Simple-to-play morra game to select m winners from n contestants. I have m apples and n people (m < n) and we need to play a fair deterministic game to decide who gets the apples.  I know how to do this if m is 1 with morra, having each player submit an integer in [0, n-1].  For example if n is 2 then each player submits 0 or 1 and we take the sum and the winner is player1 if sum=1 or player2 otherwise.
Is there a way to generalise this to situations where m is greater?  It needs to be reasonably simple so that players can verify the game simply with pencil and paper.
Edit: to the commenter who asked if I could involve a random choice, no, the game must be determined just from the entries of the players.  That's what I meant by "deterministic", above.
 A: You could play 'morra' as you described and then say that the sum (mod $n$) gives the starting winner, and then the other winners are the next $m-1$ people.  For instance if there are $5$ apples for $8$ people, with the people numbered $0,1,...,7$, and the morra number is $6$, then the winners are numbers $6, 7, 0, 1, 2$.  
I think each player is equally likely to win; although each possible subset is not equally likely.
A: Well the simplest solution would be to assign everyone a number, let's call that number i, where everyone has a unique i  from 0 to n-1. Then, like you suggested, everyone will put a number pi to do this pooling of numbers m times, and then take the % of these numbers (remainder when divided) by n, and reward the apple to the person with the corresponding number.
Generalizing this, you could say
$$f(p) = (\sum_{s=0}^{n-1} p_s)$$
Then, the person for whom i = f(p) % n receives the apple.
For example, let's say n is 10, and that m, for simplicity, is 1. Everyone would put in a number from 0 to 9, let's say you get p = {3, 6, 4, 7, 0, 7, 7, 5, 1, 9} f(p) would then be 49. Divide that by 10 and you get a remainder of 9, so the apple would go to person 9.
Now, there's several problems that this could cause. For example, one person might get more than one apple, which, while not expressly against the rules, might feel unfair to some of the others. Another potential problem is that if m starts to get big, you have to repeat the process for each apple one by one. Let's say there's 63 people and 50 apples, by the time you get to apple 22, half the group will have gotten tired and left (though admittedly, this might leave you with enough apples to give one to everyone).
The simplest way then that I can think of off hand to solve these potential issues would be to set some sort of rule for who gets an apple if a certain number results from our randomization process. For example, if you have 12 apples, and the number 15 comes out as your resuld, then the apples go to person 15, 16, 17, ... to thee first 12 people in order (maybe wrapping back around to 0). Now, this also might not seem fair because the apples aren't assigned independently of one another, or because the pattern seems to take away from the randomness, but on an individual scale, each person has the same chance as anyone else of receiving an apple.
Or if you want you could assign everyone a set of numbers i0, i1, ... im-1 where no two people have the same ix for any x, and also where within any one person's set of given i's there are no repeated values either (now, assigning these values would be another algorithm in itself, but if you're really dedicated to this game....)
I hope I've explained myself clearly, I'm no expert on math notation, and most of my variable names were chosen pretty arbitrarily, but I hope you understood my point and that I didn't overlook anything. There's probably a better way of doing this, and if someone comes up with something better, please throw it out there, or if you find something wrong with my answer tell me so I can fix it. :)
A: Assuming you want only one apple per person, assign a number to each $m$-subset of the players. Then play morra where each player picks a number from $1$ to $n \choose m$. The disadvantage is that they probably will not have enough fingers; they can use scraps of paper instead to write down their choices.
If there are few apples compared to the number of people, then you can successively cut the number of people (roughly) in half until you reach a reasonable size by having them play "evens-odds": each player shows either one or two fingers, and those in the minority are eliminated. (Don't eliminate anyone if it happens that doing so would reduce the number of people to fewer than the number of apples.)  Repeat as necessary.
