The Monty Hall Problem

Transcription

1 The Monty Hall Problem The Monty Hall Problem Door 2! 3? (a) Choosing a door. (b) Changing one s choice or not? Figure: The Monty Hall problem J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 1 / 29

2 The Monty Hall Problem You are the goat! (Glenn Calkins, Western State College, Colorado) J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 2 / 29 The Monty Hall Problem Letters to Marilyn vos Savant regarding her solution (Change!): You blew it! [...] Please help by confessing your error and, in the future, being more careful. (Prof. Robert Sachs, George Mason University in Fairfax, Virginia) You are utterly incorrect. [...] How many irate mathematicians are needed to get you to change your mind? (Prof. E. Ray Bobo, Georgetown University, Washington D.C.) Our math department had a good, self-righteous laugh at your expense. (Prof. Mary Jane Still, Palm Beach Junior College, Florida) Maybe women look at math problems differently than men. (Don Edwards, Sunriver, Oregon)

3 The Monty Hall Problem The Monty Hall Problem: vos Savant s Explanation Table: All cases of the Monty Hall problem Door 1: Goat Door 2: Goat Door 3: Car Strategy Gain Game 1 chosen door stay goat Game 2 chosen door stay goat Game 3 chosen door stay car Game 4 chosen door change car Game 5 chosen door change car Game 6 chosen door change goat J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 3 / 29

4 The Monty Hall Problem The Monty Hall Problem: vos Savant s Explanation Choosing a door door with goat door with goat door with car same door other door same door other door same door other door J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 4 / 29

5 The Monty Hall Problem Some Basic Notions From Probability Theory A (finite) probability space is given by a finite set E ={e 1,e 2,...,e k } of atomic events, where event e i occurs with the probability w i = P(e i ) and we have k i=1 w i =1. Such a correspondence between probabilities w i and atomic events e i specifies a probability distribution. If all atomic events occur with the same probability (w i = 1 /k for each i, 1 i k), we have the uniform distribution. This can be extended to every subset E E, which represents an event, as follows: P(E)= w i. e i E P(E) denotes the (total) probability for event E to occur. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 5 / 29

6 The Monty Hall Problem Some Basic Notions From Probability Theory For example, for the uniform distribution on E, P(E)= E k simply gives the frequency that among all possible atomic events some from E occurs, where E denotes the cardinality of E, i.e., the number of elements in E. For all events A,B E, we have the following basic properties: 1 0 P(A) 1, where P(/0)=0 and P(E)=1. 2 P(A)=1 P(A), where A=E A is the event complementary to A. 3 P(A B)=P(A)+P(B) P(A B). J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 6 / 29

7 The Monty Hall Problem Some Basic Notions From Probability Theory Definition (conditional probability) Let A and B be events with P(B)>0. 1 The conditional probability for event A to occur under the condition of event B occurring is defined as P(A B)= P(A B). P(B) 2 A and B are said to be (stochastically) independent if P(A B)= P(A) P(B). J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 7 / 29

8 The Monty Hall Problem Some Basic Notions From Probability Theory Theorem (Bayes) 1 If A and B are two events with P(A)>0 and P(B)>0, then P(A) P(B A)=P(B) P(A B). 2 If A and B 1,B 2,...,B l are events with P(A)>0 and P(B i )>0, 1 i l, where l i=1 B i = E is a partition of the finite probability space E in disjoint events, then for all i, 1 i l, P(B i A)= P(A B i) P(B i ) l j=1 P(A B j) P(B j ) = P(A B i) P(B i ). (1) P(A) J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 8 / 29

9 The Monty Hall Problem The Monty Hall Problem: By the Law of Total Probability The first statement of this theorem is the special case of its second statement for l=1. The proof of this theorem follows immediately from the definition of conditional probability. The equality P(A)= l j=1 P(B j A)= l j=1 used in (1) is also called the law of total probability. P(A B j ) P(B j ) (2) Using (2), we can solve the Monty Hall problem: The success probability is 1/3 if one stays with the initially chosen door and 2/3 if one changes the initially chosen door. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 9 / 29

10 The Monty Hall Problem The Monty Hall Problem: Is It a Game? Yes, because there is a player who wants to maximize her gains by choosing the right strategy. There is only one player: It s a game against nature. In some sense, the player is playing against the host, who sometimes does not content with the passive role of the doorkeeper and opener, but seeks to actively influence the strategy chosen by the player and thus the outcome of the game: On the first trial, the contestant picked Door 1. That s too bad, Mr. Hall said, opening Door 1. You ve won a goat. But you didn t open another door yet or give me a chance to switch. Where does it say I have to let you switch every time? I m the master of the show. Here, try it again. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 10 / 29

11 The Monty Hall Problem The Monty Hall Problem: Is It a Game? Or Hall offered money to make players change their strategy: On the second trial, the contestant again picked Door 1. Mr. Hall opened Door 3, revealing a goat. The contestant was about to switch to Door 2 when Mr. Hall pulled out a roll of bills. You re sure you want Door No. 2? he asked. Before I show you what s behind that door, I will give you $3,000 in cash not to switch to it. I ll switch to it. Three thousand dollars, Mr. Hall repeated, shifting into his famous cadence. Cash. Cash money. It could be a car, but it could be a goat. Four thousand. I ll try the door. Forty-five hundred. My last offer: Five thousand dollars. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 11 / 29

12 The Monty Hall Problem The Monty Hall Problem: Is It a Game? Let s open the door. You just ended up with a goat, he said, opening the door. As one can see, games are not only subject to chance, but also to the psychology of the players involved, in particular in games with incomplete information where opponents can bluff. The game where bluffing and other psychologic feints are part of the strategy repertoire is poker. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 12 / 29

13 Bayesian Games: Games with Incomplete Information Borel and von Neumann were fascinated by poker that motivated them to explore the laws of gambling by mathematical models. Poker is different from all previously considered games. It is a game with incomplete information: There is uncertainty not only about which hands your opponents hold, but also about which types of players there are. One gets to know one s opponents and their gambling tactics better and better during a long poker night. Conversely, one also gives away information about one s own way of gambling, which might help the other players to better assess them. When playing poker, we constantly observe our opponents and try to learn from their behavior. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 13 / 29

14 Bayesian Games: Games with Incomplete Information That is why poker is a so-called Bayesian game: Players make assumptions about the behavior of their opponents and intuitively determine the probabilities of their future actions. To poker successfully and to maximize one s gains, if one is a great bluffer, it is not at all necessary to have a lucky hand. All that matters is to be able to bluff well enough. I think my opponent will bluff. But because she thinks that I know she will bluff, she doesn t bluff. But since I think that she knows this, I do not actually believe she will bluff. But because she in turn thinks that I think that she thinks that... J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 14 / 29

15 Bayesian Games: Games with Incomplete Information Harsanyi developed a method for how to deal with such an infinitely nested chain of thoughts about one s opponents: Interrupt this chain by the assumption that all players know the roles or types of their opponents after a certain number of moves. They are then common knowledge. He proposed a transformation to obtain games with complete, even though imperfect, information from games with incomplete information by letting nature (randomness) come into play as an additional player and by letting the gains of the other players depend on the unknown random moves of nature. By this method players can learn during a game by observing the actions of their opponents. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 15 / 29

16 Games with Complete vs. Games with Perfect Information The notions of complete information and perfect information are similar, but not identical. In games with complete information, every player has complete knowledge of the structure of the game (whose turn it is when to move and which strategies are then available to this player) and of the gain functions of all players, yet the players do not necessarily know which actions the other players have performed so far or are performing now. In games with perfect information (e.g., chess or Tic-Tac-Toe), every player in addition knows also in each of her moves which actions the other players have performed so far, i.e., in each of her moves she knows her exact position in the game tree. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 16 / 29

17 Bayesian Games: Games with Incomplete Information Bayesian games like poker are games with incomplete information. The players don t know the hands of their opponents. This concealed information requires them to estimate the probabilities about the expected moves of their opponents. But the special challenge in poker (as in every Bayesian game) is that the players initially don t even know which types of players are sitting at the table: a risk-averse player, a risk-neutral player, or a risk-loving player? J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 17 / 29

18 Von Neumann s Simplified Poker Variant: Rules Two players, Belle and David, randomly draw a card from an infinite deck of cards, each card showing an arbitrary real number between 0 and 1. We assume that each card can be drawn independently and with the same probability. $ 1 $ 1 Before the game, both players pay one dollar into the jackpot. Card between 0 and 1 J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 18 / 29

19 Von Neumann s Simplified Poker Variant: Rules Now the betting stage starts. Belle moves first. She can: 1 either pass on, that is, she pays no further money into the jackpot, 2 or raise, which means that she pays another dollar into the jackpot. Then it is David s turn. His actions depend on Belle s strategy: 1 Suppose that Belle has passed on. Then David has no choice and must call, which means he doesn t have to pay money into the jackpot, as Belle hasn t paid anything either. Now it comes to the showdown: Both players show their cards and whoever has the card with the greater number wins the jackpot. 2 Suppose that Belle has raised. Then David does have a choice: 2(a) either David folds with a bad hand in this case Belle wins the jackpot without her or David needing to show their cards, 2(b) or David calls in this case he pays the same amount as Belle (another dollar) into the jackpot, and it comes to the showdown: The greater card wins. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 19 / 29

20 Von Neumann s Simplified Poker Variant: Rules passes on $ 1 raises folds calls $ 1 must call showdown greatest card wins J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 20 / 29

21 Analyzing a Simplification of von Neumann s Poker Variant Binmore proposed an even simpler variant of von Neumann s original simplified poker: Belle and David draw their cards not from an infinite deck of cards, but they can choose among exactly three cards with the values 1, 2, and 3. Except for this change, the rules of the game are identical. We assume that Belle and David both are risk-neutral players and they both believe that the other player is risk-neutral. The three cards (1, 2, and 3) are now shuffled, and we obtain the following six possibilities for arranging the cards in the deck: J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 21 / 29

22 Analyzing a Simplification of von Neumann s Poker Variant Every arrangement occurs with a probability of 1 /6. Belle obtains the card on top and David the card in the middle. We may assume that David in case he has a choice because Belle raises will fold with the 1 on his hand, since he would only lose more money by calling. It is also clear that he will call whenever he wins for sure, i.e., whenever he has the 3 on his hand. Therefore, we may further assume that Belle passes on (and does not raise) if she has the card 2 on her hand. Of course, Belle raises with the 3 on her hand. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 22 / 29

23 J. Rothe Figure: (HHU Düsseldorf) Game tree for the Algorithmische simplified Spieltheorie poker variant due to Binmore 23 / 29 Analyzing a Simplification of von Neumann s Poker Variant The game tree is too small! Shuffling and Dealing /6 1/6 3 1/6 1 1/6 3 1/6 1 1/ Belle: 1 Belle: 2 Belle: 3 passes passes passes passes passes passes on on on on on on raises raises raises raises raises raises David: 3 folds calls folds calls David: 1 folds calls folds calls +2 2 David: 2 folds calls folds calls

24 Analyzing a Simplification of von Neumann s Poker Variant The game tree is too big! Shuffling and Dealing /6 1/6 1/6 1/6 1/6 1/6 Belle: 1 Belle: 2 Belle: 3 passes on passes on passes on passes on passes on passes on raises raises raises raises raises raises J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 24 / 29

25 J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 25 / Shuffling and Dealing /6 1/6 1/6 1/6 1/6 1/6 Belle: 1 Belle: 2 Belle: 3 passes passes passes passes passes passes on on on on on on raises raises raises David: 3 folds calls folds raises raises calls 2 +2 David: 1 folds calls folds calls raises David: 2 folds calls folds calls Figure: Game tree for the simplified poker variant due to Binmore Analyzing a Simplification of von Neumann s Poker Variant Full House: Games with Incomplete Information

26 Analyzing a Simplification of von Neumann s Poker Variant draws a card with value 2 1/6 1/6 raises with prob. p 1 passes on with prob. 1-p 3 raises with prob. 1 folds calls must call folds calls gain: -1 gain: +2 gain: +1 gain: -1 gain: -2 Figure: Belle bluffs with probability p so as to make David indifferent J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 26 / 29

27 Figure: David calls with probability q so as to make Belle indifferent J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 27 / 29 Analyzing a Simplification of von Neumann s Poker Variant raises 1 passes on folds with prob. 1-q 1/2 1/2 2 calls with prob. q 3 calls with prob. 1 gain: -1 gain: +1 gain: -2 gain: -2

28 Bayesian Nash Equilibrium Recall that a strategy profile is in a Nash equilibrium if every strategy in this profile is a best response to the other strategies in the profile, i.e., they guarantee each player a highest possible gain, provided the other players stick to their strategies from the equilibrium profile. In a Bayesian game, the influence of randomness supervenes: Rational players seek to maximize their expected gain, depending on their beliefs about the other players (viewed as a probability distribution over the possible player types). In a Bayesian game, a Bayesian Nash equilibrium for risk-neutral players is defined to be a strategy profile that maximizes the expected gains of all players depending on what each player believes about the strategies chosen by the other players. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 28 / 29

29 Bayesian Nash Equilibrium Bayesian Nash equilibrium for the simplified poker variant: Belle bluffs with probability 1 /3 whenever she holds the 1, and David calls with probability 1 /3 whenever he holds the 2 and she raises. In principle, this result is meaningful for the more general variants of this game as well, such as for von Neumann s simplified poker game or the more common poker variants with Full House, Flush, etc. For example, beginners often make the mistake to bluff only with a relatively good hand. Moreover, it is important for poker to bluff frequently enough, or otherwise one will stay way below one s expected gains. On the other hand, one should not exaggerate bluffing. J. Rothe (HHU Düsseldorf) Algorithmische Spieltheorie 29 / 29