OST
Ostin Technology Group Co., Ltd. Class A
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This solution may look complicated, but once you understand this "Gambler martingale" method, you can solve many such problems very quickly. Suppose at every throw we have a gambler come in and bet one seeing the pattern 3,4,5, except he does not bet on the whole pattern at once, but in stages. So he first bets x on seeing a 3. If he wins he has 6x (fair payout) and bets on the next throw being 5 etc. Now let us do it so that the gamblers entering at an even turn bet a, while the others bet b. Let X\_n be the total profit of all gamblers after n throws. Let T be the stopping time which stops when we see the 3,4,5. What is X\_T? It is the total money remaining minus the total initial buy-in of each gambler. If T is odd, the total buy-in is (T-1)/2 \* a + (T + 1)/2 \* b, and the total remaining money is 6\^3 \* b. If T is even, the total buy-in is T/2 \* a + T/2 \* b, and the total remaining money is 6\^3 \* a. We could now choose a = 1, b = -1 for example to eliminate T (otherwise we would need to know E\[T\] although this should be well known to be 6\^3). Plugging in these numbers and denoting by p the desired probability gives by OST: 0 = E\[X\_0\] = E\[X\_T\] = (-6\^3 + 1) \* p + (6\^3) \* (1-p). Solving for p gives p = 216/431

This solution may look complicated, but once you understand this "Gambler martingale" method, you can solve many such problems very quickly. Suppose at every throw we have a gambler come in and bet one seeing the pattern 3,4,5, except he does not bet on the whole pattern at once, but in stages. So he first bets x on seeing a 3. If he wins he has 6x (fair payout) and bets on the next throw being 5 etc. Now let us do it so that the gamblers entering at an even turn bet a, while the others bet b. Let X\_n be the total profit of all gamblers after n throws. Let T be the stopping time which stops when we see the 3,4,5. What is X\_T? It is the total money remaining minus the total initial buy-in of each gambler. If T is odd, the total buy-in is (T-1)/2 \* a + (T + 1)/2 \* b, and the total remaining money is 6\^3 \* b. If T is even, the total buy-in is T/2 \* a + T/2 \* b, and the total remaining money is 6\^3 \* a. We could now choose a = 1, b = -1 for example to eliminate T (otherwise we would need to know E\[T\] although this should be well known to be 6\^3). Plugging in these numbers and denoting by p the desired probability gives by OST: 0 = E\[X\_0\] = E\[X\_T\] = (-6\^3 + 1) \* p + (6\^3) \* (1-p). Solving for p gives p = 216/431