Question

A person has 6 cards A, K, 2, Q, 10, 9. The person randomly draws the cards one by one with replacement till he gets 3 consecutive A. If $P_n$ represents the probability that atleast n cards are drawn, then $216P_n - 5P_{n-3} - 30P_{n-2} - 180P_{n-1}$ is (For n > 3)

• A. 0

Answer 1

Answer: (A)

0

Answer 2

Let the process be such that a card is drawn with replacement until three consecutive A’s appear. Define the state probabilities after n draws as follows:

• Let $u_n$ = probability the process is active and ends in “non-A” (0 consecutive A’s),
• $v_n$ = probability the process is active and ends in “A” (1 consecutive A),
• $w_n$ = probability the process is active and ends in “AA” (2 consecutive A’s),

with initial conditions $u_0$ = 1, $v_0$ = 0, $w_0$ = 0.

At every draw:

• If a non-A card is drawn (with probability 5/6), the process goes to state 0 regardless of the previous state.
• If an A is drawn (with probability 1/6):
  • From state 0, it goes to state 1,
  • From state 1, to state 2,
  • From state 2, drawing A would complete three A’s so the process stops.

Thus, the recurrences are:

$\begin{cases} u_n = \frac{5}{6}(u_{n-1}+v_{n-1}+w_{n-1}) = \frac{5}{6}f(n-1),\\[1mm] v_n = \frac{1}{6}u_{n-1},\\[1mm] w_n = \frac{1}{6}v_{n-1}. \end{cases}$

where $f(n) = u_n+v_n+w_n$ is the probability that no three consecutive A’s are obtained in n draws.

Then,

$f(n)=\frac{5}{6}f(n-1)+\frac{1}{6}u_{n-1}+\frac{1}{6}v_{n-1}.$

Since $u_{n-1}=\frac{5}{6}f(n-2)$ and $v_{n-1}=\frac{1}{6}u_{n-2}=\frac{5}{36}f(n-3)$, we have:

$f(n)=\frac{5}{6}f(n-1)+\frac{5}{36}f(n-2)+\frac{5}{216}f(n-3).$

Multiplying both sides by 216 yields:

$216\,f(n)=180\,f(n-1)+30\,f(n-2)+5\,f(n-3).$

Now note that "at least n cards are drawn" means that in the first $n-1$ draws, the process has not stopped. Define:

$P_n = f(n-1).$

Substitute $n-1$ for n in the recurrence:

$216\,f(n-1)=180\,f(n-2)+30\,f(n-3)+5\,f(n-4),$

or, writing in terms of $P$,

$216\,P_n=180\,P_{n-1}+30\,P_{n-2}+5\,P_{n-3}.$

Rearrange to obtain:

$216P_n - 5P_{n-3} - 30P_{n-2} - 180P_{n-1} = 0,$

for $n > 3$.