David McAdams sends this along:

I’ve created a fun and simple game-theory problem that I thought you might enjoy … This is the sort of problem you could give undergrads to find out who are the really bright ones. It might also be fun to mention (or play) in class.Problem: Find the (unique) symmetic equilibrium of “The World’s Simplest Poker Game”, played as follows:**0** two players**1** each player pays ante of $100**2** each player receives ONE card, which we can think of as independent random numbers on [0,1]**3** each player SIMULTANEOUSLY decides whether to “raise” $100 or “stay”**4A** if one player raises and the other stays, the raiser wins the pot, for net gain +$100**4B** if both raise, the players show their cards and whoever has the highest card wins for net gain +$200

**4C** if both stay, the players show their cards and whoever has the highest card wins for net gain +$100If you decide to solve this problem, please let me know how long it takes you … I’m curious how immediately obvious the answer is to you :) I have solved it myself and, I can tell you, the answer is simple and elegant.Cheers,David

N.B. My answer based on 5 minutes of thinking was wrong. I will post David’s solution over the weekend.

**Update: **As promised, here is David’s solution. Looks like Keith was the first to post the correct answer in the comments and thanks to Nicolas for pointing out that this example appeared in von Neumann and Morgenstern.

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November 20, 2012 at 3:03 am

JanI think I have it. I’d like to post it, but I don’t want to spoil anyone. Took me around 3-4 minutes, during which I drew the game both in normal and tree form. I look forward to the official solution – what throws me off is that you were wrong… and I think you would arrive at the same solution.

November 20, 2012 at 9:00 am

José Antonio Espín SánchezThis is just an entry game with asymmetric information, right? The first $100 play no role here. The second $100 are the cost of entry. Your payoff in the second stage depends on your opponent’s type, but in a discontinuous form (winner takes all).

I will not spoil the solution either, but Alvarro Parra and I wrote a paper showing that all the equilibria in entry games of this type have the same form.

BTW. The game is way more interesting when one of the players know his opponent card (type), this being common knowledge. Think of this game as one in which one player is a cheater, and this is common knowledge.

November 20, 2012 at 9:29 am

NicolasThis game of poker is exactly the one analyzed by Von Nerumann and Morgenstern in the Theory of Games and Economic Behavior 1953. (p 203 for the solution) The method they use is very interesting for historical reasons.We would not look at the problem this way nowadays.

November 20, 2012 at 6:51 pm

RajivNicolas, thanks for this pointer. The discussion in von Neumann and Morgenstern is definitely worth a look. They consider a more general game in two respects: the high and low bids are arbitrary (a and b instead of 200 and 100) and the low bidder does not automatically lose but can choose to raise after the initial bids have been revealed. They show that such delayed raises never occur in equilibrium. Most interestingly, as the high bid gets large relative to the low bid, the probability of bluffing goes to zero. This can’t be seen in the example here because the bids are fixed at 200 and 100 respectively. Partial preview available here:

http://books.google.com/books?id=_aIGYI-jGEcC

November 22, 2012 at 7:27 pm

JonFull text of Von Neumann and Morgenstern is online at the Internet Archive / Million Books Project http://archive.org/details/theoryofgamesand030098mbp.

November 20, 2012 at 10:57 am

JasonWell, not sure what the comments are for if NOT for discussing the solution, so reader-beware, spoilers and/or confusing errors ahead.

.

.

I don’t know the solution, but here are my thoughts:

Can we “evolve” towards the right answer from symmetric extremes?

Always fold is an obvious loser. Always call is another obvious loser. Always raise? If I detect my opponent is always raising, I can never call against him, but it might make sense to fold if my hand is bad. If I’m holding x, my opponent (who always raises) has a (1-x) probability of beating me if I also raise.

If I sometimes fold, my opponent will detect that he’s winning a small pot when I fold, and disproportionately losing large pots when I raise. The expected return for the “always raise” (AR) is x*(payout 1) + (1-x) * (2*payout 1). x+2-2x = 2-x. But x is [0,1], so the expected payout for AR is always >1.

Therefore I think the symmetric solution is to always raise.

Thoughts?

November 20, 2012 at 11:08 am

erikIf my opponent’s strategy is always raise, and if my draw is x, then my payoff from staying is zero with certainty, while my expected payoff from raising is 200x – 100(1-x) = 100(3×-1), which is negative for x<(1/3). Thus always raise is not a symmetric equilibrium strategy.

November 20, 2012 at 11:12 am

JasonCan the symmetric strategy not be negative? You’re both starting with 100 in the pot

November 20, 2012 at 11:32 am

DRDRJason, I don’t follow you, but always raise is clearly dominated. If you hold card x, and play against an opponent who always raises, the expected payout for raising is (+2)*(x)+(-2)*(1-x) = 4×-2, while the payout for staying is 0, so always raising is dominated by a strategy where you stay when your card x<.5.

November 20, 2012 at 11:35 am

DRDR(Erik, when you lose when both raise, the payoff is -200 not -100. Also, note I omitted the 100s in my reply)

November 20, 2012 at 11:43 am

erikDNDR, you are right about my error, but the payoff from staying is -1 when your opponent always raises, not 0. Thus it’s strictly better to raise iff

4×-2>-1, or x>(1/4).

November 20, 2012 at 11:51 am

DRDRoops, thanks.

November 20, 2012 at 12:29 pm

ELFun question!

~~~~~~~~~~~~

A hint for people:

A tractable way to express a symmetric equilibrium is through a function F on [0,1], where F(t) is interpreted as the probability that your opponent is of type below t AND will raise. So F weakly increasing and weakly contractive (informally, of “slope never exceeding 1″), and F(0) = 0.

More of a spoiler, on pinning down the form of F:

Can it have slope zero anywhere? What happens if it ever has slope 1? More generally, what slopes can it have for “more than an instant”?

November 20, 2012 at 1:03 pm

DRDRFor a first step, I’d suggest characterizing the best response if you draw a card on either endpoint. Then think about what the equilibrium must look like in the neighborhood of each endpoint, and so on.

November 20, 2012 at 7:41 pm

AntonioNice problem and thanks for posting!

Perhaps this is nitpicking but it would be more appropriate to say “uniform independent random numbers on [0,1]” rather than just “independent random numbers on [0,1]”. I personally found that confusing.

November 21, 2012 at 5:44 am

Peter PanPr(Raise/type=t) = t

November 21, 2012 at 9:40 am

AnonymousNot an equilibrium. Types t > 0.99 would have strict incentive to raise.

In fact, types t>1/3 would have strict incentive to raise, and types t<1/3 would have strict incentive not to.

November 21, 2012 at 9:51 am

KeithRaise w/ probability 1/3 if t<1/2, always raise otherwise. My derivation isn't anything close to "simple and elegant", though (which may indicate that I'm wrong…)

November 21, 2012 at 12:32 pm

AnonymousConfirmed.

I guess this is only essentially unique, since you could arbitrarily change the action for a measure zero set of types <= 1/2.

November 21, 2012 at 2:21 pm

RajivYes, this is the von-Neumann Morgenstern solution for the parameters specified here. And as the high bid rises (holding the low bid constant) the threshold above which you raise with probability 1 rises, and the probability of raising below the threshold (bluffing) falls. The limiting solution is a bit counterintuitive: all types bid low with probability 1.

November 22, 2012 at 2:44 am

DRDRRajiv, I find the limiting case intuitive. You clearly bluff less as the cost of bluffing rises. Also, in the limiting case you’re close to the game restricted to pure strategies, in which case you always benefit from raising slightly less aggressively than your opponent (you save a first-order amount from avoiding losing when raising, while forgoing the wins when raising is 2nd order.)

And to be clear, you do still raise when you draw a 1, though it’s a measure zero outcome.

November 22, 2012 at 5:26 am

RajivActually, you are indifferent between the two actions if you draw a 1 (in the limiting case). You get 100 either way, since nobody else is raising.

But you’re right that it’s intuitive once you think about it. I just wouldn’t have guessed it. Von Neumann and Morgenstern argue that there are bounds to raising in real world poker because unbounded raising would eliminate bluffing. But it turns out that unbounded raising also eliminates raising.

November 26, 2012 at 2:14 am

Danny LynchIt was counterintuitive to me that you would just raise randomly below 1/2, instead of bluffing the worst cards in the range (i.e. everything below 1/6).

November 26, 2012 at 9:07 am

JasonSince bets are made simultaneously, bluffing is far less valuable.

November 21, 2012 at 12:17 pm

EnriqueI am going to review the Von Neuman-Morgenstern solution — in the meantime, I wonder if the correct solution depends on the value of one’s hole card and thus involves a mixed strategy or a probabilistic strategy?

November 22, 2012 at 8:52 pm

MoPositiveWhat happens on a tie?

November 23, 2012 at 6:41 am

How would you play the World’s Simplest Poker Game? « Close to the Edge[…] to Jeff Ely and David McAdams – I’m still thinking about this one. Anyone have a good […]

November 24, 2012 at 7:51 pm

BerangStay if F^-1(number)<1/6

raise o.w.

November 26, 2012 at 4:35 pm

EnriqueDavid’s beautiful solution (and mathematical notation) is certainly more elegant (i.e. more simple) than von Neumann & Morgenstern’s, but I personally prefer how von Neumann & Morgenstern abstract the problem, using the variables a and b to represent “high” and “low” bids respectively, and also, their model allows the low bidder to change his bid to high after the players have made their opening moves