Average Times Coin Combination Appears in 10 Letters

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SUMMARY

The average occurrence of the sub-word "CCLLCC" in a sequence of 10 letters generated by flipping a fair coin is calculated to be 0.0781. This is derived from the total number of combinations of 10 letters, which is 1024, and the 80 valid configurations that include "CCLLCC". The analysis shows that "CCLLCC" can occupy 5 distinct positions within the 10-letter sequence, with 16 variations for the remaining letters. This calculation provides a clear understanding of the probability involved in this specific coin-flipping scenario.

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Flipping a fair coin 10 times. This creates a word of length 10 of the letters C and L. On average how many times it appears in this sub-word CCLLCC?
C=crown and L= the other side of coin

i think that if i count CCLLCC in 10 letters of C and L we will have 5 times but how i will find the average?
 
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If 6 of the letters are "CCLLCC" then the other four letters can be either "C" or "L". There are 2^4= 16 ways to do that. But you also need to allow for where in the set of 10 letters those letters occur.

I assume by "sub-word", you require that the letters "CCLLCC" occur together. Do you see that there are 5 places the initial C can be?
 
Hello, I'm not quite sure if I'm right as I'm still a novice in the world of math. It's been a while since I have taken my prob/stats class.

__ __ __ __ __ __ __ __ __ __

There are 2 ways either C or L for each of the spaces above. So, the total number of ways for spelling out a 10 letter word is 2*2*2*2... = 2^{10} = 1024

There are 5 ways for CCLLCC to occur inside some of the 10 letter words. One of them looks like this:

C C L L C C __ __ __ __

The spaces containing CCLLCC are all "locked in". So, there are 4 spaces left for choices.

2*2*2*2 = 2^{4} = 16
There are 5 ways in which CCLLCC can occur because you can just shift CCLLCC one space to the right each time to get a new word that contains CCLLCC.

Possible ways of getting a word with CCLLCC in it: 5*16 = 80

I'm not sure what you meant by "average", so I guess you want the probability as well. \frac{80}{1024} = 0.0781

Once again I'm not a 100% sure, but I hope it helps you a bit. :P
 
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