Probability of getting a streak

[Biosyn]

Homework Statement

If a coin is tossed 200 times, what is the probability of getting 5 heads in a row?

Homework Equations

The Attempt at a Solution

I know that the probability of getting 5 heads in a row for the first 4 flips is 0. I don't know what to do with that information.

1 - [(0.5)^5 * (0.5)^195] ?

I used this streak simulator:
http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html#ev-runs

And the probability is 0.965903381259682.

 

[haruspex]

Are you familiar with Markov chains?

 

[Biosyn]

No, I am not.

We are starting binomial distributions in our stats class this week.

 

[haruspex]

Then this rather an advanced problem. See e.g. http://math.stackexchange.com/quest...ngth-of-the-longest-run-in-n-bernoulli-trials, http://mathworld.wolfram.com/Run.html.
You can get a crude lower bound by chopping the 200 into 22 blocks of 9 (and 2 left over). (9, because 9 = 5*2-1, so there cannot be two separate runs of 5 in the same block.) This allows us to break it into 5 disjoint cases:
- HHHHHxxxx
- THHHHHxxx
- xTHHHHHxx
- xxTHHHHHx
- xxxTHHHHH
for a tot prob of 3/32.
The prob that none of the 22 blocks have such a run is therefore (1-3/32)²² ≈ 0.11. But 0.89 is some way short of the 0.96 you found.

 

[Biosyn]

Then this rather an advanced problem. See e.g. http://math.stackexchange.com/quest...ngth-of-the-longest-run-in-n-bernoulli-trials, http://mathworld.wolfram.com/Run.html.
You can get a crude lower bound by chopping the 200 into 22 blocks of 9 (and 2 left over). (9, because 9 = 5*2-1, so there cannot be two separate runs of 5 in the same block.) This allows us to break it into 5 disjoint cases:
- HHHHHxxxx
- THHHHHxxx
- xTHHHHHxx
- xxTHHHHHx
- xxxTHHHHH
for a tot prob of 3/32.
The prob that none of the 22 blocks have such a run is therefore (1-3/32)²² ≈ 0.11. But 0.89 is some way short of the 0.96 you found.

Oh you posted this before I was going to edit my first post.
I asked this question earlier today and someone answered it. Would you mind explaining his solution to me?
http://answers.yahoo.com/question/i...BRxA.2Lty6IX;_ylv=3?qid=20121217171124AAPB2E5

Okay - this is actually a REALLY complex probability problem. It involves state-based probabilities. Here's a whole math paper dedicated to the topic:

http://www.askamathematician.com/2010/07…

First of all, the probability of having 5 heads in a row after 1, 2, 3, or 4 flips is 0. Moving forward, you need to look at the probability of being in a certain "state", based on how many heads have been flipped so far. Then you can calculate the probability. In "Steady State", the probability of a preceding run of 0 heads is:

P(0) = 1/2
P(1) = 1/4
P(2) = 1/8
P(3) = 1/16
P(4) = 1/32
...

So - from flip 5 through flip 200 is a total of 196 flips. And the probability of NOT getting 5 heads in a row will be approximately (1 - (1/32) (1/2))^196 = 4.56%

What is (1/32)(1/2).
Is this the probability of getting 5 heads for the first 5 flips multiplied by the probability of getting a head the next flip?

 

[haruspex]

Oh you posted this before I was going to edit my first post.
I asked this question earlier today and someone answered it. Would you mind explaining his solution to me?

It considers the last 5 toss outcomes at any point. If you've not had a run of 5Hs yet, the mutually exclusive possibilities can be categorised as:
xxxxT
xxxTH
xxTHH
xTHHH
THHHH
the probs being 1/2, 1/4, 1/8..., 1/32. It treats this as a 'steady state', i.e., so long as you've still not had a run of 5Hs this always represents the current state. (It's clearly not quite right because it does not add up to 1.)
The prob of completing a run of 5Hs at the next toss is therefore 1/64. If we do not complete such a run, either because we toss a T or were not yet in the '4H' state, the assumption is that we remain in the 'steady state' distribution. So the prob of getting no run is (1-1/64)¹⁹⁶ ≈ 4.6%

 

[Biosyn]

It considers the last 5 toss outcomes at any point. If you've not had a run of 5Hs yet, the mutually exclusive possibilities can be categorised as:
xxxxT
xxxTH
xxTHH
xTHHH
THHHH
the probs being 1/2, 1/4, 1/8..., 1/32. It treats this as a 'steady state', i.e., so long as you've still not had a run of 5Hs this always represents the current state. (It's clearly not quite right because it does not add up to 1.)
The prob of completing a run of 5Hs at the next toss is therefore 1/64. If we do not complete such a run, either because we toss a T or were not yet in the '4H' state, the assumption is that we remain in the 'steady state' distribution. So the prob of getting no run is (1-1/64)¹⁹⁶ ≈ 4.6%

Ah I see. Thank you.

[STRIKE]
So let me summarize what I get out of this. And then would you correct me?

If we have not completed a run of 5H, then it can be assumed that we remain in 'steady state' distribution meaning it can be either of those mutually exclusive categories (P(0)-P(4)).

And 1/64 is the probability of completing a run. So to find out the probability of not completing a run, we subtract 1/64 from 1. And take it to the power of 196 because we subtract the first 4 flips which have a probability of 0 of flipping 5 heads consecutively.
[/STRIKE]I'm confused here:

xxxxT_ The probability of getting a head next: 1/2
xxxTH_ The probability of getting a head next: 1/4
xxTHH_ The probability of getting a head next: 1/8
xTHHH_ The probabilit of getting head next: 1/16
THHHH_ The probability of getting a head next: 1/32

And the probability of getting a run of 5H is 1/64.

But isn't the probability of getting 5 heads .5^5? Or is it because you flipped a Tails first.

 

[haruspex]

I'm confused here:

xxxxT_ The probability of getting a head next: 1/2
xxxTH_ The probability of getting a head next: 1/4
xxTHH_ The probability of getting a head next: 1/8
xTHHH_ The probabilit of getting head next: 1/16
THHHH_ The probability of getting a head next: 1/32

No, those are the probabilities of being in those states. In each case, the prob of a head next is 1/2.