This is a pretty mundane and mostly pointless general question.
But,
How do you play that game where one person calls odds, the other evens and then they show a number with their fingers? We never used this when I was a kid.
Never played it myself, but I believe it goes like this: I’ll call odds, which leaves you with evens (or vica-versa). Then on the count of 3, we display a number of fingers. If your and my finger count totals an odd number, whoever picked odds wins; if it’s even, even wins.
At least that’s how I think it goes.
This is my understanding as well. Note that if you are not allowed to display zero fingers, evens have more numbers that may come up:
Even numbers available on a one-hand, non-zero showing:
{2, 4, 6, 8, 10}
Odd numbers available on a one-hand, non-zero showing:
{3, 5, 7, 9}.
1 isn’t available for odds because the minimum “throw” by each party is 1, resulting in a sum of 2.
Note that if you include zero as an even, even still has better chances:
{0, 2, 4, 6, 8, 10}
vs
{1, 3, 5, 7, 9}.
Even if you count zero as odd can the chances aren’t equivalent:
{2, 4, 6, 8, 10}
vs
{0, 1, 3, 5, 7, 9}
Summary: the game is always in someone’s favor, because in any case there are an odd number of possible results to split up between two people as winning conditions; someone will always have more chances to win than the other person.
Play standard rock paper scissors instead.
erislover, your analysis is true only if each number has an equal probability of appear. This is not the case, since the number is the sum of two independent events (just like rolling dice). Assuming no fingers (i.e., 0) is valid, then all probable results would be:
Player1
| 0 1 2 3 4 5
P --------------------------------
l 0 | 0 1 2 3 4 5
a 1 | 1 2 3 4 5 6
y 2 | 2 3 4 5 6 7
e 3 | 3 4 5 6 7 8
r 4 | 4 5 6 7 8 9
2 5 | 5 6 7 8 9 10
where each column heading indicates the number of fingers Player 1 holds, each row heading indicates the number of fingers Player 2 holds, and each cell contains the result of that combination of fingers. If you count the number of odd and even results you’ll see that there are 18 odd outcomes and 18 even outcomes. The game is indeed fair.
An explanation that’s possibly more intuitive. Say you throw a number. The opponent can throw one of 6 numbers, obviously 3 of which will have the same parity as yours, so it’s fair.
And if anyone cares, this is often how ultimate frisbee players decide who pulls (throws off) first. They flip two discs and someone calls the parity.
Why make things so complicated? The rules I always used is that you can either show one finger or two fingers. That’s it. If both sides show the same, even wins. If both sides show differently, odd wins.
Haj
I agree with hajario. All five fingers makes things needlessly complicated. We always used one or two fingers when I was a kid. (Zeroes not allowed.)
Uh… what if only one side shows differently?
Zero isn’t a valid throw, just one or two. For decisions I prefer Rock, Paper, Scissors.
I’ve been meaning to start a thread about this every time I see that commercial with the girls playing volleyball in the winter.
?I have ne’er seen this ad…Volleyball uniforms are tight…it’s winter and it’s cold…there’s girls playing…I need to see this ad… 
Of course, if only one or two fingers are allowed, then the odds:evens ratio is still 50:50.
http://www.ifilm.com/?sctn=collections&pg=superbowl2004
Scroll down to near the bottom and look for “Kerri & Misty Play Snow Volleyball”
That’s what prompted me, too! Funny.
This is a simplified version of an Italian game called Morra. The first point is that this is a problem of game theory, not probability. The question is not what is likely if the players choose their strategies randomly, but what strategies they ought to adopt. Using the rules described by Chandeleur, the second player’s best approach is clear. He should determine randomly (say by tossing a coin) whether to show 1 finger or 2 fingers. Any departure from this strategy gives Player 1 the advantage. Player 1’s options are slightly more complicated. He should choose his strategy in such a way that:
half of the time he either calls evens and shows 1 finger or calls odds and shows 2
half of the time he either calls evens and shows 2 fingers or calls odds and shows 1.
The obvious way of doing this is to choose each strategy with probability 1/4. If both players follow their optimum strategy, there is no advantage to either player.
The original game is more complicated. Each player simultaneously shows 1 finger or 2 fingers and guesses how many fingers the other player will show. If both players guess correctly, or neither player guesses correctly, there is no payoff. If one player guesses correctly, he wins a payoff equal to the total number of fingers shown by both players. In this game, each player’s best strategy is to play “show 2-guess 1” 5/12 of the time, and “show 1-guess 2” 7/12 of the time (and never to use the other two strategies).
In the full version of the game, each player can show (and guess) 1, 2 or 3 fingers. The analysis of this is naturally more complicated. Those who are interested can find it in J. D. Williams, The Complete Strategyst.
I will indeed see that. I see it is also fair if only {1,2} are acceptable throws. Huh! Thanks.