Mr. Markov either wins or loses at a card table. His result on a given day is related to his result the day before. If he loses today, there is a 50% chance he would lose tomorrow. If he wins today, there is a 30% chance that he would lose tomorrow. How often would he lose on two consecutive days? Assume an infinite horizon so the first day result is irrelevant. (Note: Losing three days in a row counts as losing on two consecutive days twice)

First find the probability of a loss; that is, the fraction of the trials (days) that result in a loss:

Let x be that probability (of a loss, not necessarily consecutive).

So, based on the conditional probabilities:

x = .3(1 - x) + .5(x)

Solving this gives x = 3/8.

But each loss is followed by another trial, and in such a trial, 1/2 the time there will be another loss, so this happens 3/16 of the overall trials. (Successes, of course, do not lead to the next day's possibility of being the second day of a two-day loss pair.) So overall, 3/8 of the time there is a loss, and in 1/2 of those times, another loss follows--3/16 of the days are loss days that follow after at least one other loss day.

Therefore 3/16 is the answer.

Simulation verification:

DEFDBL A-Z
RANDOMIZE TIMER
DO
 trials = trials + 1
 r = RND(1)
 IF prevLoss THEN
    IF r < .5 THEN loss = 1:  ELSE loss = 0
    IF loss THEN dblLoss = dblLoss + 1
 ELSE
    IF r < .3 THEN loss = 1:  ELSE loss = 0
 END IF
 prevLoss = loss
 IF loss THEN losses = losses + 1
 PRINT losses, dblLoss, trials, dblLoss / trials
LOOP

results, as the last line, when stopped, in:

 losses      cons losses     trials             prob
 161143        80562         428453        .1880299589453219

Posted by Charlie on 2010-06-26 14:58:43