Probability of how many times one need to throw a coin to get a head twice in a row?

#1
Hi!


New here, first thread!

My issue is to explain why the probability function of how many times one need to throw a coin to get a head twice in row. The function should be 1/2^(k-1) k>2. I do not understand why it says k-1 and not simply k?


Best,
Filippa
 
#2
Re: Probability of how many times one need to throw a coin to get a head twice in a r

Just to make sure that I have things quite clear. You toss a coin repeatedly and stop as soon as you get two heads in a row. This has taken a total of k throws. For example HTTTHTHH would have k = 8. The question is what is the probability you will need k throws to get your double head.
Let's look at k = 5 for example. There are 2^5 = 32 possibilities for 5 throws. How many of these need all five throws to get two heads in a row? We have TTTHH, THTHH and HTTHH. 3 possibilities. Ones like THHTTT are not counted because the HH is too soon. It has HH after only 3 throws. So for k = 5, p = 3/32.
So, if I have interpreted the question correctly, 1/2^(k-1) doesn't work (nor does 1/2^k).
However, you have obviously got the question from somewhere, and so it is quite likely that I have misinterpreted what you wrote. Perhaps you could make it clearer with an example. kat
 
#3
Re: Probability of how many times one need to throw a coin to get a head twice in a r

On the other hand, the question may have been the probability of getting two of the same in a row.
k = 3 HTT or THH p = 2/8 = = 1/4 =1/(3-1)^2
k = 4 HTHH or THTT = 2/16 = 1/8 = 1/(4-1)^2 etc.