Lets say that it has been established that you have a 50/50 chance of each flip turning out to be either heads or tails. Should you consider the law of averages when predicting the result of the 10th try? Or would you say that no matter what the result of the previous 9 flips, it is still a 50/50 chance of heads or tails?
2007-12-29 03:25:47 · 12 answers · asked by lloyd.salimbao 1 in Science & Mathematics ➔ Mathematics
so which makes more sense to bet on, the fact that it is an isolated event or that its not likely to turn up heads 10 out of 10 times?
2007-12-29 03:50:56 · update #1
The probability of getting a tails on the 10th flip is 50/50 yes but as it is an isolated event. However the probability of achieving 9 heads in a row folloed by a tails on teh 10th flip is found by setting up a binomial distribution 10C9*(1/2)^9*(1/2). Following the general binomial equation of
nCr*p^r*q^(n-r)
The C being on your calculator as nCr
n=number of trials
r=number of successes (achieveing a heads)
p=probability of success
q=probability of failure
Although I recently learnt that the probability of achieving a heads is not quite 50/50 as the heads side is heavier and so the it is more likely toland heads down showing a tails :) (theres a useless bit of information for you).
2007-12-29 03:40:33 · answer #1 · answered by Anonymous · 0⤊ 0⤋
its still going to b a 50/50 change because a coin has two sides and your not trying to get 9 heads in a row then a tail, your onlying trying to get a tail on the next flip
2007-12-29 03:32:54 · answer #2 · answered by Tommy 2 · 1⤊ 0⤋
The coin has absolutely no memory; therefore, the odds on the next flip are still exactly 50/50.
Failure to understand this most fundamental of concepts is the reason people believe they can devise a system to beat the tables in Vegas. It can't be done for the simple reason that Vegas does not offer a single bet with odds better than 49/100. Of course, you can get lucky, but combinations and formulas and sequences are worthless because every single bet in the sequence has its own negative bias, and again, no memory.
2007-12-29 03:36:08 · answer #3 · answered by MVB 6 · 1⤊ 0⤋
Each flip would be individually rated at a 50/50 chance of heads or tails.
2007-12-29 03:30:45 · answer #4 · answered by Whitest of Trash 2 · 2⤊ 0⤋
As long as the coin is a "fair" one, that is, it is not weighted, then the 10th flip is independant of the preceeding 9. The odds are still "50/50". The "Law of Averages" refers to long-term tendencies - not particular flips. If you continute flipping this fair coin 10,000 times, the "Law of Averages" means you'll get about 5,000 heads and 5,000 tails. It says nothing about the 10th flip being heads or tails.
2007-12-29 03:33:13 · answer #5 · answered by Drew F 2 · 1⤊ 0⤋
As far as I can tell, it's still a 50/50 chance of heads or tails... Nothing has changed, and the 10th toss is a completely new (independent) event, independent of all the other previous events.
2007-12-29 03:32:19 · answer #6 · answered by vu 2 · 3⤊ 0⤋
WOW! A whole string of right answers in a row! What do you think the probability of THAT was ?!!!
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As far as what to bet on -- I'm sure you'd get a lot of people who'd be willing to bet against it coming up heads a tenth time even though that's no more likely than coming up heads again. Your problem would be the long wait for a string of nine heads in a row!
2007-12-29 03:36:54 · answer #7 · answered by Steve H 5 · 3⤊ 0⤋
Its still 50/50 no matter how many times you flip it
2007-12-29 03:33:04 · answer #8 · answered by Anonymous · 1⤊ 0⤋
it is a 50/50 chance no matter how many times you have previously flipped the coin.
2007-12-29 03:32:44 · answer #9 · answered by Anonymous · 2⤊ 0⤋
Still 50/50 as long as it's a fair coin.
2007-12-29 03:29:12 · answer #10 · answered by ironduke8159 7 · 2⤊ 1⤋