1. ## Probability of 7th Head - confusion

This question may seem fundamental, I am confused can someone help me with the below problem

Probability of getting 7 continuous heads, if I understand it right is 1/2 pow 7, which is 0.007813, Can I say probability of not getting heads on 7th time is 0.992188.

But, each time I toss the probability of getting heads is 0.5 as H/T is an independent events.

So the confusion I am having is, when I am tossing for the 7th time after 6 continuous heads, what is the probability of head is it 0.007183 or 0.5

looking forward

Thanks
Max

2. Originally Posted by Umaxx
This question may seem fundamental, I am confused can someone help me with the below problem

Probability of getting 7 continuous heads, if I understand it right is 1/2 pow 7, which is 0.007813, Can I say probability of not getting heads on 7th time is 0.992188.

But, each time I toss the probability of getting heads is 0.5 as H/T is an independent events.

So the confusion I am having is, when I am tossing for the 7th time after 6 continuous heads, what is the probability of head is it 0.007183 or 0.5

looking forward

Thanks
Max
Pr(Getting a head on the 7th toss) = 1/2. The coin has no memory of what happened in previous tosses.

0.007813 is simply the probability of not gettng 7 heads in a row.

3. Originally Posted by Umaxx
This question may seem fundamental, I am confused can someone help me with the below problem

Probability of getting 7 continuous heads, if I understand it right is 1/2 pow 7, which is 0.007813, Can I say probability of not getting heads on 7th time is 0.992188.

But, each time I toss the probability of getting heads is 0.5 as H/T is an independent events.

So the confusion I am having is, when I am tossing for the 7th time after 6 continuous heads, what is the probability of head is it 0.007183 or 0.5

looking forward

Thanks
Max
A coin has no memory. Assuming the coin is fair there will be a 50% chance of getting heads on any toss of the coin

4. Thanks for the reply. I understand the Coin has no memory(When I said its independent it addresses this) and each trail the chance of occurance of heads is 0.5.

Now Lets say I got 14 continuous heads and 15 th time to get a head is 0.0000305176?, if so doesnt tht mean probability of getting a head for 15th time is very less?(Can I comment as the rare probabilities occur, the coin is baised? )

How do I corelate these 2 scenarios and understand this rare probability(combination of independent events, combinedly looking at the nature of randomness it need to flow) and in that event(15th trail) the chance of 0.05 as getting a head is independent.

These 2 different probabilities in the 15th trail makes things confusing for me

I am here in this forum to learn. I appreciate if you can explain me in detail

Looking Forward
Max

Originally Posted by e^(i*pi)
A coin has no memory. Assuming the coin is fair there will be a 50% chance of getting heads on any toss of the coin

5. Originally Posted by Umaxx
Now Lets say I got 14 continuous heads and 15 th time to get a head is 0.0000305176?, if so doesnt tht mean probability of getting a head for 15th time is very less?(Can I comment as the rare probabilities occur, the coin is baised? )
From what I can understand the $\displaystyle 2^{-15}$ number is the chance of getting 15 heads in a row which would be true if you were calculating the chance from before the first toss. If it is given that the previous 14 tosses were heads then the chance of the 15th being heads is $\displaystyle 0.5$ as each toss is independent.

How do I corelate these 2 scenarios and understand this rare probability(combination of independent events, combinedly looking at the nature of randomness it need to flow) and in that event(15th trail) the chance of 0.5 as getting a head is independent.
The chance you have of getting a head on the next toss depends on whereabouts in the sequence you are.

For example from the first toss the chance of $\displaystyle n$ heads is $\displaystyle 2^{0-n}$

On the second toss it will be $\displaystyle 2^{1-n}$ because the first toss is not counted

On the $\displaystyle k$ th toss it will be $\displaystyle 2^{(k-1)-n}$ where $\displaystyle k \leq n$

Using that formula if we look at the chance of getting 15 heads in a row (n=15):

First toss (k=1): $\displaystyle 2^{(1-1)-15} = 3.05 \times 10^{-5}$

Second toss (k=2): $\displaystyle 2^{(2-1)-15} = 6.10 \times 10^{-5}$

Eight toss (k=8): $\displaystyle 2^{(8-1)-15} = 3.906 \times 10^{-3}$

15th toss (k=15): $\displaystyle 2^{(15-1)-15} = 0.5$