KASSALMD wrote:
This is not a hard question. Bear in mind that there could be 2 or 3 or 4 or 5 Heads but no more.
You've left out the possibility that there is one head, or zero heads. If you add those in, you should find there are 144 possibilities, and you'll get the correct answer: 144/2^10 = 9/64.
You can also do the problem inductively, which demonstrates an interesting connection between this problem and Fibonacci numbers. If you flip two coins, there are 3 sequences which do not have two consecutive heads:
TH
HT
TT
No matter how we get to 10 flips with no consecutive heads, we have to start with one of these three 'words'. If we flip another coin (i.e., if we add an H or a T to the end of one of the three words above), and we must not have two consecutive heads, we can:
-only add a T to the end of a word that ends in H;
-add either a T or an H to the end of a word that ends in T.
So, if we have:
x + y words in total with n letters, consisting of
x words that end in H
y words that end in T
we can make
x + 2y words in total with (n+1) letters (because from each word ending in T we can make two new words, and from each ending in H we can only make one), and of these, we'll have:
y words that end in H
x + y words that end in T
and then can make
2x + 3y words in total with (n+2) letters
x + y words that end in H
x + 2y words that end in T
and so on.
Notice, though, if you let
\(a_1 = x \\
a_2 = y \\
a_n = a_{n-2} + a_{n-1} \\
\text{then} \\
a_3 = x+y \\
a_4 = x + 2y \\
a_5 = 2x + 3y\)
and so on. That is, the number of words you can make are just numbers from the Fibonacci sequence, since a_1 = 1 and a_2 = 2. If you did this problem for any number of coins, the numerators would all be Fibonacci numbers. So for 2, 3, 4, 5, 6, 7, 8, 9, and 10 coins, the answers will be:
3/4; 5/8; 8/16; 13/32; 21/64; 34/128; 55/256; 89/512; 144/1028...
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