Understanding the Coin Toss Problem

The question asks for the probability that it takes fewer than 6 tosses of a fair coin to get two consecutive heads (HH).

Let \(P(\text{tosses} < 6)\) denote this probability. This means the sequence HH must occur for the first time on toss 2, 3, 4, or 5.

A common approach for this type of problem is to use states to track the progress of the sequence. We can also use the complementary probability: \(P(\text{tosses} < 6) = 1 - P(\text{tosses} \ge 6)\). The event "tosses \(\ge 6\)" is equivalent to the event that HH has not occurred in the first 5 tosses.

Modeling the Coin Toss Sequence with States

We define the states based on the recent sequence of tosses:

• State S0: The initial state, or the state where the last toss was Tails (T). This state signifies that the sequence HH has not yet been formed and the previous toss does not help in forming HH.
• State S1: The state where the last toss was Heads (H), but the sequence HH has not yet occurred. This state means we are one step away from achieving HH.
• State E: The success state, where the sequence HH has occurred. The process stops here.

Assuming a fair coin, the probability of tossing Heads (H) is \( P(H) = \frac{1}{2} \) and the probability of tossing Tails (T) is \( P(T) = \frac{1}{2} \). The transitions between states are:

• If in state S0:
  • Toss T (prob 1/2): Remain in state S0.
  • Toss H (prob 1/2): Transition to state S1.
• If in state S1:
  • Toss T (prob 1/2): Transition back to state S0.
  • Toss H (prob 1/2): Transition to the End state E (HH achieved).

Calculating Probability of Not Reaching HH by Toss 5

Let \(\pi_n\) be the probability of being in state S0 after \(n\) tosses without HH occurring.

Let \(\rho_n\) be the probability of being in state S1 after \(n\) tosses without HH occurring.

The probability of the process ending exactly at toss \(n\) (i.e., achieving HH for the first time at toss \(n\)) is given by the probability of being in state S1 after \(n-1\) tosses and then tossing H: \( p_n = \rho_{n-1} \times P(H) = \rho_{n-1} \times \frac{1}{2} \).

The probability of remaining in the game (not achieving HH yet) after \(n\) tosses is \(\pi_n + \rho_n\). The recurrence relations for \(\pi_n\) and \(\rho_n\) are:

\( \pi_n = (\pi_{n-1} + \rho_{n-1}) \times P(T) = (\pi_{n-1} + \rho_{n-1}) \times \frac{1}{2} \)

\( \rho_n = \pi_{n-1} \times P(H) = \pi_{n-1} \times \frac{1}{2} \)

We start at \(n=0\) with \(\pi_0 = 1\) (certainty of being in the initial state) and \(\rho_0 = 0\). Let's calculate the probabilities up to \(n=5\):

The probability that the process has not stopped (HH has not occurred) by the end of toss 5 is the sum of the probabilities of being in state S0 or S1 after 5 tosses.

\( P(\text{tosses} \ge 6) = \pi_5 + \rho_5 \)

From the table, \(\pi_5 = \frac{1}{4}\) and \(\rho_5 = \frac{5}{32}\).

\( P(\text{tosses} \ge 6) = \frac{1}{4} + \frac{5}{32} = \frac{8}{32} + \frac{5}{32} = \frac{13}{32} \)

Calculating the Final Probability

We want the probability that the number of tosses is less than 6, which is the complement of the probability that it is 6 or more.

\( P(\text{tosses} < 6) = 1 - P(\text{tosses} \ge 6) \)

Substituting the value we calculated:

\( P(\text{tosses} < 6) = 1 - \frac{13}{32} = \frac{32}{32} - \frac{13}{32} = \frac{19}{32} \)

Alternatively, we could sum the probabilities of stopping at tosses 2, 3, 4, and 5:

\( P(\text{tosses} < 6) = p_2 + p_3 + p_4 + p_5 \)

\( P(\text{tosses} < 6) = \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{3}{32} = \frac{8}{32} + \frac{4}{32} + \frac{4}{32} + \frac{3}{32} = \frac{8+4+4+3}{32} = \frac{19}{32} \)

Conclusion

Both methods yield the same result. The probability that the number of tosses required is less than 6 is 19/32.