How Many Flips to Get 10 Heads/Tails in a Row?

[cAn]

Ok well here goes.
what is the chance of flipping a coin 10 times and getting either 10 heads or 10 tails. Once this is known, how many times would you expect to flip a coin before you get 10 in a row(I of course don't mean that after this number you would be guaranteed to get 10 in a row, i just mean statistically).



here is what I'm thinking:
the chance of flipping a coin 10 times and getting all heads or all tails would be 1 in 2^9. Since the first of the 10 flips doesn't really matter so it is only the remaining 9 that count.
so you have a 1 in 512 chance that you will flip 10 of the same in a row.

Now, to figure out how to how many flips it should take to get this result, i used the idea that you would need 512 groups of 10 flips to achieve this.
to get 512 groups of 10 i would think that it would take 521 flips to achieve this, since:
11flips is essentially 2 chances at a group of 10(1-10, 2-11)
12 is 3 chances
13 is 4 chances
14 is 5 chances
.
.
.
521 is 512 chances.

Am I right in my logic?

 

[rachmaninoff]

Are the 512 groups you listed independent or dependent?

That is, given the first ten flips aren't enough, does the next one flip give you another chance, independent of the results of the first ten?

 

[Rogerio]

The event 'E' (flipping 10 times, and get all heads or tails) has probability p=1/2^9

The chance of 'not E' ( flipping 10 times, and get at least 1 head and 1 tail) is q=1-p

The chance of 'n' consecutive 'not E' events = q^n

When q^n < 50% , it will be more probabable you have at least one event 'E'.

Then, you need n > (ln 1/2) / (ln (1-1/2^9) )

n > ln 2 / ln(512/511) , or n > 354.5

So, n=355 , and you will need to flip the coin about 3550 times .

 

[cAn]

Are the 512 groups you listed independent or dependent?

That is, given the first ten flips aren't enough, does the next one flip give you another chance, independent of the results of the first ten?

i dunno, you tell me. I would think that since all flips are random, any group of 10 would an independent chance at 10 in a row.

 

[cAn]

The event 'E' (flipping 10 times, and get all heads or tails) has probability p=1/2^9

The chance of 'not E' ( flipping 10 times, and get at least 1 head and 1 tail) is q=1-p

The chance of 'n' consecutive 'not E' events = q^n

When q^n < 50% , it will be more probabable you have at least one event 'E'.

Then, you need n > (ln 1/2) / (ln (1-1/2^9) )

n > ln 2 / ln(512/511) , or n > 354.5

So, n=355 , and you will need to flip the coin about 3550 times .

thanks for the reply, why did you say q^n <50% was the percentage that it was probable for at least one E to occur. Also, why did you multiple n by 10 to get your final answer.
thanks

 

[Rogerio]

If q^n < 50% , then ¨The chance of one event 'E' ¨ > 50% .
This means ¨the chance of getting an event E is bigger than the chance of not getting an event E¨ .
Each event ( type 'E' and type 'not E' ) is associated to 'flipping a coin 10 times' .

 

[Chronos]

The first flip is always free, so after any given flip there is a 1 in .5^9 [1/512] chance the next nine flips will be the same. This simplifies matters. The odds of any single flip not being followed by 9 more of the same is 1-1/512 [%99.8] [it is often simpler to figure probabilities when you calculate the odds of failing, then subtract 1]. You simply take that to the power of number of flips you intend to make and check the odds. Think gaussian distributions. Rogerio had it right [I only take exception to the %100 probability, it is never %100, just too close to tell the difference]. There is a %50 chance of flipping 10 consecutive heads or tails at least once if you toss a coin 354 times. After 512 flips, the odds rise to %63.3. After 3550 flips, they rise to an impressive %99.9. In fact, after 3550 flips, there is a %50 chance you will toss 10 in a row 5 times.