If I toss a coin once, the probability of a “heads” is 50%
What is the probability of a “heads” if I toss it twice?
Three times?

How is this calculated?

pesce.del.giorno said:

If I toss a coin once, the probability of a “heads” is 50%
What is the probability of a “heads” if I toss it twice?
Three times?

How is this calculated?

do you mean the probability of heads coming up twice in two coin tosses?

If you are looking at each coin flip as an independant event, the probability of it coming up heads each time is 50% (well, just a little under)

No, I mean what is the probability of throwing at least one head in two throws.
I seem to recall that there is an equation to solve this. Might have been Poisson. Can’t remember.

pesce.del.giorno said:

No, I mean what is the probability of throwing at least one head in two throws.
I seem to recall that there is an equation to solve this. Might have been Poisson. Can’t remember.

for two throws
1 – probability of two tails = 0.75

For three throws
1 – probability of 3 tails = 0.875

and so on.

pesce.del.giorno said:

No, I mean what is the probability of throwing at least one head in two throws.
I seem to recall that there is an equation to solve this. Might have been Poisson. Can’t remember.

take the number of possible outcomes (in this case 2, heads or tails) tothe power of the number of events.

If you are looking at only heads coming up, there is only 1 of these outcomes that shows all heads, therefore

2 heads is 25%
3 heads is 12.5%

(roughly as there is still a chance the coin will land on edge)

yeah, Rev has it right, 1-(what i said) for the probability of at least 1 head coming up, not all heads coming up

Yes, thanks Rev. I think that’s it.

So what I’m working towards is this. Many clinical studies are quoted with a P level of 0.05. (i.e. the probability that the observed outcome is due to chance is 5%, or 1 in 20.) If there are 20 such studies, what is the likelihood that at least one of them will show a fallacious outcome due to chance?

pesce.del.giorno said:

Yes, thanks Rev. I think that’s it.

So what I’m working towards is this. Many clinical studies are quoted with a P level of 0.05. (i.e. the probability that the observed outcome is due to chance is 5%, or 1 in 20.) If there are 20 such studies, what is the likelihood that at least one of them will show a fallacious outcome due to chance?

That’s just like rolling 20 20-sided dice. Let’s say side ‘1’ of each die represents that the observed outcome is due to chance. So we want to know the probability that at least 1 die shows a ‘1’. Which is 1 – (the probability that no die shows a ‘1’).

So p = 1 – (0.95)^20
~= 0.6415 or 64.15%

FWIW,
lim n → ∞ of (1 – 1/n)^n = 1/e = 0.36787944…, and this converges fairly quickly.

So if the probability of a single test result being due to chance is 1/n and we conduct n independent tests, the probability that at least one of our results is due to chance is roughly 1 – 1/e = 0.63212, for sufficiently large n.

PM 2Ring said:

pesce.del.giorno said:

Yes, thanks Rev. I think that’s it.

So what I’m working towards is this. Many clinical studies are quoted with a P level of 0.05. (i.e. the probability that the observed outcome is due to chance is 5%, or 1 in 20.) If there are 20 such studies, what is the likelihood that at least one of them will show a fallacious outcome due to chance?

That’s just like rolling 20 20-sided dice. Let’s say side ‘1’ of each die represents that the observed outcome is due to chance. So we want to know the probability that at least 1 die shows a ‘1’. Which is 1 – (the probability that no die shows a ‘1’).

So p = 1 – (0.95)^20
~= 0.6415 or 64.15%

Thanks for that. It’s simpler than I thought, when you put it that way.

No worries, Mr Fish.

What’s the probability that it will be one of the ones that is published?

SCIENCE said:

What’s the probability that it will be one of the ones that is published?

Almost certain. :)