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.

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

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

- katastrofa
**Posts:**8772**Joined:****Location:**Alpha Centauri

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

- Ultraviolet
**Posts:**1655**Joined:**

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!

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:**

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:**

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:**

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.

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:**

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.

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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