B is equal to the probability of B given A times the independent probability of A all over the independent probability of B. If you plug in the dice example:

• P(3 6s given 6 6s) = 1;
• P(6 6s) = (1/6)^6;
• P(3 6s) = (1/6)^3;
• Therefore: 1(1/6)^6 /(1/6)^3 = (1/6)^(6-3) = (1/6)^3

Again, trivial, but shows that prior events can be used in the calculations of future events (even if they are independent and don't actually impact the result).

As I said, the probability of any single roll being a 6 is 1/6, because each roll is a fully independent event.

However, the probability of rolling 6 6s in a row varies based on the prior events. After 5 6s are rolled, the final roll is still a 1/6 event, but after, say, 3 6s, the probability is (1/6)^3. P(6 6s) ≠ P(6 6s given 5 6s). The given part is the piece that looks backwards.

>Probability only looks forward

-ish. Bayes' theorem shows how probability of an expected scenario changes based on what's happened so far.

Your 6 6s scenario is a great, if somewhat trivial, example of this: the odds of rolling 6 6s is (1/6)^6, but the odds of rolling 6 6s given that you already rolled 5 6s is (1/6).

Things get slightly more complex when events are not fully independent, though the fundamentals of "probability of X given Y" is still basically the same.