No I think you might be misreading something.
.
The probability of obtaining 100 heads as a result of flipping a fair coin 100 times is 1/(2^100) = 1/1267650600228229401496703205376 (1 on 1 nonillion 267 octillion 650 septillion 600 sextillion 228 quintillion 229 quadrillion 401 trillion 496 billion 703 million 205 thousand 376).20 May 2019
1 Like
No, I think you are moving the goal post and claiming I can’t see straight.
99-1 and 100-0 both could have 99-0 prior to the last flip, so nothing new there as far as 99-1 is more likely than 100-0.
I don’t understand this wording. “Both could have”.
99-1 and 100-0 are final results. They both could have stemmed from a score of 99-0.
Yes, that’s right, they could.
Right, so what gives you the idea to change your original answer of 99-1 is more likely than 100-0?
Your question isn’t worded clearly enough for me.
What new information do you have at 99-0 that makes you change your statement from the start that 99-1 is more likely than 100-0? What made you change your mind? Because at the start you knew it could be 99-0 with 1 flip left, and yet you still gave the answer that 99-1 is more likely than 100-0.
It’s the same reasoning as before : permutations being cancelled out, falsified.
It’s my position that all specific permutations are equally likely. 99 heads in a row and then one tails is one specific permutation. 100 heads is another one permutation. Since they are both individual specific permutations, they’re equally likely
At the start, there are 100 different possible permutations of 99-1. At the end, there’s only one possible one remaining.
For 2 flips how many permutations are there?
four possible permutations.
So you think it is equally likely to get 2 heads in a row as it is to get 1 head and 1 tails?
You aren’t comparing specific permutations to specific permutations. One of those things is one specific permutation, and the other thing you said can be achieved by mulitple permutations. So no.
Could you link to where you found this? Thank you.
Right, they are not the same thing, which is my point. I want to know how likely it is for 2 heads in a row, not the likelihood of ending up on one of the 4 permutation. You give equal likeliness for each of the 4 permutation, but in fact it is less likely to have 2 heads in a row compared to 1 heads and 1 tails.
Well I could tell you, but I don’t know how useful my answer is for you.
1/4 of 2-coin flips will be HH, 1/4 will be TT, 1/4 will be HT and 1/4 will be TH. Every permutation is equal. That’s my position.
You are entitled to your position, but not your own facts. It is a fact that HH is less likely than HT. Magnifying that concept, it is much harder to get 10 heads in a row, and exponentially harder to get 100 heads in a row. You seem to be claiming that it is equally likely to have 50 heads and 50 tails then it is to get 100 heads in a row. That is simply not true.
Can you prove it in any way? Do you have a reliable source of information?
Note again that HT can’t be treated as interchangable with TH - those are two different permutations.