Solutions to In Class Problems Week 15, Wed.

Transcription

1 Massachusetts Institute of Technology 6.04J/18.06J, Fall 05: Mathematics for Comuter Science December 14 Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld revised December 14, 005, 1404 minutes Solutions to In Class Problems Week 15, Wed. Gamblers Ruin A gambler aims to gamble until he reaches a goal of T dollars or until he runs out of money, in which case he is said to be ruined. He gambles by making a sequence of 1 dollar bets. If he wins an individual bet, his stake increases by one dollar. If he loses, his stake decreases by one dollar. In each bet, he wins with robability > 0 and loses with robability q ::= 1 > 0. He is an overall winner if he reaches his goal and is an overall loser if he gets ruined. In a fair game, = q = 1/. The gambler is more likely to win if > 1/ and less likely to win if < 1/. With T and fixed, the gambler s robability of winning will deend on how much money he starts with. Let w n be the robability that he is a winner when his initial stake in n dollars. Problem 1. (a) What are w 0 and w T? Solution. w 0 = 0 and w T = 1. (b) Note that w n satisfies a linear recurrence w n+1 = aw n + bw n 1 (1) for some constants a, b and 0 < n < T. Write simle exressions for a and b in terms of. Solution. By Total Probability w n = Pr {win game win the first bet} Pr {win the first bet} + Pr {win game lose the first bet} Pr {lose the first bet} = w n+1 + q Pr {w n 1 }, so () w n+1 = w n qw n 1 w n+1 = w n qw n 1. (3) Coyright 005, Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld.

2 So 1 q a =, b =. (c) For n > T, let w n be defined by the recurrence (1), and let g(x) ::= n=1 w n x n be the generating function for the sequence w 0, w 1,... Verify that Solution. w 1 x g(x) = q. (4) (1 x)(1 x) g(x) = w 0 + w 1 x + w x + w 3 x 3 + xg(x)/ = w 0 x/ + w 1 x / + w x 3 / + (q/)x g(x) = (q/)w 0 x + (q/)w 1 x 3 + so But ( ) xg(x) qx g(x) g(x) = w 0 + w 1 x w 0 x/ = w 1 x, ( ) x qx g(x) 1 + = w 1 x. (5) x qx q 1 + = (1 x)(1 x) (6) Combining (6) and (5) yields (4). (d) Conclude that in an unfair game ( ) n q w n = c + d (7) for some constants c, d. Solution. In an unfair game /q = 1, so from (4), we know that there will be c, d such that c d g(x) = + q (8) 1 x 1 x so w n will be the corresonding combination of the coefficients of x n in 1/(1 x) and 1/(1 (q/)x), namely, (7).

3 3 (e) Show that in an unfair game, Solution. Given (4), we want c, d such that So c, d satisfy Letting x = 1 gives Letting x = /q gives So lugging into (7) gives (q/) n 1 (q/) T 1 w 1 x c d q = (1 x)(1 x) 1 x + q. 1 x q w 1 x = c(1 x) + d(1 x). w 1 c =. 1 q/ w 1 /q w 1 d = = 1 /q q/ 1 = c. (( ) n ) w 1 q w n = 1. (9) q/ 1 Now we can solve for w 1, by letting n = T in (9): ( ( )T ) w 1 q 1 = w T = 1 q/ 1 so Combining this with (9) yields (q/ 1) w 1 =. T (q/) 1 ((q/) n 1) (q/) T 1 (f) Verify that if 0 < a < b, then Conclude that if < 1/, then a a + 1 <. b b + 1 ( ) T n w n <. q

4 Solution. a a(1 + 1/b) a + a/b a + 1 = = <. b b(1 + 1/b) b + 1 b + 1 So from the revious art, we have ( ) n n n T ( ) T n (q/) 1 (q/) q w n = < = =. (q/) T 1 (q/) T q 4 Problem. Show that in a fair game, Hint: Use equation (4) again. w T Solution. This time = q = 1/so from (4), Now we need a, b such that so we will have Solving for a, b, we have from (10) w 1 x g(x) =. (1 x) w 1 x a b = (10) (1 x) 1 x + (1 x), w n = a + b(n + 1). w 1 x = a(1 x) + b. Letting x = 0yields a = b and x = 1yields b = w 1, so w n = w 1 + w 1 (n + 1) = w 1 n. Also, so and hence 1 = w T = w 1 T 1 w 1 = T n T

5 5 Problem 3. Now suose T =, that is, the gambler kees laying until he is ruined. (Now there may be a ositive robability that he actually lays forever.) Let r be the robability that starting with n > 0 dollars, the gambler s stake ever gets reduced to n 1. (a) Exlain why r = q + r. Solution. By Total Probability But r = Pr {ever down $1 lose the first bet} Pr {lose the first bet} + Pr {ever down $1 win the first bet} Pr {win the first bet} = q + Pr {ever down $1 win the first bet} Pr {ever down $1 win the first bet} = Pr {ever down $} = Pr {being down the first $1} Pr {being down another $1} = r. (b) Conclude that if 1/, then r = 1. Solution. r r + q has roots q/ and 1. So r = 1or r = q/. But 1 r, which imlies r = 1 when q/ 1, that is, when 1/. In fact r = q/ when q/ < 1, namely, when > 1/, but this requires an additional argument that we omit. (c) Conclude that even in a fair game, the gambler is sure to get ruined no matter how much money he starts with! Solution. The gambler gets ruined starting with initial stake n = 1recisely if his initial stake goes down by 1 dollar, so his robability of ruin is r, which equals 1 in the fair case. The recurrence (1) will also hold in this T = case if we interret w n as the robability of not being ruined, that is, the gambler wins if he can gamble forever. So w 1 is the robability he is not getting ruined starting with a 1 dollar stake, that is w 1 = 1 r = 0. Since w 0 = 0 = w 1, the recurrence imlies that w n = 0 for all n 0.

6 6 (d) Let t be the exected time for the gambler s stake to go down by 1 dollar. Verify that t = q + (1 + t). Conclude that starting with a 1 dollar stake in a fair game, the gambler can exect to lay forever! Solution. By Total Exectation But t = E [#stes to be down $1 lose the first bet] Pr {lose the first bet} + E [#stes to be down $1 win the first bet] Pr {win the first bet} = q + E [1 + #stes to be down $1 win the first bet]. E [#stes to be down $1 win the first bet] = E [#stes to be down $] = E [#stes to be down the first $1] + E [#stes to be down another $1] = t. This imlies the required formula t = q + (1 + t). If = 1/ we conclude that t = 1 +t, which means t must be infinite.