Tossing Coins: 5 consecutive heads out of 10 attempts?

[DannyCPA]

Homework Statement

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.

Homework Equations

Basic permutations, combinations, and counting rules.

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.

 

[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) ?

 

[DannyCPA]

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) ?

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?

 

[njama]

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?

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.

 

[DannyCPA]

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.

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.

 

[njama]

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).

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.

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.

I guess you are right. Good spot.

Now let's 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?

 

[DannyCPA]

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

Great approach, btw.

 

[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...