Gstamsharp t1_jaewst6 wrote

>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.

This is exactly what you do. If you want to know the probability of rolling 6 6s in a row, you calculate it for that. If you've rolled 3 out of 6 and want to know the probability of making it to 6 straight, you do the very basic 6 - 3 = 3, and then crunch the probability for only 3 rolls, because you are only actually calculating the probability of 3 rolls.

Your "preservation of information" is a simple subtraction. It's not some mystical connection influencing future rolls. All those future rolls are still entirely independent of the ones you've already made.

The probability of rolling the same on 6 dice, and the probability of rolling 3 of the same having already rolled 3 are identical, because you're still rolling 6 dice.

The probability of rolling 3 more of the same after any arbitrary amount is not the same question. This is where your confusion is coming from. Here, you only need the probability of 3 rolls, no matter how many you've rolled previously.


Gstamsharp t1_jacan52 wrote

It's trivial because they're independent. Bayes theorem applies to dependent events. Dice rolling isn't dependent. You're talking in circles to justify using incorrect logic.

"The 1998 Superbowl championship isn't related to how many Doritos I ate yesterday, but if it was here's how I'd twist this logic diagram to explain it."

That's what you're doing here.


Gstamsharp t1_jab4xlc wrote

>However, the probability of rolling 6 6s in a row varies based on the prior events.

I mean, sort of, but that's not what you're actually calculating.

>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.

Yes, but no. That's still just the odds of rolling 3 in a row. That calculation does not care about or consider the previous 3 rolls at all. It's indistinguishable from just checking the probability of 3 rolls, because that's all it's doing.

You're not wrong in removing the previous 3 rolls from the calculation, but that's exactly what they mean by "only looking forward." You literally need to ignore the earlier rolls to calculate the future outcome. You don't look back.


Gstamsharp t1_ja9kqlz wrote

>example of this: the odds of rolling 6 6s is (1/6)5, but the odds of rolling 6 6s given that you already rolled 5 6s is (1/6).

Your example supports the previous assertion, though. The probability of 1/6 is the probability of only that final roll. There's no difference between "the odds after 5 sixes" and "the odds after zero rolls" and "the odds after 6000000 sixes in a row." It's not looking back at all.