pushing_wins
Hall of Fame
whats the answer?
thanks
thanks
stats: avg number of time you have to flip a coin to ensure a head
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pushing_wins
Hall of Fame
zero percent there is no head. stupid question?
pushing_wins
Hall of Fame
i dont know
but average number of flips or expected number of flips?
but average number of flips or expected number of flips?
goran_ace
Hall of Fame
when you flip a coin there are two outcomes, heads or tails.
let x = # of coin flips (or since the outcome of a coin flip has no effect on any other coin flips you can also just throw a bunch of different coins all at once)
probability of not throwing a head, or essentially the probability of throwing all tails in x flips = 1/(2^x)
so the probability of throwing at least one head in x flips = 1 - (1/(2^x))
let x = # of coin flips (or since the outcome of a coin flip has no effect on any other coin flips you can also just throw a bunch of different coins all at once)
probability of not throwing a head, or essentially the probability of throwing all tails in x flips = 1/(2^x)
so the probability of throwing at least one head in x flips = 1 - (1/(2^x))
pushing_wins
Hall of Fame
the law of large number says - if we flip a coin 100 times, we will have 50 heads.
similarly, if we flip a coin twice, we should get one head. 2 is the lowest number of flips to flip a head. is 2 the answer?
pushing_wins
Hall of Fame
what about looking at it from the law of averages angle?
goran_ace
Hall of Fame
the law of large numbers does not state that out of 100 flips you will have 50 heads. it only states that you can expect to have approximately 50 heads. at a larger numbers of flips you will can expect even closer to that 50%.
law of averages isn't really a statistical term and doesn't apply. it's known by another term - gambler's fallacy.
if you threw 99 tails in a row, would you bet on the next throw being heads or tails? according to the law of averages you'd bet heads thinking things should even out and you are about due for a head any time now.
in theory those 99 throws have no affect on that 100th throw so the probability of throwing a head is still 1/2 = 50%. but the thinking man would know that odds of throwing 99 tails in a row are ridiculous so its likely not a fair coin flip and the coin is likely biased towards heads
hope this helps
law of averages isn't really a statistical term and doesn't apply. it's known by another term - gambler's fallacy.
if you threw 99 tails in a row, would you bet on the next throw being heads or tails? according to the law of averages you'd bet heads thinking things should even out and you are about due for a head any time now.
in theory those 99 throws have no affect on that 100th throw so the probability of throwing a head is still 1/2 = 50%. but the thinking man would know that odds of throwing 99 tails in a row are ridiculous so its likely not a fair coin flip and the coin is likely biased towards heads
hope this helps
ramseszerg
Professional
expected waiting time is 2. But to ensure with 100% certainty that you have a head.. can't be done. Unless you flip an infinite number of times.
pushing_wins
Hall of Fame
whats that?
pushing_wins
Hall of Fame
the law of large numbers does not state that out of 100 flips you will have 50 heads. it only states that you can expect to have approximately 50 heads. at a larger numbers of flips you will can expect even closer to that 50%.
law of averages isn't really a statistical term and doesn't apply. it's known by another term - gambler's fallacy.
if you threw 99 tails in a row, would you bet on the next throw being heads or tails? according to the law of averages you'd bet heads thinking things should even out and you are about due for a head any time now.
in theory those 99 throws have no affect on that 100th throw so the probability of throwing a head is still 1/2 = 50%. but the thinking man would know that odds of throwing 99 tails in a row are ridiculous so its likely not a fair coin flip and the coin is likely biased towards heads
hope this helps
the probablity of getting 50heads and 50 tails in 100 flips is 0.079589.
however, the law of large numbers says you should expect the number of heads to be 50.
how do you reconcile the two?
GetBetterer
Hall of Fame
I skipped Statistics and went directly to Calculus and never took it in College, but just by going off a guess:
There's a 1/2 chance it would work the first time, another 1/2 chance it would work the second, and a 1/2 chance the third and so forth.
This leads us to 1/infinity using basic probability. However, if we find the limit of 1/infinity, we get 0.
There's a 1/2 chance it would work the first time, another 1/2 chance it would work the second, and a 1/2 chance the third and so forth.
This leads us to 1/infinity using basic probability. However, if we find the limit of 1/infinity, we get 0.
Because 100 is not a "large" number in the scope of this scenario.
Steady Eddy
Legend
Yep, let A be the average # of flips it takes to get a heads, if you miss on your first try, your average has just grown by 1, so A = .5(1) + .5(A + 1), which implies that A = 2
There's no contradiction. By "expectation" all that is meant is that this is the average result, not a guaranteed result, hence, both are true.
pushing_wins
Hall of Fame
in a driving test center, 50% of the drivers tested walk away with a license.
from that statistic, can we find out the average number of times a driver has to take the driving test before passing?
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thats the question that lead me to think coin question.
from that statistic, can we find out the average number of times a driver has to take the driving test before passing?
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thats the question that lead me to think coin question.
i'll have to go with two on that one too.
Wolfram Alpha says the integral does not exist (does not converge)
pushing_wins
Hall of Fame
maybe so
but could you help with the real life question?
the passing rate at driving test center is 50%.
can you infer from that the average number of attempts to pass at that test center?
if there is not enough info, u can make some reasonable assumptions.
the pass rate is 100% ------> everyone pass first attempt
0% --------> never pass
50% --------> ????????
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RyanRF
Professional
If you flip a coin 1 time the chance of getting at least one head is 50%
If you flip a coin 2 times, the chance of getting at least one head is 75%
So, if you flip the coin 2 or more times, it is more likely that you will get at least one head than no heads at all.
However, you used the word 'ensure'. It is impossible to ensure anything when it comes to chance. Even flipping the coin an infinite number of times could result in an endless string of tails.
If you flip a coin 2 times, the chance of getting at least one head is 75%
So, if you flip the coin 2 or more times, it is more likely that you will get at least one head than no heads at all.
However, you used the word 'ensure'. It is impossible to ensure anything when it comes to chance. Even flipping the coin an infinite number of times could result in an endless string of tails.
RyanRF
Professional
My answer above applies to this too
F-T-S
Rookie
would be twice if that's all testers
pushing_wins
Hall of Fame
doesnt the law of large numbers rule out that scenario?
Steady Eddy
Legend
There's a difference, though. We're justified in thinking of each coin flip as being independent of the others, but it harder to assume that with drivers. For example, a driver who fails the test her first time, might not have a 50% chance of passing her second time because the class of people who don't pass the test might not be as good at taking the test. Also, someone who keeps retaking the test might be getting better each time, so his probability may not be staying at 50%.
Go ahead and model it by thinking of a coin flip, just remember that the reality might be different.
What integral would that be? I see this as the sum of an infinite series.
Yes, since the chance of missing 10 times in a row is 1/1024, and 1024 > 1000, the chance is smaller than 1-in-a-thousand. So the chance of missing 20 times in a row is smaller than 1-in-a-million, 30 in a row smaller than 1-in-a-billion. We can say that as the number of flips approaches infinity the chance of getting a head approaches 100%, but for any finite number of flips, it is still possible to not get a single head.
goran_ace
Hall of Fame
the law of large numbers doesn't rule out anything. it only states that you can expect that your results will look more and more like your predicted with large numbers of trials. it doesn't influence nth coin flip in any way.
you can't apply your knowledge of probability to your real life situation because it is not a completley random and discrete event like a coin flip. with a coin flip, it doesn't matter what your previous result is, your probability is the same on the next coin flip. with a driving test, there's a learning/experience curve. if you've failed the test on your first try, your previous experience makes it more likely you'll be able to pass the second time around. you've been there before. you know what to expect, you know what the difficulty level of questions of the written/computer portion (and will likely see repeat questions), you know the route of the road test, you know what you need to work on for your second test, maybe you're less nervous this time. if you fail the second, you've got less to work on and can concentrate your efforts on those areas to pass the third time around.
also, it's not a level playing field. some people are just smarter than others. with a coin flip if you flip that coin seven times you have a really good chance of getting a head. if you've failed a driving test 6 times already.... well, you're probably going to fail it a 7th and probably shouldn't be on the road even if you do eventually pass. might want to take the bus.
you can't apply your knowledge of probability to your real life situation because it is not a completley random and discrete event like a coin flip. with a coin flip, it doesn't matter what your previous result is, your probability is the same on the next coin flip. with a driving test, there's a learning/experience curve. if you've failed the test on your first try, your previous experience makes it more likely you'll be able to pass the second time around. you've been there before. you know what to expect, you know what the difficulty level of questions of the written/computer portion (and will likely see repeat questions), you know the route of the road test, you know what you need to work on for your second test, maybe you're less nervous this time. if you fail the second, you've got less to work on and can concentrate your efforts on those areas to pass the third time around.
also, it's not a level playing field. some people are just smarter than others. with a coin flip if you flip that coin seven times you have a really good chance of getting a head. if you've failed a driving test 6 times already.... well, you're probably going to fail it a 7th and probably shouldn't be on the road even if you do eventually pass. might want to take the bus.
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I was wrong. The expected value as integral of x*p(x) works only if p(x) is a probability distribution, not just a probablilty of something.
Polaris
Hall of Fame
I assume that "Number of tosses needed to ensure a head" is the same as "number of tosses required for the first head".
The probability of ensuring a "Head" in exactly 1 trial is p
The probability of ensuring a "Head" in exactly 2 trials is (1-p)*p
...
The probability of ensuring a "Head" in exactly n trials is ((1-p)^(n-1))*p
This is a geometric random variable, so the average number of tosses is 1/p.
For a fair coin, the average number of times to ensure a head is 2.
The probability of ensuring a "Head" in exactly 1 trial is p
The probability of ensuring a "Head" in exactly 2 trials is (1-p)*p
...
The probability of ensuring a "Head" in exactly n trials is ((1-p)^(n-1))*p
This is a geometric random variable, so the average number of tosses is 1/p.
For a fair coin, the average number of times to ensure a head is 2.
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pushing_wins
Hall of Fame
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pushing_wins
Hall of Fame
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ramseszerg
Professional
No it does not because now the question is worded differently (average number of attempts at the test centre). The new answer I think is exactly 2.
pushing_wins
Hall of Fame
i do understand the difference.
i feel it is necessary to be able the answer the coin problem b4 i tackle the real life problem
ramseszerg
Professional
Just something I learned in grade 12 stats. It's the number of tries you can expect to have before getting the event based on the probability of the event. For example, if the probability of the event is 10%, you can expect that after 10 tries you will get the event.
Claudius
Professional
You might know more about this than I do, but from what I understand, the number of heads in exactly n trials is a binomial random variable, with probability of k successes (heads) being:
P[X=k]= n!/[k!(n-k)!] * p^k*(1-p)^(n-k)
or more clearly:
where p,r = .5 if the coin is fair.
So P[ensuring a head in exactly n trials] = n!/(n-1)!*p*(1-p)^(n-1)
where p = .5
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Steady Eddy
Legend
We can't know from the data we have. We don't know if the probability of passing remains at .5 for subsequent testing. It might get higher (the test taker gets better), or lower (anyone who fails it at all has a lower chance). Leave it as being a constant .5, but just remember not to be too dogmatic about what the results will be.
Yeah, for example, suppose you must roll a six on a die to begin a game, how many rolls does this take on average? A cute method is to realize that if you miss on your first try, you average will be one more than the true average. So, avg = (1)(1/6) + (avg + 1)(5/6)
avg = 1/6 + (5/6)avg + 5/6
avg = 1 + (5/6)avg)
(1/6)avg = 1
avg = 6
Generally, if your chance of success is 1/n, it will take n trials on average for the first success.
Polaris is right, this is a geometric distribution. You don't have to bother with finding the binomial coefficient, because a geometric distribution stops after the first success. Since the "success" always occurs on the last spot only, it's not like a binomial distribution in that way.You might know more about this than I do, but from what I understand, the number of heads in exactly n trials is a binomial random variable, with probability of k successes (heads) being:
P[X=k]= n!/[k!(n-k)!] * p^k*(1-p)^(n-k)
or more clearly:
where p,r = .5 if the coin is fair.
So P[ensuring a head in exactly n trials] = n!/(n-1)!*p*(1-p)^(n-1)
where p = .5
ramseszerg
Professional
I don't understand this proof. What does each term in "avg = (1)(1/6) + (avg + 1)(5/6)" represent?
Steady Eddy
Legend
Two outcomes are possible on the first trial. Either a success, or a miss. If it's a success, then we score "1", (for one trial), and that will happen 1/6 of the time. If it's a miss, (which will happen 5/6 of the time), we have to try again. What is our expectation after a miss? Since the die has no memory, it is simply 1 more than than the average was previously. This sounds like circular thinking since we use what we're looking for in the right side of the equation, but it works!
Here's an alternative, (longer), way.
Avg = (1)(1/6) + (2)(5/6)(1/6) + (3)(5/6)(5/6)(1/6) + ...
Avg(5/6) =------(1)(5/6)(1/6) + (2)(5/6)(5/6)(1/6) + ...
By subtracting the bottom row from the top row, we get...
Avg(1/6) = (1)(1/6) + (1)(5/6)(1/6) + (1)(5/6)(5/6)(1/6) +...
Multiply each side by 6 and sum the infinite series and ya get...6!
mightyrick
Legend
You can "expect" all you want. You can't ensure it. This whole question is ridiculous. At what point does 50% probability become 100% probablity? Never. Geez, let this thread die, already.
OTMPut
Hall of Fame
this is one nice example which illustrates how "casting in the language of mathematics" quickly leads to a solution.
i wish they make this point early enough in life. the point that math is a language and using it to express certain problems helps solve those problems relatively easily (just as language itself helps solve problems!).
OTMPut
Hall of Fame
You might know more about this than I do, but from what I understand, the number of heads in exactly n trials is a binomial random variable, with probability of k successes (heads) being:
P[X=k]= n!/[k!(n-k)!] * p^k*(1-p)^(n-k)
or more clearly:
where p,r = .5 if the coin is fair.
So P[ensuring a head in exactly n trials] = n!/(n-1)!*p*(1-p)^(n-1)
where p = .5
the random variable he referred to was the number of throws before you need before you get your first head. this random variable has a geometric distribution.
