# Thread: Is the probability of 2 consecutive heads = 1 - probability(otherwise) ?

1. ## Is the probability of 2 consecutive heads = 1 - probability(otherwise) ?

Coin Flips
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 ?

2. Originally Posted by scalpmaster
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?
We must be careful in answering this. But I think the answer is no to both questions.

Why careful? $1-0.140625$ is the probability that heads appear in succession at least once in ten tosses.
You understand that includes the possibility of all heads.

3. 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)?

4. I thought that I gave you a definitive answer to your question.
The probability that there are no consecutive heads in ten tosses is $0.140625$.

Thus, the probability of at least one pair of heads( two consecutive heads) is $1-0.140625$.

What is your problem with that reply.

5. Originally Posted by Plato
I thought that I gave you a definitive answer to your question.
The probability that there are no consecutive heads in ten tosses is $0.140625$.
Thus, the probability of at least one pair of heads( two consecutive heads) is $1-0.140625$.
What is your problem with that reply.
Your reply to ques(2) is no also, but why?

6. Originally Posted by scalpmaster
Your reply is no to (2) also, but why?
That response makes absolutely no sense whatsoever.
The converse of none is at least one.
I don’t think you understand the question.

7. 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?

8. Originally Posted by scalpmaster
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?
I have absorptivity no idea what the sentence could mean.
Can you put into intelligible English?

9. Originally Posted by scalpmaster
(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?
The probability of at least 2 consecutive heads in one or more trials of 10 tosses is $1-0.140625^4$

CB

10. Originally Posted by scalpmaster
(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?
This is the probability of at least 2 consecutive heads in each of the four trials of 10 tosses.

The probability of at least 2 consecutive heads in one or more of 4 trials of 10 tosses is $1-0.140625^4$

CB

11. Thanks CaptainBlack, got it.