Submitted by Reason-Local t3_11de5ag in explainlikeimfive
boytoy421 t1_ja8asfv wrote
Reply to comment by hinoisking in ELI5: why does/doesn’t probability increase when done multiple times? by Reason-Local
This. The odds of rolling 6 6s in a row are pretty n low (like less than 1/1000). But once you've rolled 5 6s in a row the odds that the next one will be a 6 are 1/6.
Probability only looks forward
rohlinxeg t1_ja8t6qa wrote
>Probability only looks forward
I like this phrase. I need to try and remember it.
Canadian47 t1_ja96qts wrote
Another way of saying this is that "dice don't have a memory"
hero_in_time t1_ja9luer wrote
It's not always true, events can be dependent
boytoy421 t1_jaa3nk4 wrote
Sometimes, poker is a good example, assuming you're playing with a 52 card deck and there's 5 people playing and you're 4 flushing and waiting for the river to hit. You'd think that since you're 4 flushing your odds of hitting the flush are 1 in 4 since the river card can be 1 of 4 suits.
But it's actually slightly less because if you're 4 flushing then that means a disproportionate number of your suit were already played.
It doesn't throw it off MUCH (so if you're 4 flushing you should probably chase it, especially if you're pocket suited) but it is LESS than the 1 in 4
hero_in_time t1_jaa6q46 wrote
That's one example. It makes quite a difference tho, 6%. Also, whether you should call or not would depend on the bet size and the pot size. Getting slightly over 4 to 1 would be break even, not considering implied odds.
spicymato t1_ja9ft0g wrote
>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.
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.
spicymato t1_jab2i7v wrote
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.
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.
spicymato t1_jabmsjj wrote
>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).
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.
spicymato t1_jaejafi wrote
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).
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.
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