Nobody is talking about “evenly split.”
The odds from the start are that it is more likely to be 99-1 than it is to be 100-0. We get to 99 heads in a row. There is no new information there to use to change your mind to “the last flip is equally likely to be heads or tails.” For all you know 99-1 is still more likely than 100-0.
For all you know, perhaps, but I have explained how I conceive of the odds - as permutations of flips - and seeing 99 heads in a row eliminates 99 possible permutations of a 99-1 split.
There’s only one possible permutation remaining of 99-1, and one permutation of 100-0.
So you’re saying from the start that reaching the permutation of 99-1 is equally likely as reaching permutation 100-0?
If be “the permutation of 99-1”, you mean specifically the one that has 99 heads in a row and then 1 tails, then yes.
All specific individual permutations are equally likely.
So you think 99-1 is equally likely as 100-0 from the start.
If by 99-1, you mean specifically 99 heads in a row and then 1 tails, yes.
It is not. It is EXPONENTIALLY less likely to get 100-0 than it is to get 99-1.
Can you confirm if you know what I mean by “specific permutation”?
Yes, as in Carleas pic earlier in this thread, each permutation is at the bottom of the pic.
Example:
HH
HT
TH
TT
Yes exactly, and HT is a different specific permutation than TH, so even though they have the same distribution of TS and HS as each other, they are counted differently when we’re counting different specific permutations. Does that all make sense?
The internet says:
Using the law of averages, one might predict that there will be 50 heads and 50 tails. While this is the single most likely outcome, there is only an 8% chance of it occurring. 20 Apr 2018
Is it really 50 50 when flipping a coin? The flipped coins, according to findings in a preprint study posted on arXiv.org, landed with the same side facing upward as before the toss 50.8 percent of the time. The large number of throws allows statisticians to conclude that the nearly 1 percent bias isn’t a fluke.
What is the coin flip probability paradox?
Every flip of a fair coin is still fifty-fifty and is not influenced by any flips before it. Take another look at the chart above. Heads appeared after heads four times and tails appeared after heads four times. Betting on tails after a heads would yield exactly the results you would expect: a fifty-fifty chance.
I understand what you are saying. I even laid this out in the other thread, I think before anyone mentioned it.
But you are moving the goal post when you get to the “red circle” way down the line. You are negating the fact that it is much tougher to get 100 heads in a row than it is to get 99 heads in a row. It is not just one more heads, it is one more heads after 99 heads.
Say you flip once, it has a 50% chance of landing on heads.
Now what’s the chance that you get 2 heads in a row?
The reason I’m asking you these questions is because, if you’re down with it, we can test it. We can write a test that checks if this is the case - but you have to accept what it means to be a different permutation first.
If we want to test it, we’d program a random number generator that generates a sequence of 4 flips - getting 4 heads in a row is much more common than 99, so we can actually get some useful statistics going.
My proposal to you is this:
I’m going to write a program that generates 4 flips and records which of the 16 possible permutations of flips it is
TTTT
TTTH
TTHT
TTHH
…
HHHT
HHHH
all 16 different possible flips.
We’re going to program it one function at a time, and at the end we’re going to run tens of thousands of tests and count how many times HHHT occurs, and how many times HHHH occurs.
This is a much more digestible test than 99 or 100 in a row.
My prediction is that HHHT (which means, not counting THHH, HTHH, HHTH – only HHHT specifically) occurs approximately as often as HHHH.
Do you want to do this experiment?
The problem is that you are giving equal likeliness for each permutation, when in fact that is not what we are calculating. We are calculating the likeliness of 100 heads in a row, regardless of the other permutations.
Does that mean you’re not interested in finding out if HHHT occurs just as often as HHHH?
Right, I want to know how likely it is for HHHH to occur, not how many times it occurs. We are not seeing how many times 100-0 occurs over multiple sets of 100 flips, we are calculating how likely it is for 100-0 to occur over ONE SET of 100 flips.
I guess that’s a rejection of the experiment then, okay.
If you can make a program to calculate the likelihood of 100-0 occuring on 1 set of 100 flips then I’d like to see it. I did just a tiny bit of C++ coding decades ago so I’m not a coder, but I’d like to look at what you have to offer. I can figure stuff out most of the time.
The answer is 1 / (2 ^ 100), which is the same answer as the answer for how likely it is to get 99 Heads in a row follwed by one tails.
So you are recanting your answer that it is more likely for 99-1 than 100-0?