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RoniNYC
Topic Author
Posts: 18
Joined: September 6th, 2011, 11:25 pm

### What is the probability to toss HH in 10 tosses?

I know that on average it takes 6 tosses (fair coin) to get 2 heads in a row... Is it possible to find the probability that we toss HH in 10 tosses? Thanks.

tagoma
Posts: 18351
Joined: February 21st, 2010, 12:58 pm

### What is the probability to toss HH in 10 tosses?

Can we have HHH, HHHH, HHHHH, .... ?Or the probability out of HHTTTTTTTT, THHTTTTTTT, TTHHTTTTTT, ... ?

karakfa
Posts: 139
Joined: May 25th, 2002, 5:05 pm

### What is the probability to toss HH in 10 tosses?

Last edited by karakfa on January 1st, 2013, 11:00 pm, edited 1 time in total.

katastrofa
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Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

### What is the probability to toss HH in 10 tosses?

where F_n is the Fibbonacci series (F_0 = 0, F_1 = 1, ...).

Ultraviolet
Posts: 1655
Joined: August 15th, 2012, 9:46 am

### What is the probability to toss HH in 10 tosses?

QuoteOriginally posted by: katastrofawhere F_n is the Fibbonacci series (F_0 = 0, F_1 = 1, ...).Yup, it has a structure of a Fibonacci sequence: P(n) = 1/2 P(n-1) + 1/2 * 1/2 P(n-2) for n >2, P(2) = 1/4, P(1) = 0+ 50 XP, katastrofa, but you've lost karma for the spelling mistake!

RoniNYC
Topic Author
Posts: 18
Joined: September 6th, 2011, 11:25 pm

### What is the probability to toss HH in 10 tosses?

QuoteOriginally posted by: UltravioletQuoteOriginally posted by: katastrofawhere F_n is the Fibbonacci series (F_0 = 0, F_1 = 1, ...).Yup, it has a structure of a Fibonacci sequence: P(n) = 1/2 P(n-1) + 1/2 * 1/2 P(n-2) for n >2, P(2) = 1/4, P(1) = 0+ 50 XP, katastrofa, but you've lost karma for the spelling mistake!Hey Ultraviolet,Thank you for the answer. Could you please explain why it's a Fibonacci sequence or perhaps direct me to a source so that I understand why it is true?Thanks@edouardI'm not sure, I guess an answer to either would be okay... it's a question I made up after doing a similar question...

Ultraviolet
Posts: 1655
Joined: August 15th, 2012, 9:46 am

### What is the probability to toss HH in 10 tosses?

Sure (katastrofa must have solved it in a similar way)It's like outrun wrote.You may want to take a look at this thread and the link to the brainteaser forum it contains, where the original problem is solved in a few different ways. In analogy to the solution using conditional expectation values as intermediate steps (to obtain a closed formula for E[n]), we can write a recursive formula for P(n) as a function of P(n-1) and P(n-2) using conditional probabilities:P(n) = 1/2 P(n|T) + 1/2 P(n|H),where P(n|X), X=T,H, is the probability of obtaining HH in n tosses conditioned on the first toss being X.We haveP(n|T) = P(n-1)andP(n|H) = 1/2 P(n|HT) + 1/2 P(n|HH) = 1/2 P(n-2) (obviously, for n > 2 P(n|HH) = 0 and P(n|HT) = P(n-2))Collecting all formulas together, we obtain P(n) = 1/2 P(n-1) + 1/2*1/2 P(n-2).

ChicagoGuy
Posts: 455
Joined: April 13th, 2007, 1:45 am

### What is the probability to toss HH in 10 tosses?

I'm not getting that answer. I think this is how you solve it:To get exactly two consecutive heads out of n tosses we have:n=2 tosses: {H, H}n=3 tosses: {H, H, T}, {T, H, H}n=4 tosses: {H, H, T, T}, {T, H, H, T}, {T, T, H, H}and so onSimilarly, to get exactly three consecutive heads out of n tosses we have:n=3 tosses: {H, H, H}n=4 tosses: {H, H, H, T}, {T, H, H, H}n=5 tosses: {H, H, H, T, T}, {T, H, H, H, T}, {T, T, H, H, H}and so onAnd for exactly four consecutive heads out of n tosses we haven=4 tosses: {H, H, H, H}n=5 tosses: {T, H, H, H, H}, {H, H, H, H, T}and so onSo to have 2 consecutive heads out of n tosses we have to add all these up2 tosses: {H, H}3 tosses: {H, H, T}, {T, H, H}, {H, H, H}4 tosses: {H, H, T, T}, {T, H, H, T}, {T, T, H, H}, {H, H, H, T}, {T, H, H, H}, {H, H, H, H}so for 2 tosses: 1/2^23 tosses: (2+1)/2^34 tosses: (3+2+1)/2^4n tosses: (n+(n-1)+...+1)/2^n=(n*(n+1))/2^(n+1)

horacioaliaga
Posts: 326
Joined: August 21st, 2005, 3:30 pm

### What is the probability to toss HH in 10 tosses?

The Probability of getting HH on the 10th toss is equal to the probability of not having it (3/4)^9 during the first 9 tosses, times the probability of having it exactly on the 10th toss (1/4):1/4 * (3/4)^9 = 1.8771%
Last edited by horacioaliaga on February 4th, 2013, 11:00 pm, edited 1 time in total.

secquant
Posts: 1
Joined: October 21st, 2016, 4:31 pm

### Re: What is the probability to toss HH in 10 tosses?

Probabilities of getting ...TT... and ...HH... out of 10 are the same, probabilites of getting HTHT...HT or THTH...TH are the same

FalsePositive
Posts: 93
Joined: March 10th, 2009, 1:12 am

### Re: What is the probability to toss HH in 10 tosses?

There are N-2+1 = 9 possible 2-streaks where HH can happen with probability 1/4.
Now we know that
probability of at least one HH = 1 - probability of no HH = 1 - (3/4)^9 = .92
The chance of getting at least one HH is high because, as you mentioned, on average one HH is expected in every 6 tosses, and now we have 10>6 tosses.

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