# Math Help - Consecutive Heads/Tails in weighted coin toss

1. ## Consecutive Heads/Tails in weighted coin toss

Hi,

Can someone please help with this. I keep getting different answers to this from different people.

Supposing you have a weighted coin which lands heads up 70% of the time and tails up 30% of the time.

Q. What is the probability that it lands on tails 20 or more times in a row if I did this 1000 times?

*Also could I have the equation on how you worked this out so that I can change the weighting, number of trials etc (I need to know this for my business!!!)

Thank you very much.

Zak

2. Originally Posted by slyone
Hi,

Can someone please help with this. I keep getting different answers to this from different people.

Supposing you have a weighted coin which lands heads up 70% of the time and tails up 30% of the time.

Q. What is the probability that it lands on tails 20 or more times in a row if I did this 1000 times?

*Also could I have the equation on how you worked this out so that I can change the weighting, number of trials etc (I need to know this for my business!!!)

Thank you very much.

Zak
Hi Zak,

I don't know how to compute the probability you seek exactly, but I think I can give a pretty good approximation.

To generalize the problem, let's suppose you have a possibly biased coin which comes up tails with probability p and heads with probability q, where p+q = 1. You flip the coin N times and look for a run of R or more successive tails. We would like to know the probability that at least one such run occurs. The "at least one" part is my interpretation of your problem statement; you didn't say that, but I'm guessing that's what you mean.

As I said, I haven't come up with a reasonable way to compute the exact probability of having at least one run of length R, but think I can solve a closely associated problem which can be used to obtain an approximate answer.

Let $\lambda$ be the expected number of runs of R or more successive tails. It can be shown that

$\lambda = p^R \,[(N - R) q + 1]$.

That is an exact result, not an approximation. (I derived this formula from scratch, but I suppose it must be a well-known result.) So in your case, for example, we have N = 1000, R = 20, p = 0.3 and q = 0.7, so $\lambda = 2.4 * 10^{-8}$.

Now comes the approximation. Let's suppose the number of runs, say X, has a Poisson distribution with mean $\lambda$. It doesn't, but for small values of $\lambda$ it should be pretty close. Then

$Pr(X > 0) = 1 - Pr(X = 0) = 1 - e^{-\lambda}$.

For small values of $\lambda$ this value can be hard to compute precisely, but we can use another approximation:

$1 - e^{-\lambda} = \lambda$ (approximately),

so the answer to your question is (once again, approximately) $\lambda = 2.4 * 10^{-8}$.

3. ## Re: Consecutive Heads/Tails in weighted coin toss

Hi, great answer from awkward. But how did you figure out the expected number of runs of R or more successive tails:
\lambda = p^R \,[(N - R) q + 1].
You derived this formula from scratch. I tried to figure by myself this result and tried to Google it and did not find an answer.
Could you share your Derivation ?
Thanks
Otto

4. ## Re: Consecutive Heads/Tails in weighted coin toss

The chance of getting 20 tails in a row in 20 tosses is 0.320. When tossing a coin 1000 times there is are (1000-20+1)=981 times where you can have 20 of an outcome in a row. Since it is 20 or more we don't care what the other 980 tosses are.
Since the chance of getting 20 in a row in one attempt is 0.320 and you get 981 attempts the chance of getting 20 in a row through the whole 1000 tosses is 981(0.320)

5. ## Re: Consecutive Heads/Tails in weighted coin toss

Originally Posted by ottof
Hi, great answer from awkward. But how did you figure out the expected number of runs of R or more successive tails:
$\lambda = p^R \,[(N - R) q + 1]$
You derived this formula from scratch. I tried to figure by myself this result and tried to Google it and did not find an answer.
Could you share your Derivation ?
Thanks
Otto
Let
$X_i = \begin{cases} 1 &\text{if there is a run of at least R tails starting with flip i}\\0 &\text{otherwise} \end{cases}$
Then $X_1 = 1$ if the first R flips are tails,
but for $i > 1$, $X_i = 1$ if the (i-1)st flip is a head and the next R flips are tails.

So
$E(X_1)=\Pr(X_1 = 1) = p^R$
and
$E(X_i)=\Pr(X_i = 1) = q p^R$ for $i > 1$

Applying the theorem that E(X+Y) = E(X) + E(Y), we have

$E \left( \sum_{i=0}^{N-R+1} X_i \right) = \sum_{i=0}^{N-R+1} E(X_i) = p^R + (N-R)q p^R = p^R [(N-R)q + 1]$