Feller's coin-tossing constants - Wikipedia, the free encyclopedia
In the above 2 links, it's mentioned that if we toss a fair coin ten times then the exact probability that NO pair of heads come up in succession (i.e. n = 10 and k = 2) is p(10,2) = = 0.140625.
(1) Does it mean that the probability of 2 consecutive heads coming up in 10 coin toss
= 1-0.14 = 0.86?
(2) If so, is the probability of this condition(i.e. 2 consecutive heads coming up in 10 coin toss =1 cycle) appearing 4 consecutive cycles = 0.86*0.86*0.86*0.86 = 0.54 ?
So, how do I calculate the probability of 2consecutive heads in 10 tosses? Is it 0.86 - probability of (all+more than 2 consecutive) heads =?
Or, to re-phrase, does this => 0.86 is the probability of At Least 2 consecutive heads in 10tosses?
Any clues to the answer for (2)?
I thought that I gave you a definitive answer to your question.
The probability that there are no consecutive heads in ten tosses is .
Thus, the probability of at least one pair of heads( two consecutive heads) is .
What is your problem with that reply.
I understand the answer to (1) is the converse of none or at least one, so to re-phrase question (2) ->
(2) If so, is the probability of this condition(i.e. Let 2 consecutive heads coming up at least once in 10 coin toss =1 cycle) appearing in 4 cycles consecutively = 0.86*0.86*0.86*0.86 = 0.54 ? Or how should it be calculated?