1. Brian Duddy

2.  Two players, X and Y, are playing a card game- goal is to find optimal strategy for X  X has red ace (A), black ace (A), and red two (2)  Y has red ace (A), black ace (A), black two (2)  On each turn, each player turns over one card  If both cards are the same color, X wins value of his card (1 for ace); if they are different, Y wins value of his card, except if both players play 2, neither player wins anything

3. AA2 A+1-2 A+1 2+20 X Y Positive numbers: units won by X (colors match) Negative numbers: units won by Y (colors do not match)

4. From what was given in the model, I found 2 possible sets of rules for the game.  The game lasts an indefinite number of turns; on each turn, either player can play any of their cards. (Game 1)  The game lasts 3 turns; players can not play the same card twice. (Game 2) These two games have very different strategies!

5. What is the optimal strategy for this game? It depends! Depending on the strategy of your opponent (Y), the optimal strategy can differ dramatically! For example, if Y was very greedy, and thus played his or her black 2 often, your optimal strategy would involve playing your black ace more often than it otherwise might. On a basic level, Y will either play essentially randomly or at least attempt to play intelligently, and this makes a big difference. AA2 A+1-2 A+1 2+20 X Y

6. If Y plays truly randomly (i.e. he plays each of his cards 1 /3 of the time), it is possible to calculate, on average, how much X will gain or lose for each card he plays-this can be done by multiplying the gain or loss of each possible outcome by how often it will appear (which in this case is always 1/3) Red A: 1( 1 / 3 )-1( 1 / 3 )-2( 1 / 3 ) = - 2 / 3 dollars Black A: -1( 1 / 3 )+1( 1 / 3 )+1( 1 / 3 ) = + 1 / 3 dollars Red 2: 2( 1 / 3 )-1( 1 / 3 )+0( 1 / 3 ) = + 1 / 3 dollars Therefore, X can play his black ace and red 2 in any combination and gain, on average, 1/3 dollars per turn AA2 A+1-2 A+1 2+20 X Y If Y does not play truly randomly (few humans do!), this may not work; for example, if X always plays the black ace, eventually Y will start playing his red ace. Therefore, if playing against a real person who seems to play semi-randomly, X should vary his choices between the two cards and try not to fall into a pattern Y will notice. However...

7. If Y is an intelligent player (and he knows that X is as well), he will likely deduce that there’s no reason for X to play his red ace. Why? The red 2 can always do better or the same; i.e., it dominates the red ace.  If Y plays a red ace, X’s red 2 will win 2 units instead of just 1.  If Y plays a black ace, X’s red ace and 2 both lose 1 unit.  If Y plays a black 2, X’s red ace loses two and his red 2 loses none! Now, Y’s red 2 is dominated by his red ace, because he knows X should never play the red ace. AA2 A+1-2 A+1 2+20 X Y We now have a simpler grid, as each player only has 2 real possibilities (and they will never gain by switching to the other. However, it is impossible to determine an optimal strategy for X without knowing what Y’s is. At this point X should use psychology; maybe Y will more often play his/her black ace, to avoid losing 2. Any specific pattern will be exploited by an intelligent human opponent, so some randomness must come into play.

8.  Because each player only has 3 cards, this game only lasts 3 turns. Also, the strategies are obviously different because each move affects the next moves, and on the 2 nd and 3 rd moves of the game, there are less choices.  Again, what X should do depends very much on what Y’s strategy is. AA2 A+1-2 A+1 2+20 X Y

9.  Assuming Y has an equal chance of playing any of his cards, we can again find the expected payout for any particular strategy of X. Let’s assume he starts by playing his red ace. AA2 A+1-2 A+1 2+20 X Y If Y responds with his own red ace, X gains one unit. Now, on the second turn, X has a choice of two cards: black ace and red two. If X plays his black ace and Y again responds with another black ace, X gains 1; on the final turn they both must play their 2 and there is no gain for either. Therefore, X will gain a total of 2. If Y instead responds with a black 2, X still gains 1; on the final turn X will lose one, making for a net gain of 1 for X. The average gain, then, for X playing the red ace is 1.5. If X plays his red 2 and Y plays a black ace, Y gains one; on the 3 rd turn X will have the black ace and Y the black 2. X will gain a total of 1. Finally, if X plays his red 2 and Y plays the black 2, X will end up with a total of +2. Again, if Y is playing randomly, X will gain 1.5 units on average with this strategy. Eight more cases of this type have to be considered!

10.  After all of that analysis, the final result… AA2 A+1-2 A+1 2+20 X Y No matter X’s strategy, if Y plays randomly, the game is, on average, a draw! AA2Total A+1.50-1.50 A-2+0. 5 +1. 5 0 2 -1.500 X’s first play Y’s first play Expected value after 3 moves

11.  Let’s look at that chart again… but this time, from the point of view of Y, assuming X is playing randomly. AA2 A+1.50-1.5 A-2+0.5+1.5 2 -1.50 Total+10 X’s first play Y’s first play Unlike X, Y does have a difference in his choices; his black ace, if played first, will on average win him 1/3 dollar per game, and his red ace will lose him 1/3 dollar per game. (Remember that + indicates scores for X and – indicates scores for Y!)

12.  An intelligent Y might then be expected to play his black ace first more often; if this is true, X should counter with his own black ace to maximize the expected value. However, Y might figure out X’s strategy and begin leading with his red ace, which gives him a large advantage. Again, there is no mathematical formula for what X should do; only analyzing the opponent and his play style will allow X to make money in the long term. AA2 A+1.50-1.5 A-2+0.5+1.5 2 -1.50 Total+10 X’s first play Y’s first play (Remember that + indicates scores for X and – indicates scores for Y!) Note: In all cases, for both X and Y, the selection of the second card is a “wash”- neither card is better then the other. Psychology might again come into play, but both players need to remember that the third card needs to be played as well!

13.  If the game is played an indefinite number of times, X should play only his black ace or red 2 and try to avoid playing them in a specific pattern  If cards cannot be played more then once, X should probably start with his black ace most of the time, but he should make sure Y does not catch on to this strategy  In either case, if a human opponent is being played against, no specific strategy is always the best; the opponent’s own strategy and psychology must be taken into account