# 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? $\displaystyle 1-0.140625$ 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 $\displaystyle 0.140625$.

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

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

6. Originally Posted by scalpmaster
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 $\displaystyle 1-0.140625^4$

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 $\displaystyle 1-0.140625^4$

CB

11. Thanks CaptainBlack, got it.