ordering Probabilty in coin tosses

#1
Hi Guys,

First post here so please accept my apologies if this question was posted before.

This is baffling me for quite a few days now and I can't figure out a way to calculate it.
What I know: Successive tosses are independent; knowing we got H on the first toss does not help us predict the outcome of the second toss

Each trial has the same probability P(H)=1=2=P(T)

What I can't figure out is this question: Knowing each Time you toss a coin it is independent of the previous toss, lets say you toss the coin 5 times and it comes (For Example: HTTTH) what is the probability of I toss the coin 5 times again and it comes exactly HTTTH ? what about 6 times? what about 7? so on and so forth... I believe there is some sort of factorial involved in this but I am not 100% sure.

a formula with some explanation would be greatly appreciated as I can't wrap my head around this... seems very simple but I am not too good at this.
 

Dason

Ambassador to the humans
#2
The previous tosses don't have any impact on the probability. The probability that the next five tosses come up exactly HTTTH is the same regardless of if the previous five tosses were HTTTH or not.
 
#3
Thanks for your quick response.

I did not quite understood the answer. so you mean the chance of having the second round of coin tosses coming exactly HTTTH is %50?

every coin toss is the same chance of the previous coin toss, but every new round of coin toss (lets say 5) to me looks like have less chance of being exactly the same as the previous round of 5? I want to know that probability...
 

rogojel

TS Contributor
#4
hi,
we have 2^6 different patterns from HHHHHH to TTTTTT. Each combibation is equally probable so the probability of getting your pattern is 1/2^6
regards