Game theory question

[MagicMon]

I'm re-reading Sklansky's GETTING THE BEST OF IT and I was wondering if I could get some clarification on a section of the book.

Pg. 126 - Poker, Bluffing, and Game Theory

"Consider a second example. Once again my opponent and I write down a 1 or a 2 without showing what we have written to the other. But this time the payoffs are a little different. If we match, I win $20. If we don't match, I pay him $10 if I wrote 1 and he wrote 2. If I wrote 2 and he wrote 1, I pay him $30. Once again this appears to be an even game, and in fact it would be if I wrote as many 1's as 2's. However, Game Theory tells me to use a different mixed strategy. It says to write 1 five-eights of the time (randomly, of course). By playing this way, I assure myself of a profit in the long run. ... Thought it may not be immediately apparent, the fact that this strategy earns me $1.25 per decision when he uses one strategy or the other exclusively will assure me of an average of $1.25 per decision no matter how he mixes up his play."

Can some expand upon this? What if the other player decides to employ the same strategy. My math is a little fuzzy, but to me, if both players employ the same optimal strategy, isn't this a zero sum game? How is the 5/8 / 3/8 calculated? Thanks in advance!

MagicMon

[Mike]

I will take a shot at this but someone will probably give you a better explanation. The extra 1/8 is your edge.

What gives you the advantage also is it doesn't matter what you opponent does. Your plays are random. If his are not that puts him at a worse disadvantage. Your opponent can not always correctly guess random events. The important idea is you have that little edge and you are more or less random in your play. Your opponent can not guess correctly more than half the time so over time you win.

[BarronVangorToth]

Let's run a few scenarios:

Let's say you wrote down

1
1
1
1
1
2
2
2

and I wrote down 1 all 8 times....

You would match the first five times, winning $100.
You would have 2 to my 1 three times, losing $90

You win $10

Same scenario....

I write 2 every time this time.

You'd mismatch the first five times, losing $50
You'd match the next three times, winning $60

You'd win $10 again

So obviously if I'm player 2, I can't write down any given number the whole way as I'm losing $10 to you no matter what.

Let's say you keep your say 1x5 and then 2x3 rundown, this time I alternate between 1,2,1,2,1,2 etc etc

This time we match 5 times, earning you $100

I get the 1-2 match-up twice, making me $20
I get the 2-1 match-up one, making me $30

You just made another $50

Next I try 2,1,2,1 etc etc

You get 3 match-ups, making you $60

I get 3 1-2's, making me $30
I get 2 2-1's, making me $60

This time ... you lost $30


So if I write the same number down the whole way, no matter which number, I'm losing $10 -- bad, Barron!

If I alternate, I have a 50% chance of losing $50 or a 50% chance of making $30

Would you flip coins for $50 vs. $30?

So I'm an overall loser randomly -- if we run the exercise twice, I lose $20 if I'm random ... and I lose $10 each time if I am not random, so I'd lose $20...

No matter what you do, you lose $10 on the run-through

/8 = $1.25 each time

Make more sense?

Barron Vangor Toth
www.BarronVangorToth.com

[Bad Lobster]

Because you don't have to use the same strategy, and the optimal strategy is not necessarily the same for both players. Your optimal strategy is the one that gives you the best expectation WHEN HE'S USING HIS OPTIMAL STRATEGY, which is not the same thing as the one that gives you the best expectation when he's using the same strategy as you.

Given the rules of this game, clearly you have an incentive to write more 1's, so it's not surprising that the results are not symmetrical.

[BarronVangorToth]

What strategy would you employ, then, if you know my optimal strategy is to have 5 of my picks be 1 and 3 of my picks be 2?

Barron Vangor Toth
www.BarronVangorToth.com

[timmer]

Game theory is a branch of logic which deals with cooperation and conflict in the context of negotiations and payoffs. The theory of games can elucidate the incentive conditions required for cooperation, can aid understanding of strategic decisions of nations or actors in conflict, and can help in the development of models of bargaining and deterrence

The Prisoner's Dilemma is the classic game in game theory literature. It centers on a game in which both actors would be better off cooperating, but both have an individual incentive to defect (not to cooperate) and as a result the likely outcome is one which is worse for both players than had they cooperated.

http://www.brunel.ac.uk/research/AI/alife/ga-axelr.htm

http://www.xs4all.nl/~helfrich/prisoner/

timmer

[jtr]

[ QUOTE ]
What strategy would you employ, then, if you know my optimal strategy is to have 5 of my picks be 1 and 3 of my picks be 2?

[/ QUOTE ]

Assuming I'm not allowed to refuse to play, I'd be utterly ambivalent about what strategy to employ as I am going to lose $1.25 per game (on average) no matter what. Once I knew that you'd committed to the 5/8 x 1, 3/8 x 2 strategy, I'd probably just write "1" all the time as it's easier. [img]/images/graemlins/tongue.gif[/img]

The beauty of it is that once I know you're using the optimal random strategy, my average payout doesn't change no matter what I do.

[pzhon]

[ QUOTE ]
If we match, I win $20. If we don't match, I pay him $10 if I wrote 1 and he wrote 2. If I wrote 2 and he wrote 1, I pay him $30. Once again this appears to be an even game, and in fact it would be if I wrote as many 1's as 2's.

[/ QUOTE ]
Side point: This is wrong. If your frequency of 2s is 50%, he should write 1 every time and average $5.

[ QUOTE ]
What if the other player decides to employ the same strategy. My math is a little fuzzy, but to me, if both players employ the same optimal strategy, isn't this a zero sum game?

[/ QUOTE ]
This is a zero-sum game. Zero-sum does not mean symmetric. It means one player's loss is the other's gain. There is no reasonable cooperation, unlike many win-win situations from real life. This game is not symmetric, as the payoffs are different: One player can only win $20, but loses different amounts, while the other player can only lose $20, but wins different amounts. It may not be obvious who has an advantage in a zero-sum game.

In this game suppose you write a 2 with probability p, and your opponent writes a 2 with probability q. On average, you win

20pq + 20(1-p)(1-q) - 10(1-p)q - 30(1-q)p = 20 - 50 p - 30 q + 80 pq.

That is not symmetric under the possible symmetries of switching p with q or switching p with 1-q, so there is no reason to suspect the players' best strategies are identical or opposites.

[MagicMon]

What great feedback...thanks, I have been enlightened!

[jdl22]

In Game Theory you are looking for an equilibrium set of strategies. What that means is that if we are playing a game I am playing a best counterstrategy to what you're playing and you are playing the best response to what I'm playing. In actually finding and calculating these things you simply find a strategy that the other player cannot exploit. For an example read the game theory bluffing chapter in TOP but the gist is that you bluff often enough so that the other player gets the same EV from folding and calling.

Let's look at this example. If we match I win 20. If we don't match I lose 10 if I wrote 1 and 30 if I wrote 2. Let's say I write 1 with probability p and 2 with probability 1-p.

The EV for him of writing 1 is: -20p+30(1-p)=30-50p
Of 2: 10p-20(1-p)=30p-20.

As I said you need to set p so that he can't take advantage of your play so set them equal.

30p-20 = 30-50p
80p = 50
p = 5/8.

His correct play is to make you indifferent between 1 and 2 as well. Suppose he plays 1 with probability q and 2 with probability 1-q.

EV of 1 = 20q-10(1-q) = 30q-10
EV of 2 = -30q+20(1-q) = 20-50q

Set them equal:
30q-10=20-50q
80q=30
q=3/8

So in equilibrium he will play 1 3/8 of the time while you will play 1 5/8 of the time.