Percentages?

Question

Someone I know claims to be able to predict the outcome of a coin toss thought not perfectly.. I know they basically have a 50% chance of guessing it right the first time...but how many times would they need to guess it right for it somewhat confirm that they can somehow predict the outcome. Like for example, would they need to guess it right 90% of the time or something like that. And what would be the odds of them guessing say 70% right? I would kind of like to be able to setup some sort of a bell curve to know what values would be out of the normal range for deviation? Thanks

Answer 1

Any given coin toss is 50-50, but we are saying, what are the chances of them getting the first one right AND the second one right. That's a different question all together.

You need to look at the combinations. Let's suppose, they always say "heads". With the second coin toss, your possiblities are:

H-H (right)
H-T (wrong)
T-H (wrong)
T-T (wrong)

It doesn't matter what they choose, one one right answer out of 4 - 25% chance of guessing correctly. That is 50% (or 0.5) times 50%. 0.5 x 0.5 = 0.25 = 25%. So, yes, in fact, it does multiply.

3rd coin toss:

H-H-H (right)
H-H-T (wrong)
H-T-H (wrong)
H-T-T (wrong)
T-H-H (wrong)
T-H-T (wrong)
T-T-H (wrong)
T-T-T (wrong)

1 in 8 chance - 12.5% chance (which happens to be 50% times 50% times 50%).

With every coin toss, it cuts the percentage chance in half. It will never be zero. The question is, how close to zero does it have to get before you say "close enough - nobody could guess correctly that many times". I would say something less than 1%.

Answer 2

The chance of guessing a fair coin toss is .5
The chance of guessing a fair coin toss twice is .25. So that 25% of the time, a random guesser would get it right twice.

The curve is a binomial distribution (right, not right) and looks like a bell curve with enough tosses.

So, to say with 95% accuracy that your friend guess right all the time, you'd need the probability of right to be less than 5%.

.5^x = .05
x(log .5) = log(.05)
-.3x = -1.3
x = 4.3
so five correct guesses out of five chances

Note that this is not the same as five correct guesses in a row if you have more than five guesses. At some point, if you have enough guesses, you will get five in a row right. Just as you will get five heads in a row.

If you want 99% accuracy, you need
.5^x = .01
x(log .5) = log(.01)
x = 6.7
so seven correct guesses out of seven chances

For 99.9% accuracy, you need 10 correct out of 10. Another way of saying that is there is only 0.1% chance that someone could guess 10 out of 10 correctly.

As for someone who guesses right .70 of the time, since the mean is .50, the variance is .5(1-.5)n = .25n, where n is the number of trials. Standard deviation is .5sqrtn

We would want .7 to be more than 3 standard deviations from .5 before we would believe it was not random.
.2n = 3(.5sqrtn)
2n = 15sqrtn
sqrtn = 7.5
n = 56.25
So, someone would have to be 70% accurate over 57 tosses before we'd believe it's not random.
If they're 90% accurate, it would take
.4n = 3(.5sqrtn)
4n = 15 sqrtn
sqrt n = 3.75
n = 14.1
so it would take 90% accuracy or better out of 15 tosses to be 3 standard deviations from normal.
90% accuracy out of 15 means 14 right.

Answer 3

I believe the odds would remain at 50% regardless of the number of tosses, I don't think the odds multiply or divide.