Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 19
  • Posts: 560
April 20th, 2020 at 2:18:51 PM permalink
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?

ďA streakĒ meaning at least 1 and ď5Ē meaning at least 5. The head and tail streaks do not have to be adjacent.

Integer answers only.
Last edited by: Ace2 on Apr 20, 2020
Itís all about making that GTA
GMan
GMan
Joined: Jan 5, 2013
  • Threads: 1
  • Posts: 12
April 20th, 2020 at 2:49:01 PM permalink
The probability of a streak of at least 5 consecutive successes in 40 trials is 46.79%
G Man
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 19
  • Posts: 560
April 20th, 2020 at 2:59:46 PM permalink
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
Itís all about making that GTA
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1317
  • Posts: 21598
April 20th, 2020 at 4:06:11 PM permalink

111,898,752,105 / 549,755,813,888 = apx. 0.203543
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 19
  • Posts: 560
April 20th, 2020 at 5:00:54 PM permalink
Quote: Wizard


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

Thatís reasonably close, but I disagree. My answer may be wrong, but Iím fairly confident itís not. I confirmed it with a simulation
Itís all about making that GTA
ssho88
ssho88
Joined: Oct 16, 2011
  • Threads: 41
  • Posts: 427
April 20th, 2020 at 5:52:35 PM permalink
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 ?
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 19
  • Posts: 560
April 20th, 2020 at 6:00:43 PM permalink
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 isnít very practical.
Itís all about making that GTA
ssho88
ssho88
Joined: Oct 16, 2011
  • Threads: 41
  • Posts: 427
April 20th, 2020 at 6:46:50 PM permalink
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.
Last edited by: ssho88 on Apr 20, 2020
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 19
  • Posts: 560
April 20th, 2020 at 6:50:45 PM permalink
Quote: ssho88


Is it correct ? I am not sure.

Is what correct?
Last edited by: Ace2 on Apr 20, 2020
Itís all about making that GTA
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1317
  • Posts: 21598
April 20th, 2020 at 9:18:57 PM permalink
Quote: Ace2

]Thatís reasonably close, but I disagree. My answer may be wrong, but Iím fairly confident itís not. I confirmed it with a simulation




I made a mistake. My new answer is 107,094,548,225 / 549,755,813,888 = apx. 0.194804
Last edited by: Wizard on Apr 20, 2020
It's not whether you win or lose; it's whether or not you had a good bet.

  • Jump to: