How many variations required so that x pairs are drawn from sample size of y with z guesses Please see the puzzle below, I'm not sure if my answer below is too simplistic or if I'm missing something, I think my main question is, does the sample size matter e.g. does it make a difference if there are 100 or 500 cards to choose from. 
Thanks in advance for any help. 
See example and my (guessed) calculation below: 
"The game is to choose pick two cards with a symbol on them, if the two cards match, you win.
There's a board with 256 cards 
There are 15 prizes
there are 60 players with one try each
Each time a player chooses a pair, that pair is removed from the board
How many different variations of symbols does there need to be so that 15 people out of 60 win? 
With 2 different symbols there's a 2/1 chance that the a person will guess correctly, therefore 60 people guessing means that 30 will win (sample size doesn't matter???)
With 3 different symbols there's a 3/1 chance that the a person will guess correctly, therefore 60 people guessing means that 20 will win 
With 4 different symbols there's a 4/1 chance that the a person will guess correctly, therefore 60 people guessing means that 15 will win"
Does this look correct, or even close? Thanks 
 A: First you have to understand that all of the following is about probabilities, which makes certain outcomes more or less likely, but doesn't prevent any completely. So in your example with 256 cards, 4 different symbols and 60 players, in theory all could win or none could win. The 15 prizes could be way too many or way to few.
Your reasoning has 2 flaws, both are probably not so big to totally distort the outcome, but it is important to understand that they exist.
The first flaw is that with 2/3/4/... different symbols in the same quantity, even the first player doesn't have a chance to win that is exactly $\frac12, \frac13, \frac14,...$.
To see this, take just 4 cards with 2 each of 2 different symbols. Whichever symbol is picked for the first card, there is now only 1 'winning' card remaining among the 3 that are candidates for the second pick. So for the first player, the winning chance is $\frac13$, not $\frac12$. The effect becomes smaller when the number of cards with the same symbol increases, for example with 20 cards and 10 copies each of 2 different symbols the probability to win is $\frac{2{10 \choose 2}}{20 \choose 2} \approx 47.37\% $ for the first player.
The second flaw is after any pair of cards is removed, the probability for the next player to win will be changed. Say you start with 20 cards, with 10 copies of 2 different symbols. The first player picks 2 different cards (he looses), so 18 cards (9 of 2 symbols) remain and next player has now a probability of $\approx 47.06\%$ to choose a pair of the same symbols (slightly less than the first player). But if the first player picks a pair of the same symbols (he wins), we now have 10 of symbol #1 and 8 of symbol#2 remaining. The probability for the next player to win now becomes $\approx 47.71\%$ (slightly more than the first player). Taking all of this into account, the second player has a slightly increased chance of winning ($\approx 47.40\%$).
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For your problem, you need to add up the probabilities for each player, to find out how many players will (on average) win. The good news is that the small differences between your assumed results ($50\%, 33.333...\%, 25\%)$  to the real values (see above) will stay small, because you are adding values that (from what I can see) are all of the same magnitude. 
In your 256 cards/4 different symbols/60 player scenario, you still have 138 cards left before the last player picks his cards. It is certainly not the same as 256 cards, but I don't assume that the values will be drastically different.
So in conclusion, your reasoning is flawed, but the effects are probably small enough for a "back of the envelope" calculation. But you need to be aware of them anyway because if the numbers change, the small changes might become too big to ignore.
