No, to be fair, a walk was in order.
Shak Mat Flannel Jesas!
Note to self: don’t leave for work if PZR is waiting for a reply.
Dang.
I actually saw a guy do this, by the way, using a bread board computer he built from scratch.
It’s about matching permutations - which you’ve said already, I think, though maybe not in those words. The reason why, at the start, 50/50 is more likely than 100-0 is because there are more ways to flip 50/50 than 100-0. You can flip 50T and then 50H, you can flip them alternating starting with H, you can flip then alternating starting with T, you can… etc etc. So many permutations for 50H 50T and only one permutation for 100H.
Just consider 2 flips:
HH
HT
TH
TT
We have 2 ways to reach 50/50, HT and TH, but only one way to reach HH and one way to reach TT.
BUT each individual permutation is as likely as any other one. once you’ve already flipped a single H, HT and HH are both equally likely because each relevant permutation is equally likely.
When we’ve flipped 99 H’s, there’s only 2 permutations of 100 flips we can be in: the one with 99 consecutive H’s and a T, or the one with 100H’s. Since each permutation is equally likely, it’s 50/50 which one it goes to.
You’re a programmer, so you might like this analogy:
A sequence of 100 coin flips is like generating a number in binary that’s 100 digits long.
So imagine all numbers in binary, with 100 0s and 1s, starting with
000…00
000…01
000…10
000…11
And increasing all the way up to
111…11
You can word mine and Carleas position as, “each of those permutations of 0s and 1s is equally likely”, and since each of those is an integer from 0 to (2^100)-1, it’s isomorphic to saying “every integer in that range is equally likely to be generated by a fair random number generator.”
Now, only one integer above can be represented in binary as 100 0s, and only one can be represented as 100 1s, but many of them can be represented by some various mixes of 0s and 1s that add up to 50 1s, or 49 1s, or 48 1s, etc.
When we’re at 99 flips, and we’ve just seen 99 heads in a row but don’t know what the last flip is, then we’re in this situation:
111…1?
The question mark could be a 0 or a 1. It’s in the future, we don’t know yet.
But either option is just a single integer in the range. If the final flip is a 1, the whole sequence is the integer (2^100)-1, and if the final flip is a 0, it’s the integer (2^100)-2.
Carleas and I are basically saying something isomorphic to, if we’re generating a genuinely random integer between 0 and (2^100)-1, there’s no reason to assume (2^100)-2 is more likely than (2^100)-1. They’re both just single individual integers in the range, there is no obvious reason (to us) to assume one of those two equally specific integers should be favoured.
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The error you keep making is getting to the 100th flip after 99 heads in a row, and then throwing away the 99 previous flips to start your calculations over. You never completed the original 100 flip scenario, you got to 99 and tossed the scenario and started from scratch so that the odds increased to 50-50. What you did was remove the fact that 99-1 is more likely than 100-0 once 99 flips were complete. Why not do that at flip 32, or flip 63? Heck, do it for every flip so that you can say at the start there is a 50-50 chance of 100-0 or 99-1, and you can show that is the case for every flip.
Again, 99-1 is more likely than 100-0, so you can’t say there is a 50-50 chance after 99 flips and the score being 99-0. It is more likely to end up 99-1 than it is to end up 100-0.
Yes, when someone asks you a question affects the reasonable answers.
If you ask me “will it rain tomorrow at noon?” I might say “I don’t think so”. If it’s tomorrow at 11:59 and it’s raining and you ask me “Do you still think it probably won’t rain at noon?”, I’m gonnna say no, I think it’s probably still going to be raining in 1 minute.
Let’s say we’re at a basketball game and, before the game starts, you and I both agree that the Lakers and the Golden State Warriors have roughly even likelihood of winning. As the game starts, we both agree that their odds of winning are 50/50.
Then, it gets to 5 seconds left, and you turn to me and say, “What’s the probability of the Lakers winning now?”
I look at the score and see, with 5 seconds left to go, the Lakers are 10 points ahead. So I turn to you and say, “basically 100%, I’m not sure it’s even possible to score 10 points in 5 seconds”.
My answer to the question changed because you asked me it at a different time. That’s… normal. If you ask me a question before the game, and you ask me the same question right near the end of the game, the reasonable answer is going to change. I don’t see a problem with that. Answers change as time passes, as you gain more information.
The question is asked at the start and the answer is not known until the end. What you continue to do is ask question A at the start, and then ask question B just before the end.
Do you agree at the start that 99-1 is more likely than 100-0? What changes after 99 heads in a row? Both 99-1 and 100-0 can get to 99-0 and have only 1 flip left, and yet you change from “99-1 is more likely than 100-0” to “It is equally likely for 99-1 or 100-0.”
Yeah, there are 100 permutations of 99-1 at the start, and only 1 permutation of 100-0 at the start. So at the start, 99-1 is roughly 100 times more likely.
Well, you eliminated 99 of those 100 possible permutations of 99-1. Now there’s only one possible permutation of 99-1 left that it could be, and also only one permutation of 100-0, which is why they have even odds now - same number of possible permutations.
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No. That is not how it works. From the start, if I asked you for 100 flips, which is more likely if it lands on heads 99 times in a row, 99-1 or 100-0?
Now you have two contradictory answers, that 99-1 is more likely, and that 99-1 and 100-0 are equally likely. You can’t do that.
If you ask me the same question at different times, of course my answers will seem contradictory. If you ask me if I need to poo, I might say no. If you wait an hour and ask me if I need to poo, I might say yes. That’s not ACTUALLY contradictory, things change over time.
Would your answer to the question not change depending on how late into the flip session I ask you the question?
You ask a different question. The first question is “which is more likely, 99-1 or 100-0?” The later question after 99 heads in a row is “Which is more likely, heads or tails?”
Yeah, you can absolutely consider them different questions, and thus it’s okay that they have different answers.
But the question is out of 100 flips which is more likely 99-1 or 100-0, not out of 1 flip which is more likely, heads or tails. That is moving the goal post.
Your original answer at the start is that 99-1 is more likely than 100-0, so find out what happens after 100 flips, not before.
I don’t know what you mean by “THE question”. There have been multiple questions asked, not just one. That’s certainly not the only question that was asked in the context of this conversation.
The question is which is more likely, 99-1 or 100-0?
Your answer is 99-1 is more likely than 100-0 and you are correct.
So when it gets to 99-0 there is still 1 flip left in order to complete that scenario. It wasn’t a question of 99 flips. The odds weren’t calculated for 99 flips, they were calculated for 100 flips. All 100 flips, not 99 flips and then change the question to 1 flip. The last flip of 100 makes it less likely to get 100-0 than 99-1 after a score of 99-0. The more flips the exponentially harder it is to achieve another heads in a row. You eliminate that when you start over at 99 and ask a different question for the last flip of 1, which is more likely out of 1 flip, heads or tails. You just eliminated the last flip from the original calculation which already took into account how much less likely it is to reach 100-0 than it is to reach 99-1.
I don’t see why it would. Coins don’t have memory.
Coins don’t do the calculations either. You already calculated from the start of 100 flips that it is more likely to end up with a score of 99-1 than 100-0. You already answered the question at the start that after 99 heads in a row it is more likely to end up tails than it is to end up heads, because you already answered that 99-1 is more likely than 100-0.
No, my answer at the start didn’t say anything about how many heads in a row.
My answer at the start includes 99 heads with 1 tail as the first flip, or one tail as the second flip, or one tail as the third flip, or … and so on