What is the probability of getting "heads" on every throw when
flipping a coin seven times?
1 in ?

Babyhayley50 --

With an "unbiased" coin your odds of a single toss are 1/2.

With two tosses, it becomes 1/2*2.

With 7 tosses, it is 1/2^7 (that 2 to the 7th power) = 1/128.

Not impossible, but it's a very low percentage (less than 1% of the tries).

Best regards,

Omnivorous-GA

thank you

1 in 2^7 or 1 in 128.

50%. Either you will or you won't.

toufaroo is right it is 1/2 to the 7th power (1/128)

a more precise answear is :
when u face the last toss then u'r chances are still 50-50 but in
probablity when u want the specific complete event of having 7 tosses
which comes up heads then the 1/128 are u'r chances.

I've always been confused by this question.
You take your coin and flip it, having a 1/2 chance of it landing on heads.
You flip it again, having a 1/2 chance of it landing on heads.
So on and so forth until your 7th flip.
Now assume you have managed to get 6 heads up to this point.
Are the odds of this next flip landing on heads 1/2?? 
Within the bounds of this single flip, of course the odds are 1/2. But
within the context of the seven flip run, are the odds 1/128? (1/2^7)?
I'm beginning to suspect the answer lies in the depreciation of the
odds with each successful flip. Can someone clarify this?? My brain
hurts.

Probability can seem confusing but it isn't really - at this level
it's just about counting and ratios. A chance of something happening
is, simply, the count (or total) of what is required as a ratio of all
possible outcomes.

When you reach the 7th throw, the chance of a head on that particular
throw is 1/2, and obviously that's because, for that throw only, a
head is the one result required out of 2 possible results - and that's
always the case for any particular throw.

So, what you may require from one throw is different from what you may
require from a series of throws. If we look at a series of throws:

For two throws, the reason why HH has a probability of 1/4 isn't
because we multiply 1/2 by 1/2 - that's just a method of counting. The
1/4 represents one required result (i.e. HH) from four possible
results (i.e. HH, HT, TH and TT).

If we throw the coin three times, the possible results are:

Three heads:		1 way
Two heads and one tail:	3 ways
Two tails and one head:	3 ways
Three tails:		1 way
Total:			8 ways

So the chances of getting three heads are 1/8 (and the method for
calculating this quickly is (1/2)^3). But note that there are 3 ways
of getting two heads and a tail, so the probability of this is 3/8.

You can count in this manner all the way up to seven throws and beyond
(if you have the time and patience to do it!) so that, for seven
throws, you get the possibility of:

7H:	   	  1 way
6H and 1T:	  7 ways
5H and 2T:	 21 ways
4H and 3T:	 35 ways
3H and 4T:	 35 ways
2H and 5T:	 21 ways
1H and 6T:	  7 ways
7T:		  1 way
Total:	128 ways

Finally, the chances for seven heads are the same if you throw a
single coin seven times, or if you throw seven coins at once, or seven
people each throw a single coin, or any other combination of seven
throws. That's because (as long as each throw is fair) the number of
possible outcomes is always 128, and the number of times you can get
seven heads is always one, so the probability of seven heads will
always be 1/128.

In brief, then, the odds don't "depreciate" with the number of throws:
the number of possibilities increases, and how that effects the odds
depends on what it is you're looking for (require), e.g. if you?re
looking for two heads from two throws, the odds are lower than two
heads from three throws, or from four throws, and so on.

Oops! That last paragraph of mine should have read "...a minimum of two heads..."

Sorry.

I've always been confused by this question.
You take your coin and flip it, having a 1/2 chance of it landing on heads.
You flip it again, having a 1/2 chance of it landing on heads.
So on and so forth until your 100th flip.
Now assume you have managed to get 99 heads up to this point. What is
the probability to get another head in the 100th toss?
Theoretically, because each toss is independent, it should be again
50%. But probability theory also tells us that in the long run, the
tendency should be half heads and half tails. So, we got 99 heads
already, but it doesn't increase the chance of getting a tail next
toss. Then, how could the probability structure of half-half be
demonstrated anyway?

For the people who are confused maybe this will help:

When I flip one coin it will either be {H} or {T} heads or tails. (50%
chance of getting heads)

If I flip it again and again it will always either be {H} or {T} heads
or tails. (50% chance of getting heads)

but if I flip 2 coins (or one coin twice) the possibilities are {HH},
{TT}, {HT} or {TH} (1 in 4 chances of flipping two heads with two
coins)

now if I flip 3 coins: {HHH} {HHT} {HTH} {HTT} {THH} {THT} {TTH} {TTT}
or a 1 in 8 chance of flipping 3 heads with 3 coins.

....but what if I flip a fourth coin what are the odds of it too being heads? 
it is still going to either be {H} or {T} regardless of what the first 3 coins are.
 
So what are the odds that all 4 coins are going to heads.....there are
the above 8 combinations plus the fourth coin being heads and the
above 8 combinations plus the fourth coin being tails for a total of
16 combinations of which 1 has all heads. So a one in 16 chance of
being all heads.

In general, with each flip there will be twice as many possible
combinations, but only 1 where there is all heads so the odds of
flipping n coins all heads will be 1 in 2^n.

A little wordy but I hope it helps.