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 Mr. Markov plays cards (Posted on 2010-06-26)
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)

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 solution Comment 1 of 1

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.

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

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