Originally posted by mtthwPoint taken.
Oh, and...
[b]In fact Bayes theorem does not apply, conditional probabilities does not apply, in fact gives the wrong answer.
Bayes' theorem is a statement of equivalence. It is simply true. It might not always be useful, but it never gives the wrong answer. If it does, you're applying it wrong.[/b]
Originally posted by PBE6You're a rather dense fellow, all things considered.
I'm telling you, he's the next Andy Kaufman. This thread will live in infamy as the breakthrough performance of the world's crappiest math comic. For posterity, I'd like to brand this type of humour as "macomicmatics", from "mathematics" and "comic" and the Latin meaning turd-gurgler. 😀
Originally posted by eldragonflyActually, the answer is 1/38 for both. The result of one roulette spin is completely independent of any other roulette spin on a fair wheel. What has happened is in the past and set in stone, remember..
(a) 1/38
(b) (1/38)^2
What IS true is that the chance of the next two spins being seven is (1/38)^2.
Originally posted by geepamoogleYou are backpedalling here, the usual. Nice try.
Also a note on probability for one-time events. Your odds will match the long term odds or repeated runs of the same identical event. When I roll a fair six-sided die once, I have a 1/6 chance to get 6, not a 1/2 chance.
When people talk about 50/50, that means more than "it could either way". It means "It's equally likely to go either way."
Also a note on probability for one-time events. Your odds will match the long term odds or repeated runs of the same identical event. When I roll a fair six-sided die once, I have a 1/6 chance to get 6, not a 1/2 chance.
When people talk about 50/50, that means more than "it could either way". It means "It's equally likely to go either way."
You are backpedalling here, the usual. Nice try.
And on which point do you disagree, and with what statement would you replace it?
Originally posted by geepamoogleThis statement:
[b]Also a note on probability for one-time events. Your odds will match the long term odds or repeated runs of the same identical event. When I roll a fair six-sided die once, I have a 1/6 chance to get 6, not a 1/2 chance.
When people talk about 50/50, that means more than "it could either way". It means "It's equally likely to go either way." ...[text shortened]... y.
And on which point do you disagree, and with what statement would you replace it?[/b]
Also a note on probability for one-time events.i am only saying that i made a similar statement(s) long ago, they was routinely "debunked" by you and the others, in fact i can find more recent examples of this. Oh and i lied, i was not an honors student, i have only taken some classes at the local jc, that is all.
Originally posted by eldragonflyi am only saying that i made a similar statement(s) long ago, they was routinely "debunked" by you and the others, in fact i can find more recent examples of this. Oh and i lied, i was not an honors student, i have only taken some classes at the local jc, that is all.[/b]You're welcome to point to the specific statement you made and I will gladly reexamine it to see if it is the same statement or whether I see some difference.
This statement:[b]Also a note on probability for one-time events.
Now it can be different if we are given information about the results of specific instance of an event.
For instance, if I knew the dice was at least a 4 in a particular roll, then the odds of a 6 double for this specific instance, because I have eliminated 50% of the possibilities, none of which are 6. It goes from 1 out of 6 equally likely possibilities to 1 of 3 equally likely possibilities.
However, in the absence of any information about the results of a one-time event, the odds will match the odds if the event is repeated a large number of times.
Originally posted by geepamoogleYour backpedalling once again geepamoogle, hiding behind your rather unfortunate embellisments, remote statements and lackluster criticisms. Let's move on..
You're welcome to point to the specific statement you made and I will gladly reexamine it to see if it is the same statement or whether I see some difference.
I added to my statement because I realized (and was typing out before you posted your response) what you may have been referring to.
If information is given on the result that eliminate some possibilities, that has a good chance of affecting the odds of an event which has already taken place, but for which the results are unstated fully.
This is where Bayes formula comes into usefulness, provided you can determine what percentage of the time the results of an event will tend to match the information supplied about the results, and what percentage of those cases meet the condition the question asks about.
However, if no information is given about the results of an event already done, then the odds you have will match the overall long term odds if the event is repeated many times.
It is, of course, impossible to know the results of an event which has not yet occurred unless the setup only allows one possible result.
However, if the event is linked with a past event, the results of the past event may affect odds of the future event. But in looking at it fromthe perspective of the future event, that is a part of setup, and not results. Results only come once the event is finished.
Let's try it this way:
There are six children arranged in three groups, according to the following model:
Group A-Sarah, Rebekah
Group B-Benjamin, Naomi
Group C-Isaac, David
A little girl walks out, and a guy says 'I bet you even money that the other member of the
group is a also a girl.' This is a bad bet, and here's why.
So, we know that it's not group C, so we can rule that out entirely -- neither Isaac nor David
is going to walk out.
Now, it could be Sarah, which means Rebekah will walk out (loser).
Now, it could be Rebekah, which means Sarah will walk out (loser).
Now, it could be Naomi, which means Benjamin will walk out (winner).
I hope this clears it up for you, Eldragonfly.
Nemesio