If you flip a coin 9 times, what are the chances you'll have at least one streak of 3 Heads or 3 Tails in a row?

He's an engineer, so he likes equations. This time his fancy binomial whatsits suggested the answer might be 86.6%, but that felt a little high to me, so I asked him to apply his equation to a set of 4 coin flips.

Since a set of 4 coin flips only has 16 possible outcomes, I was able to write them all out and ascertain there's a 6 in 16, or 37.5% chance of a '3 straight' streak. His fancypants math produced an incorrect number of 43.75%.

So I have two questions:

1. What's the correct answer to the Streak of 3 Heads or Tails in 9 coin flips question? and,

2. What's the proper equation to use to arrive at that number? (Teach a man to fish, and all that.)

Thanks again.

http://wizardofodds.com/ask-the-wizard/253/

http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html

Note that to get the probability of either 3 heads *or* 3 tails, you need to double the reported probability.

The trick while using that streak calculator repaired by BruceZ is to useQuote:hmmm23So I have two questions:

1. What's the correct answer to the Streak of 3 Heads or Tails in 9 coin flips question? and,

n-1 for n and

k-1 for k

this trick only works when p = 0.50 (1/2)

n=8 and k = 2 = 0.785156250000000

pacomartin in another post also showed this trick

I think in the link provided for the math of streaks where the calculator lives

I also have this in a simple three-state Markov Chain using Excel

A1,B1,C1 as shown D1 = 1-C1

A2: =0.5*SUM(A1:B1)

B2: =0.5*A1

C2: =C1+0.5*B1

D2: =1-C2

fill down from row 2 to how far you want.

300. Fine

ColA is probability after N flips that you are at 1 in a row

ColB is probability after N flips that you are at 2 in a row

other columns are shown

easy stuff and very clean

Look at C9.

the answer to the probability that in 9 flips of a fair coin, one sees a Head or Tail run of at least 3

there is some R code by BruceZ here for Heads or Tails Runs where p does not have to equal 0.50

http://forumserver.twoplustwo.com/25/probability/flipping-coins-getting-3-row-1233506/#post34329689

BruceZ is 'fantabulous'

a note: # length of run (must be > 1 and <= N/2)

R returns for 4 flips and either run of 3

Error in h[n] = ((1 - p) - t[n - m] - prob[n - m - 1] * (1 - p)) * p^m :

replacement has length zero

>

> # m heads OR m tails

> prob[N]

[1] 0.375

So that looks right for 4 flips

9 Flips

> # m heads OR m tails

> prob[N]

[1] 0.7851562

One can use

http://www.compileonline.com/execute_r_online.php

to run the R code below

##############################################

# Biased/unbiased recursion of heads OR tails

##############################################

N = 9 # number of flips

m = 3 # length of run (must be > 1 and <= N/2)

p = 0.5 # P(heads)

prob = rep(0,N)

h = rep(0,N)

t = rep(0,N)

h[m] = p^m

t[m] = (1-p)^m

prob[m] = h[m] + t[m]

for (n in (m+1):(2*m-1)) {

h[n] = (1-p)*p^m

t[n] = p*(1-p)^m

prob[n] = prob[n-1] + h[n] + t[n]

}

for (n in (2*m):N) {

h[n] = ((1-p) - t[n-m] - prob[n-m-1]*(1-p))*p^m

t[n] = (p - h[n-m] - prob[n-m-1]*p)*(1-p)^m

prob[n] = prob[n-1] + h[n] + t[n]

}

# m heads OR m tails

prob[N]

There are recursive methods and Markov chain methods.Quote:hmmm232. What's the proper equation to use to arrive at that number? (Teach a man to fish, and all that.)

Thanks again.

a bit more challenging than just adding a few #s together

Good Luck and have some fun

added: IMO best reading on coin flips is

The Coin Toss: The Hydrogen Atom of Probability

Stefan Hollos, J. Richard Hollos

https://play.google.com/store/books/details?id=5ouN9QhFvYIC

http://www.abrazol.com/books/cointoss/

The Coin Toss: Probabilities and Patterns

Summary of coin toss probability formulas (pdf, 8 pages) in this book.

http://www.abrazol.com/books/cointoss/cointossruns.pdf

added:

markov chain matrix here

raised to the 9th power (the 9th coin flip)

fractions

Quote:7crapsThe trick while using that streak calculator repaired by BruceZ is to use

n-1 for n and

k-1 for k

this trick only works when p = 0.50 (1/2)

n=8 and k = 2 = 0.785156250000000

Wow, thanks very much to everyone who replied, especially 7craps, who definitely gave of his time and brain power. I'd also like to thank Tringlomane, and MathExtremist, who's helped me on some other things, as well.

You guys rock!

Sweet, now I get to surprise my buddy by beating him to the punch in his own area of strength (cue evil laughter), so thanks again.

Many are excellent at math but yet can stumble with probability. They are in the majority.Quote:hmmm23Sweet, now I get to surprise my buddy by beating him to the punch in his own area of strength (cue evil laughter), so thanks again.

Dr. Maths

"In the short run, chance may seem to be volatile and unfair.

Considering the misconceptions, inconsistencies, paradoxes and

counter intuitive aspects of probability, it is not a surprise that as

a civilization it has taken us a long time to develop some methods

to deal with this."

It gets easier when we first deal with the sample space.

added: for your 9 flip example the exact answer is

402/512

512 = 2^9 (this should be intuitive)

The challenge (and fun IMO) is the 402

The exact formula in this case

(a run of at least 3 Heads or Tails in 9 flips) is

(2^9) - (2*(F2[9])) / (2^9)

where F2[9] is the 9th Fibonacci 2-step number.

(For a run of 3 we use k-1. There is proof of this on the net. I do not have a link handy)

The regular Fibonacci numbers are obtained by summing the previous 2 numbers (0,1,1,2,3,5,8,13,21,34,55...).

The First F#

F2[1]= 1 (0+1) (some do not use the 0, I do, in that case use the 8th F#)

F2[2]= 2 (1+1) and so on to the 9th F#

F2[9]=55

You want a run of Heads or Tails we multiply 55*2 = 110

This is the # of sequences in the 512 sample space that do NOT have a run of at least 3.

That must leave 512-110 that do (402)

Hey, it worked.

This method can get crazy and very unstable for large N so I use it for flips of 100 or less in a spreadsheet)

of course, I do check my answers with a sim or two

I used a BruceZ R code

##################################

# Biased/unbiased simulation of

# Heads OR Tails

# Heads AND Tails

# Neither heads NOR tails

# Code by BruceZ

##################################

N = 9 # number of flips

m = 3 # length of run

p = 0.50 # P(heads)

sims = 0

count1 = 0

count2 = 0

while (1) {

flips = ifelse(runif(N,0,1) < p, "H", "T")

runs = rle(flips)

heads = runs$lengths[runs$values == "H"] >= m

tails = runs$lengths[runs$values == "T"] >= m

if ( sum(heads) | sum(tails) ) {count1 = count1 + 1}

if ( sum(heads) & sum(tails) ) {count2 = count2 + 1}

sims = sims + 1

}

p1 = count1/sims

error1 = 3.29*sqrt(p1*(1-p1)/sims)

p2 = count2/sims

error2 = 3.29*sqrt(p2*(1-p2)/sims)

# m heads OR m tails

p1

p1 - error1

p1 + error1

# m heads AND m tails

p2

p2 - error2

p2 + error2

# neither m heads NOR m tails

1 - p1

1 - p1 - error1

1 - p1 + error1

sims

results

> # m heads OR m tails

> p1

[1] 0.7845934

> p1 - error1

[1] 0.7828374

> p1 + error1

[1] 0.7863494

> # m heads AND m tails

>>> this is a bit more advanced, but still can be done with Fibonacci numbers (answer 74/512 or 0.14453125)

> p2

[1] 0.1442022

> p2 - error2

[1] 0.1427017

> p2 + error2

[1] 0.1457027

> # neither m heads NOR m tails

> 1 - p1

[1] 0.2154066

> 1 - p1 - error1

[1] 0.2136506

> 1 - p1 + error1

[1] 0.2171626

> sims

[1] 593278

>

Good Luck

I gots to disagree here and add some.Quote:MathExtremistNote that to get the probability of either 3 heads *or* 3 tails,

you need to double the reported probability.

Maybe you were in a hurry to finish typing on your way out with your wife and family for an extended weekend?

BruceZ

"The probability of getting 3 heads OR 3 tails in a row in 14 flips is

the same as the probability of getting 2 heads in a row in 13 flips.

To see this, think about a sequence of changes rather than a sequence of heads and tails.

We need 2 consecutive no changes.

There are only 13 trials because we don't care about the result of the first trial."

Either/Or would be using the streak calculator again

n = n-1 (8)

k = k-1 (2)

We get exactly 402/512 or 0.78515625

as agrees with a previous post

Now if we are looking for both Head AND Tail runs of at least 3 in length,

ME is correct that we start by adding both together.

OP Question is 9 flips and both H and T run of 3

0.46484375 + 0.46484375 = 0.9296875

We should know this horribly over-counts the successful sequences

and need to use the inclusion-exclusion principle to remove that over-count

P(3 H) + P(3 T) - P(3 H or 3 T)

0.9296875 - 0.78515625 = 0.14453125 (or 74/512)

Looks like my sim data is close to this

(and I used Excel to list all 512 sequences and counted exactly 74 of them that did indeed have both runs)

and the markov chain method

p=

v=[1,0,0,0,0,0,0,0,0,0,0,0,0]

v*p^9 =

0, 17/256, 21/512, 47/256, 47/512, 23/512, 17/256, 21/512, 47/256, 47/512, 23/512, 37/512, 37/512

yep (37/512 + 37/512)

p^9 =

This thread really has enough info in it for one to become a world class expert on

everything you want to know about a coin toss and more.

Pool is open for the weekend

Quote:7crapsThere are recursive methods and Markov chain methods.

a bit more challenging than just adding a few #s together

I tried calculating a non-recursive formula using matrix algebra.

P(3 in a row) = P(3 heads in a row) + P(3 tails in a row).

Since every streak has a "partner streak" where the Hs and Ts are switched (e.g. HHTH <--> TTHT), P(3 in a row) = 2 x P(3 heads in a row).

Let P(n) be the probability of getting 3 heads in a row in n tosses.

P(n) = 1/8 [for HHH] + 1/8 x P(n-3) [for HHT] + 1/4 x P(n-2) [for HT] + 1/2 x P(n-1) [for T].

This is rewritten as:

1/8 = -P(n) + 1/2 P(n-1) + 1/4 P(n-2) + 1/8 P(n-3)

Also, when calculating P(n-1):

1/8 = -P(n-1) + 1/2 P(n-2) + 1/4 P(n-3) + 1/8 P(n-4)

Thus, -P(n) + 1/2 P(n-1) + 1/4 P(n-2) + 1/8 P(n-3) = -P(n-1) + 1/2 P(n-2) + 1/4 P(n-3) + 1/8 P(n-4)

which becomes, P(n) = 3/2 P(n-1) - 1/4 P(n-2) - 1/8 P(n-3) - 1/8 P(n-4)

However, the eigenvalues of the recursion matrix are the roots of a 4th-degree polynomial, so I stopped there...