Consecutive Tails in Coin Flips

Solution & Explanation

Understanding the Problem:

• We are given a scenario where a fair coin is flipped 10 times.
• We need to calculate the probability of getting exactly three tails (T) in a row, anywhere in the sequence of 10 flips.

Step-by-Step Solution:

1. Total Possible Outcomes:

   • Each flip has two possible outcomes: Heads (H) or Tails (T).
   • With 10 flips, the total number of possible sequences is $2^{10} = 1024$.

2. Favorable Outcomes:

   • We need sequences where exactly 3 tails appear consecutively.
   • Consider the sequence as a 10-character string.

3. Identifying Positions for Consecutive Tails:

   • The three consecutive tails can start at any position from 1 to 8 in the sequence, ensuring they fit within the 10 flips.
   • For example, if the tails start at position 1, the sequence is TTT followed by 7 heads (TTTHHHHHHH).
   • If they start at position 8, the sequence is 7 heads followed by TTT (HHHHHHHTTT).

4. Counting Valid Sequences:

   • The possible starting positions for the consecutive tails are:
     • Position 1: TTT followed by 7 heads
     • Position 2: H + TTT + 6 heads
     • Position 3: HH + TTT + 5 heads
     • Position 4: HHH + TTT + 4 heads
     • Position 5: HHHH + TTT + 3 heads
     • Position 6: HHHHH + TTT + 2 heads
     • Position 7: HHHHHH + TTT + 1 head
     • Position 8: HHHHHHH + TTT
   • There are 8 such sequences.

5. Calculating Probability:

   • The probability of obtaining any one specific sequence is $\frac{1}{1024}$ since each sequence is equally likely.
   • Therefore, the probability of any of the 8 valid sequences occurring is: $\text{Probability} = \frac{8}{1024} = \frac{1}{128} = 0.0078125$

Bonus: Generalizing the Problem

General Case:

• Calculate the probability of getting exactly $t$ tails in $n$ flips, with all tails consecutive.

1. Total Possible Outcomes:

   • For $n$ flips, there are $2^n$ possible sequences.

2. Favorable Outcomes:

   • The sequence of $t$ consecutive tails can start at any position from 1 to $(n-t+1)$.
   • This gives $(n-t+1)$ valid sequences.

3. Calculating Probability:

   • Probability of any one specific sequence occurring is $\frac{1}{2^n}$.
   • Therefore, the probability of any of the $(n-t+1)$ sequences occurring is: $\text{Probability} = \frac{n-t+1}{2^n}$

Conclusion:

• For the specific case of 10 flips and 3 consecutive tails, the probability is $0.0078125$.
• For the general case, use the formula $\frac{n-t+1}{2^n}$ to find the probability of $t$ consecutive tails in $n$ flips.