Coin Flip Problem

If you flip a fair coin 40 times, what is the probability of getting BOTH a streak of 5 heads and a streak of 5 tails?

“A streak” meaning at least 1 and “5” meaning at least 5. The head and tail streaks do not have to be adjacent.

Integer answers only.

The probability of a streak of at least 5 consecutive successes in 40 trials is 46.79%

Quote: GMan

The probability of a streak of at least 5 consecutive successes in 40 trials is 46.79%

I disagree with the answer and the non-integer format

Quote: Wizard

Wiz answer

111,898,752,105 / 549,755,813,888 = apx. 0.203543

That’s reasonably close, but I disagree. My answer may be wrong, but I’m fairly confident it’s not. I confirmed it with a simulation

What do you mean by "BOTH a streak of 5 heads and a streak of 5 tails?" ?

Could you please give some examples that comply and some example that not comply to the above rules ?

Quote: ssho88

What do you mean by "BOTH a streak of 5 heads and a streak of 5 tails?" ?

Could you please give some examples that comply and some example that not comply to the above rules ?

In 40 flips you must see at least one instance of 5 or more consecutive heads and also at least one instance of 5 or more consecutive tails. Does that make it clearer?

Listing 40 character examples isn’t very practical.

Prob = [ 2* ( 40!/31!/1! + 40!/30!/2! + 40!/29!/3! + 40!/28!/4! + 40!/27!/5! + 40!/26!/6! + 40!/25!/7! + 40!/24!/8! + 40!/23!/9! + 40!/22!/10! + 40!/21!/11! + 40!/20!/12! + 40!/19!/13! + 40!/18!/14! + 40!/17!/15! ) + 40!/16!/16! ] * 0.5^40. Please read the [ and ( carefully.

To simply above, Prob = [ 2 * Σ(40!/(32-x)!/x!) + 40!/16!/16! ]* 0.5^40, where x varies from 1 to 15, Σ is summation symbol

Is it correct ? I am not sure.


Edited.

Sorry, I don't think my above answer is correct.

Quote: ssho88

Is it correct ? I am not sure.

Is what correct?

Quote: Ace2

]That’s reasonably close, but I disagree. My answer may be wrong, but I’m fairly confident it’s not. I confirmed it with a simulation

Quote: Wizard

Correction

I made a mistake. My new answer is 107,094,548,225 / 549,755,813,888 = apx. 0.194804

Through combinations analysis or simulation ?

Quote: Ace2

I disagree with the answer and the non-integer format

Results through a simulation(10 million trials), prob = 19.489%.

Still thinking how to solve it arithmetically.

Quote: Wizard

Correction

I made a mistake. My new answer is 107,094,548,225 / 549,755,813,888 = apx. 0.194804

That is the correct answer. Please show your method

Quote: Ace2

Please show your method

Thanks, Wizard. I didn't realize you could get to an integer solution for this problem from a Markov chain. I found another way to solve this.

In 40 flips (2^40 = 1,099,511,627,776 permutations) we know that the number of strings with no streaks of 5+ heads is given by the (40 +2) 42nd "Pentanacci" number which is 585,029,621,920. Pentanacci meaning the Fibonacci sequence of the 5th order, where the previous 5 (not 2) terms are summed. Subtract that from total permutations to get 514,482,005,856 strings where there is at least 1 streak of 5+ heads. You can say the same thing about tails.

I found a pattern for strings with no streaks of 5+ heads OR tails. This number of strings is obtained by taking the (40+1) 41st "Tetranacci" number which is 142,368,356,257 and multiplying by 2 to get 284,736,712,514. Tetranacci meaning the Fibonacci sequence of the (5 - 1) 4th order, where the previous 4 terms are summed. Subtract that from total permutations to get 814,774,915,262 strings where there is at least 1 streak of 5+ heads OR tails.

Since we know 514,482,005,856 strings have at least 1 streak of 5+ heads and 514,482,005,856 strings have at least 1 streak of 5+ tails, sum those two figures to get 1,028,964,011,712 strings where there is at least 1 streak of 5+ heads OR tails…however the strings with BOTH are double counted here.

Therefore, 1,028,964,011,712 minus 814,774,915,262 (no double counting) equals 214,189,096,450 strings out of 1,099,511,627,776 (19.48%) with BOTH streaks.

Ace2, what is your math background? My initial guess would have been a much different method, but I trust your answer over mine.

Here is my solution in PDF form.