Lol 0.02% ( I haven’t looked it over yet but I’m trusting it for now) fits
?
Lol 0.02% ( I haven’t looked it over yet but I’m trusting it for now) fits
?
Anyway, maybe you’re right Carleas.
You call that directly?
Are you suggesting the flips I’m testing aren’t tail-heavy? If they are all tail-heavy, why are they evenly split?
I’m saying if you had proposed “my program tests this question,” that might have been honest, but adding “directly” was not.
In my findings, the more twists and turns you add to a path, the more convoluted you make it, the more opportunity you give yourself to have it conveniently end up where you want it.
For myself, I only ever trust the straight cut.
But that’s just me, and I may be wrong.
Maybe the boundary effect is real and my program gives a skewed result.
And maybe you found a way to solve that boundary effect.
Maybe the objections and difficulties I found in making a program to test this weren’t really relevant much, or at all.
That’s probably what happened.
Salute Carleas.
You guys are funny. Divine intervention does happen on occasion. People gamble on coin tosses to see if they are loved by existence or not. To validate themselves above others.
It is actually possible that everyone who throws a coin gets heads every-time.
We don’t live in that universe. If we did, we’d make it a law of the universe and ponder it.
So just as a reminder:
This whole debate was settled already in the first thread after my suggested amendment to the program provided.
The suggestion was that only cases where hhh followed a perfect 50 50 be counted. This would, almost by definition, give a higher likelihood of t for that one flip. “All else being equal.” Substantialists have a hard time with this, but a statistitian is quite comfortable stating it this way: “if all you know about it is…” If you know 0 about previous flips, and all you know is hhh, you have to guess a 50 50 previous composition and thus assign a greatr likelyhood of t. So selecting only instances where the sample previously gives 50 50 simulates an hhh off the bat.
The problem with a program like that is that you begin to select for increasingly specific circumstances, more and more likely to give few occurances per sample and thus less statistically reliable proof. What you want is a program that takes a good chunk of the overall data, that has consistently many thousands of repetitions per sample.
As a thought experiment, however, it was already settled.
The answer is no. This is basic statistics.
The belief that past probablities affect future probabilities is the gamblers fallacy.
All the people unwaveringly spouting nonsense about statistics should have all opinions they hold scrutanised, as it shows they hold no value for basic research.
Welcome to ILP Ourora.
Thank you for your welcome.
Obviously if someone has already somehow coin flipped 100 “tails” in a row then you would be wise to bet on the next coin flip outcome being “heads”. Or else you’re just silly.
This is the same deal as the whole “should you change your guess based on new information” when playing that game show of what’s behind the three doors. One has a prize, the other two don’t. You guess one of the three options, but before they reveal the door the show you a different door which has nothing, then ask if you want to change your guess.
Hell yes you should change your guess. Why? Because NEW INFORMATION matters.
This is the same reason why some people don’t understand that a dozen monkeys randomly pounding on keyboards will NEVER, regardless of infinite time, produce a Shakespearean drama. The odds of the next step in the sequence at any given point being correct to the meaningful outcome receded exponentially faster than do the permutations themselves. This is easy math but weirdly non-intuitive for, well let’s say umm hard-headed “I BeLieVe iN ScIenCeE” type people.
Obviously if someone has already somehow coin flipped 100 “tails” in a row then you would be wise to bet on the next coin flip outcome being “heads”. Or else you’re just silly.
If someone flips a coin tails 100 times in a row, your best bet is to bet on tails.
Either it’s a fair coin and it’s 50/50, and we just saw a miracle run of 100 tails in a row, OR the coin is explicitly weighted towards tails, physically.
Obviously. And yet I was explicitly assuming a fair 50/50 flip. But thanks for playing, Capt. Obvious.
If it’s obviously 50/50, then what makes tails silly? Why is it silly not to bet on heads?
All of your words put together don’t give me the impression that what I said was obvious to you.
50/50 is due to the fact that there is 2 possible outcomes. It would still be 50/50 even if one side of the coin was weighted. There would still be 2 possible outcomes, heads or tails…rigged coin or not.
So watching a coin land on heads 100 flips in a row would have a 50/50 chance of heads or tails the next flip, but regardless of that there is more of a chance than not that the coin is not a fair coin due to the coin landing on heads 100 flips in a row, so my bet is on heads for the next flip. Something is obviously going on other than it just being an even chance of landing on either of the two possible outcomes that make it 50/50.
50/50 is due to the fact that there is 2 possible outcomes…rigged coin or not.
I don’t think you’re using 50/50 in the way other English speakers use it. For everyone else, 50/50 doesn’t just mean “2 possible outcomes”, it specifically means “2 possible outcomes where each one has 50% probability of happening”. That’s what the 50s in 50/50 refer to - probability.
So if a coin is rigged somehow and that changes the probabilities, and I know it’s rigged and I know it comes up heads a lot more often than tails, then it’s not 50/50.
The only thing calculated is how many possible outcomes. One flip divided by two possible outcomes means each possible outcome has 1/2 a chance, or 0.50 chance (50%), or we say 50/50 heads or tails.
A six sided dice means each side has a 1/6 chance. Weight is not part of that equation. The only thing considered is how many possible outcomes for one throw of the dice, which is 1 divided by 6, or each side has a 0.1666… chance. We could say the chances are 16.6/16.6/16.6/16.6/16.6/16.6 but who does that?