Raise the Stakes:

Play for crackers, raisins, etc.

Even and Odd Numbers:

Fingers are a great tool for thinking about what makes a number even or odd. An even number of fingers can be grouped into pairs, with none left over. An odd number of fingers can be grouped into pairs but will have one left over.

Ask your child what happens when both players choose an even number. Say one player holds up 4 fingers, and the other 2 fingers. 4 fingers can be grouped into 2 pairs with none left over. 2 fingers can be grouped into 1 pair. Together, all 6 fingers are grouped into pairs, so 6 is an even number.

Now ask what happens when both players choose an odd number. Careful! This can be a bit of a trick question. Say one player holds up 5 fingers and the other 3 fingers. 5 fingers can be grouped into 2 pairs with 1 left over, and 3 fingers can be grouped into 1 pair with 1 left over. Both numbers are odd, and so both numbers have a "left-over" (these left-overs are called "remainders"). But when you add the numbers on both hands, the left-overs become a new pair. Together, the 5 fingers and 3 fingers make 8 fingers, an even number.

So how can an odd number occur? The only way for there to be 1 left-over is if one player holds out an even number, and the other player an odd number.

Shortcut:

It doesn't really matter whether you hold up a 5 and a 4 or a 1 and a 0. In both cases, an odd number plus an even number equals an odd number. Let your child notice that, generally, even + even = even, odd + odd = even, and even + odd = odd. Instead of holding up fingers, try playing Odds vs. Evens by writing any number on pieces of paper. If one player writes 7,844,213 and the other player writes 4,722, there isn't any need to actually add the two numbers. All we need to do is recognize that the first number is odd, the second is even, so the sum will be odd.

Probability:

Two even numbers add to an even number, and two odd numbers add to an even number. The only way to get an odd sum is for one player to hold up an even number of fingers, and the other an odd number of fingers. Does this mean the Evens player has an advantage? It can seem that way at first glance. To see why this isn't true, we use a table to organize the possible outcomes:

2 of the possible outcomes are even, and 2 are odd, so both outcomes are equally likely.

Strategy:

The table above shows that both outcomes are equally likely if each player is equally likely to hold up an even or odd number of fingers. Is that always the case?

First, notice that if you are only thinking in terms of 1, 2, 3, 4, or 5 fingers, an odd number is more likely (1, 3, and 5 are odd, while only 2 and 4 are even). But this leaves out one possible even number: 0. (The number 0 is even because it has no "left-overs.") So each player has 3 even numbers (0, 2, 4) and 3 odd numbers (1, 3, 5) to choose from.

Older players may notice while playing, though, that their opponent seems to hold up a certain number more often. For example, maybe the "Evens" player notices that their opponent holds up an odd number most of the time. If that's true, the "Evens" player should also hold up an odd number of fingers most of the time (since an odd number plus an odd number equals an even number). This shows, too, that the best strategy is, in general, to try to hold up odd half the time and even half the time. Otherwise a player's bias towards one or the other can be used against them.

So does this mean it's best to just alternate odds and evens each turn? Of course not, since that would also be easy for an opponent to take advantage of! Playing odds half the time and evens half the time, but doing so in an "unpredictable" way is its own puzzle.