Tossing Coins: 5 consecutive heads out of 10 attempts?

1. Sep 11, 2009

DannyCPA

1. The problem statement, all variables and given/known data

Suppose you will toss a coin 10 times.

A) Compute the probability of event A that the longest run of consecutive Heads has length 5 (i.e., within your 10 tosses there is a run of 5 Heads in a row, but not 6 consecutive Heads.)

B) Compute the probability of the event B that the longest run of consecutive Heads has length 5 or more.

2. Relevant equations

Basic permutations, combinations, and counting rules.

3. The attempt at a solution

Okay, so, I am pretty sure I have solved this in a very long and cumbersome way. There has to be a simpler way to do it.

My approach: I essentially counted the "possible" outcomes. For instance, one example that satisfies P(A) = H,H,H,H,H,T,2,2,2,2. Thus, there are 24 possible outcomes for that specific sequence/position that satisfies A. I continued from there, moving to the right, placing a "T" before and after my sequence of 5 consecutive heads to act as a placeholder allowing me to compute the number of ways such a sequence, at such a position, could occur.

I did a similar approach for B. Suprisingly enough, my results were a probability of.0625 for both of them.

My question is this: Is there a better way than to sit there and try to chart out the possible placements and outcomes?

What makes this hard is the consecutive placement.

I have taken Calculus I, Calculus II, and various other math classes below it, however, this is my first serious "Probability" class.

Thanks in advance for any help.

2. Sep 11, 2009

njama

Welcome to Physics Forums!

Here is my approach:

A) You got . . . THHHHHT

Let a=THHHHHT (because there must be 5 consecutive heads, tails must be from the begging and the end to avoid 6 or 7 consecutive heads)

Now you got . . .a

A dot . means that there is open space for T or H. Do you figure it out now?

Does this gave you idea for B) ?

Last edited: Sep 12, 2009
3. Sep 12, 2009

DannyCPA

Hmm, your method pretty much is the same as what I utilized to approach this, less different signifiers.

Where I used "2," you used a "." This method does seem to work, but I refuse to believe there is no other way other than literally "counting" the possibilities. Combinations/permutations have to be utilized, but how do these two take into account consecutive streaks?

4. Sep 12, 2009

njama

You literally do not count anything.

You got a . . . where . could be H or T

You can have a . . . , . a . . , . . a ., . . . a

And that is $4*P_2^3=2^5$
I don't find this method messy.

Last edited: Sep 12, 2009
5. Sep 12, 2009

DannyCPA

Njama,

I don't see where the permutation comes in. By your definition, "a..." represents
= "THHHHHT222," where 2 is the number of possible outcomes at that position that satisfies this sequence, correct?

If so, isn't that 23, which is not P(3,2).

And lastly, what about the possibility of "HHHHHT2222," or what I assign, the 24. Your DPS (discrete probability space) does not capture that possibility, nor the one of "2222THHHHH," or the other edge.

Anyways, I appreciate your effort to assist me, and I believe we are getting closer to computing this tedious problem.

6. Sep 12, 2009

njama

Ok I assume we are using different notations.

$$P_n^k=n^k$$

We got n=2 (T, H) and k (three places) of putting T and H with repetitive manner.

I guess you are right. Good spot.

Now lets suppose a=HHHHH

a . . . . .

Here H can't be on second place so

aT. . . .

H can't be on first or third either

TaT. . .

and so on....

. TaT. .

. . .TaT

. . . . Ta

Am I right now, or I missed something?

7. Sep 12, 2009

DannyCPA

You are missing ..TaT., I believe. Thus, computing this should lead to (after simplification): 26 / 210 = 1 / 24 = .0625

Great approach, btw.

Last edited: Sep 13, 2009
8. Sep 12, 2009

njama

$$4*2^3+2*2^4=2^5+2^5=2^6$$

$$2^6 / 2^{10} = 1/2^4 =.0625$$

I guess you are right.

For B) is much harder.

Consider HHHHH..... 25

THHHHH.... 24

.THHHHH... 24

and so on...

Last edited: Sep 12, 2009