class: center, middle, inverse, title-slide .title[ # Puzzles in Decision Theory ] .subtitle[ ## Fernando Alvear & Connor Copanas ] .date[ ### MSA 2024 ] --- # Agenda 1. Pascal's Wager 2. Monty Hall Problem 3. Sleeping Beauty Problem 4. Newcomb's Problem --- # Pascal's Wager (1670) .pull-left.w35[ <img src="assets/pascal.jpg" alt="" width="500"/> ] .pull-right.w60[ > <mark-red>Yes, but you must bet. It is not voluntary; you are already involved</mark-red>. Which will you choose then? Look: since you must choose, let us see which is the less profitable option. (...) Let us weigh up the gain and the loss in choosing heads, that God exists. Let us figure out the two results: <mark-blue style="animation-delay: 2s;">if you win, you win everything, and if you lose, you lose nothing. So bet that he exists, without any hesitation</mark-blue>. ] --- # Pascal's Wager - Two potential __actions__: - Believe in God or do not believe in God - Two possible __states of affairs__: - God exists or he does not. - Four possible __consequences__ of the actions, given states of affairs. - If you bet that God exists and he does, "_you win everything_." If you bet that God exists and he doesn't, "_you lose nothing_." --- # Pascal's argument (summary) > If God exists and I believe in God, I’ll go to heaven, which is infinitely good. If God exists and I don’t believe in God, I may go to hell, which is infinitely bad. If God does not exist, then whether I believe in God or not, whatever I’d gain or lose would be finite. So, I should believe in God. - If you believe in God... - ...__and God exists__, you go to heaven (positive infinite value). - ...__and God doesn't exist__, you lose a little (time going to Church, effort to not sin, etc.). - If you _don't_ believe in God... - ...__and God exists__, you go to hell (negative infinite value). - ...__and God doesn't exist__, you gain a little (time not going to Church, pleasure from sins, etc.). Given this information, _which action is the most reasonable to undertake_? --- # Should you bet? > Bet 1: If the coin comes up heads, you win $10. If you lose, you lose $1. Should you bet? > Bet 2: If the coin comes up heads, you win $1. If you lose, you lose $10. Should you bet? --- # Expected utility .shadow[ .emphasis[ __Expected utility__: The expected utility of an action is the sum of the action's utilities, weighted by the probabilities of each possible outcome. ] ] The _utilities_ are a measure of the consequences (negative or positive) of our actions. The _probabilities_ are a measure of how certain an outcome is. > Bet 1: If the coin comes up heads, you win $10. If you lose, you lose $1. $$EU(\text{bet 1}) = 0.5(\$10) + 0.5(-\$1) = \$5 - \$0.5 = \$4.5$$ $$EU(\text{no bet 1}) = 1(\$0) = $0$$ This means that, on average, for each play, if you bet, you should expect to win $4.5. --- # Expected utility > Bet 1: If the coin comes up heads, you win $10. If you lose, you lose $1. $$EU(\text{bet 1}) = 0.5(\$10) + 0.5(-\$1) = \$5 - \$0.5 = \$4.5$$ $$EU(\text{no bet 1}) = 1(\$0) = $0$$ This means that, on average, for each play, if you bet, you should expect to win $4.5. Suppose you bet ten times. You should expect to win $4.5 x 10 = 45. - Maybe you win 5 times, so you win $50 in total (5 times $10). - Maybe you lose 5 times, so you lose $5 in total (5 times $1). - Net gain would be: $45. --- # Expected utility > Bet 2: If the coin comes up heads, you win $1. If you lose, you lose $10. $$EU(\text{bet 2}) = 0.5(\$1) + 0.5(-\$10) = \$0.5 - \$5 = -\$4.5$$ $$EU(\text{no bet 2}) = 1(\$0) = $0$$ This means that, on average, for each play, if you bet, you should expect to _lose_ $4.5 Suppose you bet ten times. You should expect to lose -$4.5 x 10 = -$45 - Maybe you win 5 times, so you win $5 (5 times $1). - Maybe you lose 5 times, so you lose $50 (5 times $10). - Net gain would be: -$45. --- # Expected utility and rational choice These examples show that, intuitively, __it is rational to do what has greater expected utility__. Since in bet 1, `\(EU(\text{bet1}) > EU (\text{no bet 1})\)`, _you should bet_. Since in bet 2, `\(EU(\text{bet 2}) < EU (\text{no bet 2})\)`, _you should not bet_. .shadow[ .emphasis[ __Expected utility rule__: A rational agent always acts so as to maximize expected utility. ] ] --- # Pascal's wager > Let us weigh up the gain and the loss in choosing heads, that God exists. Let us figure out the two results: <mark style="animation-delay: 2s;">if you win, you win everything, and if you lose, you lose nothing</mark>. So bet that he exists, without any hesitation. `$$EU(\text{believe}) = 0.5(\infty) + 0.5(0) = \infty + 0 = \infty$$` - "_If you win, you win everything_" Why? - Idea that if you believe in God, behaving in accordance with God's commands, and God does exist, _you go to heaven_, and thus you experience infinite gain. - "_If you lose, you lose nothing_" Why? - Idea that if you believe in God, behaving in accordance with God's commands, and God does not exist, _nothing happens_. But maybe you lose something: you waste time behaving in accordance with the commands of a God that does not exist! --- # Pascal's wager - "_If you lose, you lose nothing_" Why? - Idea that if you believe in God, behaving in accordance with God's commands, and God does not exist, _nothing happens_. But maybe you lose something: you waste time behaving in accordance with the commands of a God that does not exist! Regardless, <mark>whatever you lose will be a finite value</mark>, which is insignificant compared with the infinite value of heaven. `$$EU(\text{believe}) = 0.5(\infty) + 0.5(10,000) = \infty - 10,000 = \infty$$` --- class: medium-font # Pascal's wager What happens if you don't believe in God? `$$EU(\text{not believe}) = 0.5(-\infty) + 0.5(0) = -\infty + 0 = -\infty$$` Suppose you don't believe in God, and thus do not behave in accordance with God's commands: - If God exists, you go to hell, which has negative infinite utility. - If God doesn't exist, nothing happens (or at worse, you lose a finite amount of utility). It seems that... `$$\begin{aligned} EU(\text{believe}) &> EU(\text{not believe}) \\ \infty &> -\infty \end{aligned}$$` So you should believe! --- # Pascal's argument (formal) 1. Believing in God has more expected utility than not believing in God. 2. A rational agent always acts so as to maximize expected utility. 3. Therefore, a rational agent would believe in God. .center[ <img src="assets/decision-table1.png" alt="" width="900"/> ] --- class: medium-font ## Objection: Many-Gods .pull-left.w35[ What if I believe in the _wrong_ God? - The possibilities are not only two, but many more: that the Christian God exists, that the Muslim God exists, that the Greek God exists, etc. - There might be a _weird_ God, one who punishes those who believe and rewards those who don't believe! - In this scenario, it's unclear how to calculate the utilities, so Pascal's strategy doesn't help to guide belief. ] .pull-right.w65[ <img src="assets/decision-table2.png" alt="" width="800"/> ] --- # Monty Hall problem > Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say A, and the host, who knows what's behind the doors, opens another door, say B, which has a goat. He then says to you, "Do you want to pick door C?" Is it to your advantage to switch your choice? --- # Answer 1: It doesn't matter `\(P(\text{prize is behind door A}) = 1/3\)` `\(P(\text{prize is behind door B}) = 1/3\)` `\(P(\text{prize is behind door C}) = 1/3\)` After host opening door 3... `\(P(\text{prize is behind door A}) = 1/2\)` `\(P(\text{prize is behind door B}) = 0\)` `\(P(\text{prize is behind door C}) = 1/2\)` --- # Answer 2: Switch! .pull-left[ - `\(2/3\)` of the time the host does not have a choice of which door to open (when the prize is not behind your first chosen door). - `\(1/3\)` of the time the host has a choice (when the prize is in behind your first chosen door). More often than not, the host must choose the only door that doesn’t have the prize. So it’s best for you (in the long run) to choose the door that the host didn’t open. So you should switch! ] .pull-right[ <img src="assets/montytree2-1.png" width="500"> ] --- class: small-font # Answer 2: Switch! .pull-left[ First, the probability that the prize is behind doors A, B, and C is `\(1/3\)` for each door, respectively. Second, the door that the host opens depends on your choice and where the prize is. Let’s say you choose door A. If the prize is in A, then the host can choose between opening doors B or C. Let’s assume the host does this randomly, so the probability that the host opens door B and that prize is in A is `\(1/3 \times 1/2 = 1/6\)`. Similarly, the probability that the host opens door C and that prize is in A is `\(1/3 \times 1/2 = 1/6\)`. If the prize is in B, then the host is obligated to open door C, so the probability that the prize is in B and host opens C is `\(1/3 \times 1 = 1/3\)`. If the prize is in C, then the host is obligated to open door B, so the probability that the prize is in C and host opens B is `\(1/3 \times 1 = 1/3\)`. ] .pull-right[ <img src="assets/montytree3-1.png" width="500"> ] --- class: small-font # Answer 2: Switch! .pull-left[ <img src="assets/montytree3-1.png" width="500"> ] .pull-right[ Now suppose you see the host opening door B. Then the only options available are that the prize is in A and host opened B (probability `\(1/6\)`) and that the prize is in C and host opened B (probability `\(1/3\)`). Since some options are eliminated, and the only open options are the two just mentioned, we need to normalize the probabilities so they add up to `\(1\)`. Doing this results in the following: - P(Prize in A given that host opens B) = `\(1/6 \times 2 = 1/3\)` - P(Prize in C given that host opens B) = `\(1/3 \times 2 = 2/3\)` Thus, you should switch! ] --- ## 100-door version .pull-left[ <img src="assets/montygrid-1.png" width="500"> ] .pull-right[ Suppose you chose door 1, and then host opened all doors except door 68. What is more probable? - That the prize is behind door 1. - That the prize is behind door 68. Should you switch to door 68? ] --- class: medium-font # Sleeping beauty problem .pull-left.w40[ > Sleeping Beauty is going to sleep for three days, during which time she will be woken up either once or twice. On Monday, a fair coin is tossed. If the coin comes up heads, she will be woken only on Monday; if tails, she will be woken on both Monday and Tuesday. If she is woken on Monday, she will be given a drug to put her back to sleep, which also causes her to forget that awakening. ] .pull-right.w60[ <img src="assets/sleeping-beauty-2.jpeg" width="700"> ] Now suppose that you are Sleeping Beauty, and you are woken up from your sleep. You know the above, and you know that you are being awoken on either Monday or Tuesday. What should you think is the chance that the coin flipped on Monday came up heads? --- # A half, of course! Sleeping Beauty knows that the coin is fair, and so also knows that there is a `\(1/2\)` chance that it comes up heads on any given throw, and a `\(1/2\)` chance that it comes up tails. She has learned nothing which makes her doubt these probabilities for the Monday coin toss; so she should still estimate that there’s a `\(1/2\)` chance that the coin came up heads. --- # A third, of course! Sleeping Beauty would be reasonable to believe that, were this experiment performed over and over again, she would have twice as many tails-awakenings as heads-awakenenings. So, given a random awakening, she should think that it is twice as likely that it be a tails-awakening as that it is a heads-awakening. So, she should think that the odds of heads having been thrown on any particular awakening of this sort is `\(1/3\)` . (Imagine us forcing Sleeping Beauty to bet on whether the coin came up heads on each awakening over a series of trials of the case. Wouldn’t she stand to do much better if she adopted as a hypothesis to guide her betting that `\(P(H1)= 1/3\)` ?) --- class: medium-font # Newcomb's problem > There are two boxes before you, Box A and Box B. You have a choice as to whether you can take only the contents of Box B, or can take the contents of Box A and Box B. The Predictor has placed $1000 in Box A. If the Predictor predicts that you will take only Box B, he has placed $1,000,000 in Box B. If the Predictor predicts that you will take the contents of both boxes, he has placed nothing in Box B. You’ve observed that, in the past, the Predictor is right every time. If your aim is to maximize your money, should you choose to take the contents of Box B alone, or the contents of both boxes? .center[<img src="assets/newcomb-1.jpg" width="700">] --- # One box, of course! You have no reason to think that your case is different than anyone else’s. So you should think that in your case as in the others, one-boxing will lead to a higher payout than two-boxing. Since the aim of the game is to maximize your money, you ought to one-box. Expected utilities: $$EU(\text{one box}) = 1(\$ 1,000,000) + 0(\$0) = \$ 1,000,000$$ $$EU(\text{two box}) = 0(\$1,001,000) + 1(\$1,000) = \$ 1,000 $$ .shadow[ .emphasis[ __Expected utility rule__: A rational agent always acts so as to maximize expected utility. ] ] --- # Two boxes, of course! __The money is already in the boxes__. If you one-box, you will get whatever is in box B, whereas if you two-box you will get this plus the $1000 that is in box A. So you know in advance that two-boxing will give you $1000 more than one-boxing, and hence you should two-box. .center[<img src="assets/newcomb-table2.png" width="700">] .shadow[ .emphasis[ __Dominance rule__: If one option dominates all others (in any circumstance the utilities of the option are never worse), then a rational agent always chooses the dominant option. ] ]