Marilyn Vos Savant's Even and Odd Game

On 2011 October 16, Marilyn Vos Savant proposed a simple game:

"As children, my siblings and I often settled a disagreement with a game called
"Evens and Odds." In this game, one side is assigned "evens" and the other is
assigned "odds". Then on the count of three, a representative of each side reveals
a number of fingers from zero to five. If the sum of the two numbers is even,
the "evens" win; if the sum is odd, the "odds" win. Is this method fair? Or
do the "evens" have an advantage? - Andy G., Cedar Hill, Mo.

The game is fair. One might guess that the "evens" have an advantage because
the sum of two even numbers is even and the sum of two odd numbers is even,
too. Only an odd number plus an even number will yield an odd sum. But there
are two ways to get an odd sum: ()1) an odd plus an even, and (2) an even plus
an odd. So there are twice as many odd sums as you might think. "Evens and Odds"
is a nice game. The odds are even!"

She puts an interesting little pun at the end. But she still has not explained
why it is fair. If one side plays evens twice as much as the other, does that
get an advantage? Also if Evens plays even, then Odds can beat him by playing
odds. But then Evens can win by playing odds too. But then Odds can beat that
by playing evens. So it goes around in a vicious cycle and it is hard to know
who has the advantage. Game theory comes to the rescue. First of all it calls
upon us to list the strategies for each side. Each side can play a number from
0 through 5. But playing a 2 is the same as playing a 4, isn't it? Because both
are even, either wins against an even number and loses to an odd number. So
that reduces the number of strategies on each side to just two: even and odd.
These can be described by a game matrix as follows:

	Even	Odd
Even	1	-1
Odd	-1	1

So for example, if Even picks the first row and Odd the second column, the
result is -1; i.e., Even wins. The game matrix is given in terms of how Even
fares; Odd fairs the exact opposite. The sum of the results is zero, as one
player wins and the other loses. We call such a game a "zero-sum game". According
to the theory, the optimal strategies form a "saddle point". That is, a pair
of strategies, one for each side, such that if either deviates from these strategies,
he can do no better (and probably will do worse) than how he is doing now. A
property of a saddle point matrix entry is that it is the minimum of its row
and the maximum of its column. However, an examination of this matrix reveals
that it does not contain a saddle point. So what do we do now? The argument
I gave earlier suggests that one should keep one's strategy to himself; if you
know what the opponent is going to do, by playing the appropriate strategy,
you will win. So how about flipping a coin? Even you won't know the result of
that. If you do that, then half the time you will win and half the time you
will lose. We say that we play the strategies in the ratio of 1:1. Of course
Odds also can play the same strategy, since the game is symmetric. The result
in the long run is that the wins and losses for Evens will cancel each other
out, and the value of the game (expected outcome) is zero; i.e., the game is
fair, verifying what Marilyn said. But why bring in this stuff with the fingers,
when all that is needed is whether the number is even and odd? It seems to me
that the number of fingers should be significant, too. So let's alter the game.
The two players each put out a number of fingers. Evens wins if the total is
even; Odds wins otherwise. But this time they win the total number of fingers
(except that we make payment in dollars instead, of course). So if Evens plays
3 and Odds 5, then the total is 8, so Evens wins 8 dollars. Now the actual number
of fingers is significant, so each player now has six strategies. So now the
matrix looks like this:

	0	1	2	3	4	5
0	0	-1	2	-3	4	-5
1	-1	2	-3	4	-5	6
2	2	-3	4	-5	6	-7
3	-3	4	-5	6	-7	8
4	4	-5	6	-7	8	-9
5	-5	6	-7	8	-9	10

One can argue like this. If Evens plays 3, Odds can maximize his outcome by playing 4, causing a 7 point loss for Evens (and gain for Odds). But if Evens knows Odds will play 4, he will play 4 also and win. But if Odds knows Evens will play 4, Odds will play 5 and win 9. But if Evens knows this, he plays 5 too, and wins 10. But if Odds knows this, he plays 4 and wins 9, bringing us back to the beginning. This suggests that some random mix of 4 and 5 fingers is optimal for both players, but if you calculate how 0, 1, 2, or 3 fingers fares against this mix, one gets better results. This game matrix is considerably more complicated.

So to find the optimal strategy, I converted the game problem into a linear program, using the standard procedure, and solved it using a program I devised to illustrate the simplex algorithm to a class at Florida Tech at Ft Lee, Virginia. The result is:

	0	1	2	3	4	5
0	0	-1	2	-3	4	-5	1
1	-1	2	-3	4	-5	6
2	2	-3	4	-5	6	-7
3	-3	4	-5	6	-7	8	4
4	4	-5	6	-7	8	-9	3
5	-5	6	-7	8	-9	10
	2	4	2

In other words, a strategy for Odds is to play Strategies 0, 1, and 2 in the ratio of 2:4:2. This is the same as 1:2:1. In other words, pick a number of
fingers between 0 and 2 at random but put higher odds (twice as high) on 2 fingers. Another way is to flip two coins, count the heads, and display that number of fingers.

How about Evens? His optimal strategy is to play 0, 3, and 4 fingers in the ratio of 1:4:3. The numbers in the odds ("oddments"?) add to 8, so that this strategy can be realized with an octahedral die. Since the game matrix is symmetric, an optimal strategy for either player is to play the other players' optimal strategy. So either player can play fingers 0, 1, and 2 according to odds 1:2:1, or to play fingers 0, 3, and 4 in the ratio of 1:4:3. A theorem of game theory
states that if A and B are any two optimal strategies, then any linear combination
of A and B is also an optimal strategy. So if you add the two and get fingers
0,1,2,3,4 according to odds 2:2:1:4:3, that is also an optimal strategy.
The value of the game is the same as before: zero, so it is also a fair game. The game resembles Three-Finger Morra.

So Marilyn is correct, but game theory explains in more detail just why the
conclusion she came to is correct.

Jim Blowers