Todd and Steven Divide the Estate Problem 8.10

Bargaining over 100 pounds of gold Round 1: Todd makes offer of Division. Steven accepts or rejects. Round 2: If Steven rejects, estate is reduced to 100d pounds. Steven makes a new offer and Todd accepts or rejects. Round 3: If Todd rejects, estate is reduced to 100d 2 pounds. Todd makes new offer and Steven accepts or rejects. If Steven rejects, both get zero.

Working backwards for SPNE In last subgame, Steven must either accept or reject Todd’s offer. If he rejects, both get 0. If he accepts, he gets what Todd offered him. If Todd offers any small positive amount ε, Steven’s best reply is to accept. So in next to last subgame, Todd would offer Steven ε and take 100d 2 -ε for himself.

Part of game tree Steven Todd Steven Propose Accept Reject

Back one more step At node where Steven has offered Todd a division, there are 100d units to divide. Todd would accept 100d 2 or more, would reject less. So at previous node Steven would offer Todd 100d 2 and would have 100(d-d 2 ) for himself.

Back once more Now consider the subgame where Todd makes his first proposal. At this point there are 100 pounds of gold to divide. Todd sees that Steven would accept anything greater than 100(d-d 2 ). So Todd would offer Steven 100(d-d 2 )+ε and keep 100(1-d+d 2 )-ε for himself.

SPNE Equilibrium strategy for Todd First node: Offer Steven 100(d-d 2 )+ε Second node: If Steven rejects Todd’s offer and makes a counteroffer to Todd: Accept 100d 2 or more, reject less. Third node: If Todd rejects Steven’s counter offer, make a new offer to Steven of a small ε.

SPNE strategy for Steven First node: Accept any offer greater than 100(d-d 2 ), reject smaller offers. Second node: If Steven rejects Todd’s first offer, then offer Todd 100d 2 Third node: Accept any positive offer.

Payoffs Suppose d =.9, then 100(1-d+d 2 )=91 If d=.5, 100(1-d+d 2 )= 75 In fact, 100(1-d+d 2 ) is minimized at d=.5. What happens with more rounds of bargaining?

Bayes-Nash equilibrium with Incomplete Information

An oil lease auction

The set-up You own an oil company. A new field has come up for lease. There are two bidders. You and another firm. Each of you has explored half of the oil field and knows the value of the half they explored. The value of each side is either $3 million or 0, which depended on the flip of a fair coin. Total value of field is the sum of the two sides You know what your side is worth, but not the other company’s side.

The Auction The lease for the entire field is up for auction. A bid must be an integer number possibly 0) of million $. There are two bidders, you and the company that explored the other side. You know what your side is worth. Entire field will be leased to the higher bidder in a sealed bid auction. If there are tie bids, winner is chosen by coin flip. If you win the auction, your profit or loss is the value of the total field minus your bid.

A strategy A strategy states the amount you will bid if your side is worth $0 and the amount you will bid if your side is worth $3 million.

What would you bid if your side is worth $0? A) $0 B) $1 million C) $2 million D) $3 million E) $4 million

What would you bid if your side is worth $3 million? A) $1 million B) $2 million C) $3 million D) $4 million E) $5 million

Some things to think about What would be your expected profit if the company you bid against uses the same profit that you do? If your side is worth $0 and you win the auction, what do you expect the total oilfield to be worth?

Finding a symmetric Bayes-Nash equilibrium Suppose both players bid higher when they see $3 million than when they see $0. What is the best bid if you see $0?

Is (0,4) a symmetric Bayes-Nash equilibrium? suppose other guy bids 0 when he sees 0 and $4 million when he see $3 million on his own side. My expected profit from the (0,4) strategy would be ½ x0+ ½(-1 x ½ +2 x ½)=1/4. If instead I played (0,1), my expected payoff would be ½ x0+ ½(0 x ½ +2 x ½) =1/2 So (0,4) is not a symmetric Bayes-Nash equilibrium.

Is (0,3) a symmetric Bayes-N.E? If other guy is playing (0,3), the my expected payoff from (0,3) is ½x0 +½ (0+½ x½x3)=3/8. If I play (0,1), my expected payoff is ½x0+ ½(0 x ½ +2 x ½) =1/2

Alice and Bob Revisited

She loves me, she loves me not? ( Bob moves before Alice) Go to A Go to B Go to A Alice Go to B Go to A Go to B She loves him Nature She scorns him Go to A Go to B Bob Alice Bob Alice

What are their strategies? For Bob – Go to A – Go to B Alice has four information sets. – I Love him and he’s at A – I Love him and he’s at B – I Scorn him and he’s at A – I Scorn him and he’s at B In each information set, she can go to either A or B. This gives her 2x2x2x2=16 possible strategies.

A weakly dominant strategy for Alice Go to A if you love he goes to A. Go to B if you love him and he goes B. Go to B if you scorn him and he goes to A. Go to A if you scorn him and he goes to B. (We write this as A/B/B/A) This is weakly dominant but not strictly dominant. Explain. Let’s look for a subgame perfect Nash equilibrium where Alice goes A/B/B/A

Checking equilibrium Suppose Alice goes where Bob is if she loves him and goes where he is not if she scorns him. (A/B/B/A) Payoff to Bob from A is 2p. Payoff from B is 3p+1(1-p)=2p+1. Since 2p+1>2p, for all p>=0, B is his best response to (A/B/B/A). Also A/B/B/A is a best response for Alice to Bob’s B. So we have a Bayes-Nash equilibrium.

Does she or doesn’t she? Simultaneous Play Go to A Go to B Go to A Alice Go to B Go to A Go to B She loves him Nature She scorns him Go to A Go to B Bob Alice Bob Alice

Alice’s (pure) strategies Alice doesn’t know what Bob did, so she can’t make her action depend on his choice. She can go to either A or B. She does know whether she loves him or scorns him when she chooses. She has 4 possible strategies – A if love, A if scorn – A if love, B if scorn – B if love, A if scorn – B if love, B if scorn

Bayes’ Nash equilibrium Is there a Bayes’ Nash equilibrium where Bob goes to B and Alice goes to B if she loves Bob, and to A if she scorns him? – This is a best response for both Alice types. – What about Bob?

Bob’s Calculations If Bob thinks the probability that Alice loves him is p and Alice will go to B if she loves him and A if she scorns him: – His expected payoff from going to B is 3p+1(1-p)=1+2p. – His expected payoff from going to A is 2(1-p)+0p=2-2p. Going to B is Bob’s best response to the strategies of the Alice types if 1+2p>=2-2p. Equivalently p>=1/4.

Is there a Bayes-Nash equilibrium in pure strategies if p<1/4? A)Yes, Alice goes to B if she loves Bob and A if she scorns him and Bob goes to B. B)Yes, Alice goes to A if she loves Bob and B if she scorns him and Bob goes to B. C)Yes there is one, where Alice always goes to A. D)No there is no Bayes-Nash equilibrium in pure strategies.

If p<1/4 We showed there is no pure strategy Bayes-Nash equilibrium where Bob goes to B. What if Bob goes to A? The only possible Nash equilibrium would have Alice go to A if she loves him and B if she scorns him. Then payoff to Bob from A would be 2p+0 and payoff from B would be 3p+1(1-p)=2p+1>2p. So if p<1/4, there can’t be a pure strategy Bayes- Nash equilibrium where Bob goes to A.

Mixed strategy equilbrium: Bob the stalker If Bob thinks it likely that Alice scorns him, then if he uses a pure strategy, he knows Alice would always avoid him. If he uses a mixed strategy, he would catch her sometimes. Let’s look for a mixed strategy for Bob such that Alice, if she scorns Bob would be indifferent between Movies A and B.

What about a mixed strategy equilibrium? If p<1/4, can we find a mixed strategy for Bob such that Alice is indifferent What if Bob knows Alice scorns him? Consider the Alice type who scorns Bob. If Bob goes to movie A with probability q, When will Alice be indifferent between going to the two movies?

The game if Alice hates Bob AB A1,23,1 B2,00,3 Bob Alice

Making Scornful Alice indifferent If Bob goes to Movie A with probability q and Alice Scorns Bob: – Alice’s payoff from A is 1q+3(1-q) =3-2q – Alice’s payoff from B is 2q+0(1-q)=2q – Alice will be indifferent if 3-2q=2q, which implies q=3/4.

When will Bob do a mixed strategy? Note that if Bob goes to A with probability ¾, and if Alice loves him, her best response is to go to Movie A. If there is an equilibrium where Bob uses a mixed strategy, he must be indifferent between going to A and going to B. Can we find a mixed strategy for Alice to use if she scorns him so that Bob will be indifferent between A and B?

Making Bob indifferent Let r be the probability that Alice goes to Movie A if she scorns Bob and suppose that Alice always goes to A if she loves Bob. Expected payoffs for Bob are – If he goes to A, 2p+(1-p)(2r+0(1-r))=2p-2pr+2r – If he goes to B, 1p+(1-p)(1r+3(1-r))=2p+3+2pr-2r – He is indifferent between A and B if these are equal. This implies r=(3-4p)/4-4p. Now r is between 0 and 1 if and only if p<=3/4

Summing up We previously found that if p>=1/4, there is a Bayes- Nash equilibrium in which Bob goes to B and Alice goes to B if she loves him and A if she scorns him. Now we found that whenever p<=3/4, there is a mixed strategy Bayes-Nash equilibrium in which Bob goes to movie A with probability ¾, Alice goes to Movie A if she loves Bob and she goes to movie A with probability r=(3-4p)/(4-4p) if she scorns him. So over the range of p between ¼ and ¾, there are two distinct Bayes-Nash equilibria.

Maybe, later?