# Mathematical puzzle - same probability of given card type for any number

Let's create a card game with any number of red and black cards.

First I'm creating any number of decks containing chosen number of red or black cards. I shuffle them.

Then I want to draw N cards from the decks I've created. I can draw multiple cards from one deck, but I cannot peek the cards I'm taking, nor look inside those decks while drawing. However after every draw I can add any number of red or black cards to a deck (and shuffle) and/or remove any number of cards from that deck (at random, without peeking).

The goal is following: for given N cards make sure, that the probability of drawing exactly X red cards is the same for every $X: X \in <0, N>$

For example:

For N = 4, I'm drawing 4 cards and I want to be sure that I have 20% chance to get zero red cards, 20% chance to get one red card, 20% chance to get two red cards ... and 20% chance to get four red cards.

What is the proper strategy to create decks and modify them while drawing to achieve this goal?

You cannot do it. The problem is that there are more ways to get evenly balanced collections than to get all black or all red. If we look at the cards as we draw them we can succeed because we can choose the probabilities going forward. For the problem as stated we can see the problem with just $N=2$. The first card is red with probability $p$ and the second card is red with probability $q$. You can choose $p$ and $q$ by choosing the composition of the deck. You are then asking that $pq=p(1-q)+(1-p)q=(1-p)(1-q)$ This has no real solution per Alpha with the problem being that the chance of one of each is too high. $$pq=p(1-q)+(1-p)q\\pq=(1-p)(1-q)\\3pq=p+q\\ q=\frac p{3p-1}\\\frac {p^2}{3p-1}=(1-p)\frac {2p-1}{3p-1}\\ p^2=-1+3p-2p^2\\0=-1+3p-3p^2$$ and the right is always negative.
• @knrumsey: this applies to multiple decks. $q$ is the fraction of red cards in the second deck. – Ross Millikan Feb 17 '17 at 19:35