(August 31, 2015 at 11:09 pm)IATIA Wrote: You have obviously missed the point.
Did I? Forgive me, but I guess I didn't understand your point, then. I suppose I will have to see if you explain it better here...
(August 31, 2015 at 11:09 pm)IATIA Wrote: Firstly, there is no coin toss in my argument.
Indeed, I used it as a example of something analogous. The coin toss is a binary example, like yours, so it should have similar distribution properties.
(August 31, 2015 at 11:09 pm)IATIA Wrote: Secondly, regardless of whether or not you pick the correct hand, there is still a coin.
That would be the reason that I used an example with two clear choices, instead of three. In a situation where "no coin" was a possibility, you would be at 33.3_/33.3_/33.3_ instead of 50/50. I would have had to use an analog with three choices to compensate.
(August 31, 2015 at 11:09 pm)IATIA Wrote: There is not a 50/50 chance of there being a coin, only a 50/50 chance of picking which hand it is in.
Indeed, and in mine it is not a 50/50 chance of there being a coin, only a 50/50 chance of choosing the correct result of the flip.
(August 31, 2015 at 11:09 pm)IATIA Wrote: Likewise, there is not a 50/50 chance that a god exists. only a 50/50 chance that one chooses correctly.
My example had a sum total of diddly-squat to do with gods existence, only the probability of the choice being correct.
(August 31, 2015 at 11:09 pm)IATIA Wrote: As to the coin toss, every time it is tossed there is still a 50/50 chance of heads. The number of tosses does not change the possibilities.
Let us see:
f(n)= pr(tails in first flip)×f(n-1) +
pr(heads in first flip, tails in second flip)×f(n-2) +
pr(heads in first 2 flips, tails in third flip)×f(n-3) +
pr(heads in first 3 flips, tails in third flip)×f(n-4) +
pr(heads in first 4 flips, tails in fourth flip)×f(n-5) +
pr(heads in first 5 flips, tails in fifth flip)×f(n-6) +
pr(heads in first 6 flips, tails in sixth flip)×f(n-7) +
pr(heads in first 7 flips) =
(1/2)×f(n-1) +
(1/2)2×f(n-2) +
(1/2)3×f(n-3) +
(1/2)4×f(n-4) +
(1/2)5×f(n-5) +
(1/2)6×f(n-6) +
(1/2)7×f(n-7) +
(1/2)7
Where:
f(n)=probability of success within n flips.
pr(x)=probability of x happening.
No, the number of flips changes the distribution curve, changing the odds of the sequence. Even if the odds of a single flip doesn't ever change, the odds of any sequence of numerous flips does. This also influences the probability of what will come next.
I would more generally advocate that one only leave one entrance into their mind(reason), and keep the rest of it rather closed, as it is one hell of a lot easier to shovel shit in than it is to get it out.
If the evidence and reason for you to believe something isn't really any better than the reason you should believe some rural farmer from Arkansas got anally probed by interstellar visitors, then you probably shouldn't.
If the evidence and reason for you to believe something isn't really any better than the reason you should believe some rural farmer from Arkansas got anally probed by interstellar visitors, then you probably shouldn't.