GEK1544 The Mathematics of Games Suggested Solutions to Tutorial 3

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1 GEK544 The Mathematics of Games Suggested Solutions to Tutorial 3. Consider a Las Vegas roulette wheel with a bet of $5 on black (payoff = : ) and a bet of $ on the specific group of 4 (e.g. 3, 4, 6, 7 ; payoff = 8 : ). What is the bettor s expectation (per unit) on this combined bet? Suggested solution. Let us consider the specific group of four numbers : 3, 4, 6, 7. The numbers 3 and 7 are also black. Then A = {outcome is black } and B = {outcome is 3, 4, 6 or 7}. P := P (A and B) = P ({outcome is 3 or 7 }) = P := P (A and [ not B]) = P ( {outcome is black but not 3 and 7 }) = 8 38, P 3 := P ([ not A] and B) = P ({outcome is 4 or 6 }) = P 4 := P ([ not A] and [not B]) = P ({ outcome is white, 0, or 00, and not 4 or 6 } ) = = 8 Denote by X(5 : [ 7]) the expectation with the specific betting pattern, namely, $ 5 on black and $ on the specific group of 4 (the number [7] represents the total amount wagered). These result X(5 : [ 7]) = P (5 + 8) + P (5 ) + P 3 ( 8 5) + P 4 ( 5 ) = 6 8 (5 + 8) + (5 ) + ( 8 5) + ( 5 ) = ( ) = 4 + ( ) That is, the bettor s expectation per unit is given by X ([ ]) =

2 Next, let us consider the specific group of four numbers : 7, 8, 0,. The numbers 8, 0 and are also black. Then A = {outcome is black } and B = {outcome is 7, 8, 0 or }. P := P (A and B) = P ({outcome is 8, 0 or }) = 3 P := P (A and [ not B]) = P ( {outcome is black but not 8, 0 and }) = 8 3 = , P 3 := P ([ not A] and B) = P ({outcome is 7 }) = P 4 := P ([ not A] and [not B]) = P ({ outcome is white, 0, or 00, and not 7 } ) = = Denote by X(5 : [ 7]) the expectation with the specific betting pattern, namely, $ 5 on black and $ on the specific group of 4 (the number [7] represents the total amount wagered). These result X(5 : [ 7]) = P (5 + 8) + P (5 ) + P 3 ( 8 5) + P 4 ( 5 ) = 3 5 (5 + 8) + (5 ) + ( 8 5) + ( 5 ) ( 8 = ) ( ) 38 = 4 That is, the bettor s expectation per unit is given by X ([ ]) = (the same). You may continue the calculation and show that the answer is the same no matter how many of the 4 numbers in black.

3 . A mythical slot machine has three wheels, each containing ten symbols. On each wheel there is JACKPOT symbol and other non-paying symbols. You put in silver dollar ($) in the slot and the payoffs are as follows : 3 JACKPOT symbols $487 in silver is returned (including your $). JACKPOT symbols $0 in silver is returned (including your $). JACKPOT symbols $ in silver is returned (you get your wager back!). Define what it would mean to say that this slot machine is fair and then show that it is indeed a fair one-armed bandit! Suggested solution. Assuming that the slot machine is random. P = P ( Jackpot symbol exactly) = = ; P = P ( Jackpot symbols exactly) = = ; P 3 = P (3 Jackpot symbols) = p N = P (No Jackpot symbol at all) = The profits are 0, and 486, respectively. We infer that = 000 ; 0 = = (P + P + P 3 ) [ please verify]. 0 0 X = P 0 + P + P p N ( ) = = 0.

4 3. The martingale strategy had the gambler double his bet after every loss, so that the first win would recover all previous losses plus win a profit equal to the original stake. Since a gambler with infinite wealth will with probability eventually flip heads, the Martingale betting strategy was seen as a sure thing by those who practised it. Of course, none of these practitioners in fact possessed infinite wealth, and the exponential growth of the bets would eventually bankrupt those who choose to use the martingale strategy. Moreover, it has become impossible to implement in modern casinos, due to the betting limit at the tables. Because the betting limits reduce the casino s short term variance, the martingale strategy itself does not pose a threat to the casino, and many will encourage its use, knowing that they have the house advantage no matter when or how much is wagered. Let one round be defined as a sequence of consecutive losses followed by a win, or consecutive losses resulting in bankruptcy of the gambler. After a win, the gambler resets and is considered to have started a new round. A continuous sequence of martingale bets can thus be partitioned into a sequence of independent rounds. We will analyze the expected value of one round. Let l be the probability of losing (e.g. betting for even for roulette has 0/38 chance of losing). Let y be the amount of the commencing bet, and n the finite number of bets you can afford to lose. Show that the expected profit per unit (a loss if the number is negative) per round is given by X ([]) = [ ( l) ] ( correction). Suggested solution. Backward thinking. You may choose anyone of the following arguments. In the sequence of numbers 0 y = y, y, y,, n y, n y, the chance that the gambler loses the first n games is l n. The total loss is y + y + y + + n y = y [ n ] = y n = y (n ). The chance that the bettor does not lose all n games is ( l n ). In this case, disregard where is the win in the round, the profit is y. Hence X([ round ]) = ( l n ) y + l n [ (y ( n )) ] = y [ ( l) n ].

5 To calculate the expected amount wagered in a round, let us draw an analog with the free odds in Craps : Expected amount of betting for a Pass Bettor, with possible free odds = ( + ) = + 3. On can understand this in the following manner. To start the game, one has to put down unit. However, there are /3 of the possibility that one can put down an additional unit. Back to the present case - to start the game, one has put down y units. If and when the first game is lost (with possibility equal to l ), the bettor puts down an additional ( y) units. And once the first two games are lost (with with possibility equal to l l = l ), one has to put down an additional ( y) units, and so on. It follows that Expected amount of betting in a round = y + l ( y) + l ( y) + + l n ( n y) st nd 3 rd n th game (amount put down on) = y [ + ( l ) + ( l ) + + ( l ) n ] = y [ ( l) n ] l ( = y n for l = ). Thus for l ; X([]) = X([ round ]) Expected amount of betting in a round = y [ ( l)n ] y ( ) ( l) n l = [ ( l) ]. We observe the simplicity and beauty of the final answer when we consider the expectation per unit. Also, as a side note, X([]) < 0 once the probability of losing l >. When l =, X([ round ]) = 0. Hence X([]) = X([ round ]) n y = 0 when l =.

6 Forward thinking. P (W ) = P (wins the first game) = ( l) ; P (W ) = P (loses the first game and wins the second game) = l ( l) ; P (W 3 ) = P (loses the first two games and wins the third game) = l l ( l) ; P (W n ) = P (loses the first n games and wins the last game) = l n ( l). As above, the chance that the gambler loses (all) the first n games is l n, and the amount lost is equal to y ( n ). Thus [ recall that on each round (barring bankruptcy) the profit is y ] X([ round ]) = P (W ) y + P (W ) y + + P (W n ) y l n [ y ( n )] = [ ( l) + l ( l) + + l n ( l) l n [ ( n )] ] y = [ ( l) + (l l ) + + (l n l n ) l n [ ( n )] ] y = [ n l n ] y (same as above).

7 Notes on geometry and arithmetic sum S k = + a + a + + a k. Here a is a number. To find the sum S k ( k =,, ), the idea is to go walk one step forward : a S k = a + a + a a k+. Hence a S k S k = a k+ = S k = ak+ a When a =, the idea does not work, but observe that k = = (k + ). As for S n = n, where n is an integer, we apply the mirror reflection idea : S n = n + (n ) + (n ) , Put them in juxtaposition S n = n, S n = n + (n ) + (n ) + + Summing them vertically we obtain S n = (n + ) + (n + ) + (n + ) + + (n + ) n teams = S n = n (n + ) = S n = n (n + ).