An application of the law of total probability to a problem originally posed by Christiaan Huygens is to find the probability of “gambler’s ruin.” Suppose two players, often called Peter and Paul, initially have x and m − x dollars, respectively. A ball, which is red with probability p and black with probability q = 1 − p,
is drawn from an urn. If a red ball is drawn, Paul must pay Peter one
dollar, while Peter must pay Paul one dollar if the ball drawn is black.
The ball is replaced, and the game continues until one of the players
is ruined. It is quite difficult to determine the probability of Peter’s
ruin by a direct analysis of all possible cases. But let Q(x) denote that probability as a function of Peter’s initial fortune x
and observe that after one draw the structure of the rest of the game
is exactly as it was before the first draw, except that Peter’s fortune
is now either x + 1 or x − 1 according to the results of the first draw. The law of total probability with A = {red ball on first draw} and Ac = {black ball on first draw} shows that
This equation holds for x = 2, 3,…, m − 2. It also holds for x = 1 and m − 1 if one adds the boundary conditions Q(0) = 1 and Q(m) = 0, which say that if Peter has 0 dollars initially, his probability of ruin is 1, while if he has all m dollars, he is certain to win.
It can be verified by direct substitution that equation (5) together with the indicated boundary conditions are satisfied by
With some additional analysis it is possible to show that these give the only solutions and hence must be the desired probabilities.
Suppose m = 10x, so that Paul initially has nine times as much money as Peter. If p = 1/2, the probability of Peter’s ruin is 0.9 regardless of the values of x and m. If p = 0.51, so that each trial slightly favours Peter, the situation is quite different. For x = 1 and m = 10, the probability of Peter’s ruin is 0.88, only slightly less than before. However, for x = 100 and m = 1,000, Peter’s slight advantage on each trial becomes so important that the probability of his ultimate ruin is now less than 0.02.
Generalizations of the problem of gambler’s ruin play an important role in statistical sequential analysis, developed by the Hungarian-born American statistician Abraham Wald in response to the demand for more efficient methods of industrial quality control during World War II. They also enter into insurance risk theory, which is discussed in the section Stochastic processes: Insurance risk theory.
The following example shows that, even when it is given that A occurs, it is important in evaluating P(B|A) to recognize that Ac might have occurred, and hence in principle it must be possible also to evaluate P(B|Ac). By lot, two out of three prisoners—Sam, Jean, and Chris—are chosen to be executed. There are
possible pairs of prisoners to be selected for execution, of which two contain Sam, so the probability that Sam is slated for execution is 2/3. Sam asks the guard which of the others is to be executed. Since at least one must be, it appears that the guard would give Sam no information by answering. After hearing that Jean is to be executed, Sam reasons that, since either he or Chris must be the other one, the conditional probability that he will be executed is 1/2. Thus, it appears that the guard has given Sam some information about his own fate. However, the experiment is incompletely defined, because it is not specified how the guard chooses whether to answer “Jean” or “Chris” in case both of them are to be executed. If the guard answers “Jean” with probability p, the conditional probability of the event “Sam will be executed” given “the guard says Jean will be executed” is
Only in the case p = 1 is Sam’s reasoning correct. If p = 1/2, the guard in fact gives no information about Sam’s fate.
This equation holds for x = 2, 3,…, m − 2. It also holds for x = 1 and m − 1 if one adds the boundary conditions Q(0) = 1 and Q(m) = 0, which say that if Peter has 0 dollars initially, his probability of ruin is 1, while if he has all m dollars, he is certain to win.
It can be verified by direct substitution that equation (5) together with the indicated boundary conditions are satisfied by
With some additional analysis it is possible to show that these give the only solutions and hence must be the desired probabilities.
Suppose m = 10x, so that Paul initially has nine times as much money as Peter. If p = 1/2, the probability of Peter’s ruin is 0.9 regardless of the values of x and m. If p = 0.51, so that each trial slightly favours Peter, the situation is quite different. For x = 1 and m = 10, the probability of Peter’s ruin is 0.88, only slightly less than before. However, for x = 100 and m = 1,000, Peter’s slight advantage on each trial becomes so important that the probability of his ultimate ruin is now less than 0.02.
Generalizations of the problem of gambler’s ruin play an important role in statistical sequential analysis, developed by the Hungarian-born American statistician Abraham Wald in response to the demand for more efficient methods of industrial quality control during World War II. They also enter into insurance risk theory, which is discussed in the section Stochastic processes: Insurance risk theory.
The following example shows that, even when it is given that A occurs, it is important in evaluating P(B|A) to recognize that Ac might have occurred, and hence in principle it must be possible also to evaluate P(B|Ac). By lot, two out of three prisoners—Sam, Jean, and Chris—are chosen to be executed. There are
possible pairs of prisoners to be selected for execution, of which two contain Sam, so the probability that Sam is slated for execution is 2/3. Sam asks the guard which of the others is to be executed. Since at least one must be, it appears that the guard would give Sam no information by answering. After hearing that Jean is to be executed, Sam reasons that, since either he or Chris must be the other one, the conditional probability that he will be executed is 1/2. Thus, it appears that the guard has given Sam some information about his own fate. However, the experiment is incompletely defined, because it is not specified how the guard chooses whether to answer “Jean” or “Chris” in case both of them are to be executed. If the guard answers “Jean” with probability p, the conditional probability of the event “Sam will be executed” given “the guard says Jean will be executed” is
Only in the case p = 1 is Sam’s reasoning correct. If p = 1/2, the guard in fact gives no information about Sam’s fate.