I threw a coin
n times, and never got three tails in a row. I calculated the odds of this event, and found out they were just about even; 50%-50%. How many times did I throw the coin?
A second question: what were the chances of having not gotten three heads in a row either?
Suppose the probability that there is no three continuous tails in n tosses is P(n). The last two of these tosses could only be the following four cases:
HH
TT
HT
TH
If they were HH, HT or TH, an additional toss will not make a three-tail (probility = 3/4); but if they were TT, you can get TTT if toss of n+1 is also tail (probilty = 1/4*1/2). Therefore, the probability that TTT does not happen in n+1 tosses is
P(n)*(3/4+1/8)=P(n) * 7/8.
Starting from the first 3 tosses. The probability of not gettting TTT is 7/8. Therefore, from the above equation, for any n>3,
P(n) = 7/8 * (7/8)^(n-3) = (7/8)^(n-2).
To get 50% odds, n = 7.
Solution?
