A game theory paradox puzzle

[ZFR]

So while analysing Mafia end game situations, I came across an interesting one that leads to a nice paradox: a player's expected payoff increases if he chooses a set of prizes all of which are smaller.

Consider the following simple game:

A box contains 3 white balls and 1 golden ball: WWWG.

Player 1 can make the following choice:

• Remove a white ball from the box OR
• Offer draw (in which case no balls are removed from the box).

Plyer 2 can then make the following choice:

• If a draw was offered he may choose to accept it. The game ends in a draw.
• If no draw was offered, or if he rejects the offer, he randomly selects a ball from the box. If he draws the golden ball he wins, otherwise Player 1 wins.

The prize is 1$ for winning, 0$ for losing and x$ for a draw, where x is in the range [0,1].



Question:
Assuming all players act rationally, for which value of x should Player 1 offer a draw?


So the solution I came up with is as follows:

• If player 1 removes a ball, his chances of winning are 2/3. Therefore, if x is larger than 2/3 he should offer draw, which he knows will be accepted. His expeted payoff will be x.
• If x is smaller than 2/3, but larger than 1/4 he should remove a white ball. He will than have 2/3 expected payoff.
• However if x is smaller than 1/4, then he should offer draw! Because he knows that Player 2 will reject the offer, and so he will have 3/4 chance of winning! Player 2 choice is simple: he's going to accept any offer where x is greater than 1/4 and reject otherwise to maximise his own expected payoff.

So, Player 1 should offer draw if x ∈ (0, 1/4) ∪ (2/3, 1]


Between 0 and 1/4 his expected payoff is 3/4. Betwwen 1/4 and 2/3, expected payoff is 2/3. Between 2/3 and 1, his expected payoff is equal to that value.


Which makes a nice paradox. Suppose the organiser of the game comes to Player 1 and tells him: You can have one of those sets of prizes:

• 1 for winning, 0.5 for drawing. OR
• 0.9 for winning, 0.2 for drawing.

Please choose one of them, the rules won't change otherwise.


Then Player 1 should choose the second option to maximise his expected payoff, despite both values being smaller!
In the first case, his expected payoff is 0.6666.....
In the second case, his expected payoff is 0.9 x 0.75 = 0.675


I was wondering if someone could go over this problem and tell me if I made any mistakes. And also if this is an example of a known paradox in game theory, and if so what is it called?

[uau]

Originally Posted by ZFR

Which makes a nice paradox.

I don't think there is anything particularly paradoxical here. This is not a zero-sum game unless x = 1/2. It's not surprising that changes which result in player 2 switching strategy may help player 1, whether those changes are generally in direction of smaller prizes or not.

[LaurV]

Say we play the following (roulette-style) game: You draw a circle using 2 colors, say red and green, it doesn't matter if the two colors are contiguous or not, you can make it red, green, red, green, etc, or just three quarters red and one quarter green, etc, up to you. So, if we draw an imaginary radius, it hits either red circle, or green circle, but not both. The proportion of red and green is up to you. Then we put a rotating arrow in the middle of the circle and play the following game: I put money on the table, you rotate the arrow. If the arrow stops on the color my money is, and the color is dominant (more than half of the circle has that color), I take back my money and you pay me the same amount of money. If the arrow stops on the color my money is, and the color is not dominant (half or less of the circle has that color), I take back my money and you pay me the double amount of money I did bet. If the arrow stops on the color my money is not, you take my money.

We can use the same circle to play many games, or you can re-draw the circle for every spin, up to you, but you must use 2 colors each time, you can't draw a monochrome circle, and you have to be careful with the splitting (continuous or not, it doesn't matter). Clearly, if you split it half-half, you will always lose.

How do you draw the circle to maximize your chances?

[retina]

Originally Posted by LaurV

How do you draw the circle to maximize your [profit]?

Keep your crayons in your pocket.

[Uncwilly]

Originally Posted by LaurV

If the arrow stops on the color my money is,

What color your money is is a big unknown.

[Dr Sardonicus]

Originally Posted by LaurV

How do you draw the circle to maximize your chances?

You draw the circle so that if the arrow lands on green, your money is suddenly in 20-baht notes, and if it lands on red, your money is suddenly in 100-baht notes.

[Uncwilly]

Originally Posted by Dr Sardonicus

You draw the circle so that if the arrow lands on green, your money is suddenly in 20-baht notes, and if it lands on red, your money is suddenly in 100-baht notes.

Canadian 1's were green and the 2's were red.

[Dr Sardonicus]

Draw the circle so that you (claim you) are red-green colorblind. Then you (claim you) win every time.

[LaurV]

Haha, this thread unexpectedly took an odd turn, due to my bad English and phrasing

The idea of my puzzle was to see how many of you think out of the box, and/or dream about "gaming the system"

Every time I tell such puzzles (or similar) at parties, etc., people start arguing that one quarter red and three quarters green is better, or two thirds red and one third green is better. Or something. Some even come with complicate theory to explain why. Common sense tells us that the chances are equal if you split it in thirds, because the area where you get double the money is half of the area where you get the money one time only. But math doesn't agree, and this is definitively a game with supra-unitary gain (I mean, for me, the one who bets, because, assuming the circle is split in some "one part red and x-1>1 parts green", with x being the whole circle, (just assume red is "one" and report green to it), I have ((x-1)/x)*1+(1/x)*2=(x+1)/x>1 "gain", so, you (the house, the one who draws the circle) always lose, the only question is how do you make it to lose less. If you draw the circle half-half, you lose the double of money I bet, no matter how I play, and on the other extreme, if you draw the circle a single color, you lose the money I bet on it (no other play possible). Somewhere in the middle, between half circle and full circle, you may be able to make it fifty-fifty chance, or lower your loss enough, but nobody is clear where the "middle" is. It is like a parabola, of which you need to find the minimum, but this minimum is always over zero, it has no roots. Or, you may be on one branch of it, but still no roots.

Unfortunately very few people think "out of the box", or try to "game the system", that is because most of the people are honest, they don't have "hacker blood" in their veins. Therefore very few people think that I could play "both sides". Nothing in the formulation of the problem forbids me to play both sides. In fact, when I wrote that above, I tried to be quite cautious to avoid hinting you in the direction that I could play both sides, but without saying that you are forbidden to do so. Maybe that's why the phrasing looks so odd (well, that was an effort, and you have to appreciate it, I am not a native speaker).

So, in this particular formulation, it totally doesn't matter how you draw your circle, I could just bet 3 bucks on the color which is dominant, and 2 bucks on the color which is less, and make 1 buck every time you spin. If it falls on the dominant, I lose the 2, but I gain 3, and if it falls on the other one, I lose the 3, but I make 4 (the double of the bet).

So, like somebody said above (in spoilers), the best play for you is to keep the crayons in your pocket.

Similar method is used by some dishonest fund managers, or money managers, they will convince you to open a trading account with them, telling you that if they make profit from your money you have to give them half (50%), and if they make loss, they will support, say, 30% of the loss from their own pocket. You are amazed by that, and say, whoaaa, what an expert, and what a honest guy... In reality, he doesn't give a sheep about what happens with your account, he just opens another account by himself and trades 40% of the money he trades for you, in the opposite direction. When he sells 100 for you, he buys 40 for him, in his account (which you don't see). When he buys 100 for you, he sells 40 in his account. If your trade ends in profit, your account grows, you have to give him 50, but he lost 40 in his account, so he is 10 in profit. If your trade ends in a loss, he will give you 30 "from his pocket", say sorry and show you a sad face (what a nice guy!), but on the backstage he gained 40 in his account, so he is still 10 in profit. But you lost 70.

Well, now back to the initial problem, we robbed the OP off his thread.
Sorry... (shows sad face to the OP).

[retina]

Originally Posted by LaurV

... the best play for you is to keep the crayons in your pocket.

Originally Posted by WarGames movie

... the only winning move is not to play.

[Dr Sardonicus]

Originally Posted by LaurV

Haha, this thread unexpectedly took an odd turn, due to my bad English and phrasing

The idea of my puzzle was to see how many of you think out of the box, and/or dream about "gaming the system"

I had no problem with your English or phrasing. In my case, there is another reason. (Tip o' the hat to retina): I merely assumed that his diagnosis "keep your crayons in your pocket" was correct.

And, consequently, that the only way to play and win was to cheat.