# “Fair” random pick with different pool size

I'm working on a little tool for drafting cards, but I'm having trouble wrapping my head around a probability problem I've run into.

The scenario is the following: (context is changed for easier explanation)

There's a deck of 10 cards, numbered 1-10. All the cards are red, except the last 2 (9 and 10), which are blue.

Then 3 players get "random" cards:

John will get a random red card

Bob will get a random red card

Annie will get a random card from the deck

In order to make sure a larger number of people can participate without their "pool" being empty when it's their turn to randomly get a card, the player with the smallest "pool" available gets to pick first. (In this case the guys who can only get red cards).

Now here's the thing.

Since John and Bob has already taken 2 of the red cards, it seems to me that the chance of Annie getting any one of the red cards is smaller, and thus, the chance of her having a blue card is greater.

Let's say I were to guess at which card Annie has. Would I be better off guessing "10", than say, "2"?

And if yes - is this possible to mitigate in some way, by for example applying a higher "Weight" on the random number generator to the pool already chosen from by others?

You are right that the probability of picking a "10" is more likely than a "2". Intuitively, this is because it's possible that John or Bob picked a "2", but not possible that they picked "10". This is called conditional probability.

More precisely, the chance that Anne picks a "10" is 1/8 since there are 8 cards remaining when she draws, and one of them is a "10".

On the other hand, in order to draw a "2" we need that John and Bob have not drawn a "2". There is a $$\frac{7}{8}$$ chance that John does not draw a "2", and a $$\frac{6}{7}$$ chance that Bob does not draw a "2" given that John did not draw a "2".

Therefore, there is a $$\frac{7}{8}\cdot\frac{6}{7} = \frac{3}{4}$$ chance that there is a "2" in the deck when Annie draws.

If there is a "2" in the deck, she has a 1/8 chance of drawing it, but if there is no "2" in the deck she has no chance of drawing a "2". So the probability she draws a "2" is $$\frac{3}{4}\cdot\frac{1}{8} + \frac{1}{4}\cdot 0 = \frac{3}{32}$$ chance of drawing a "2".

Note that $$\frac{1}{8} = \frac{4}{32}$$ is larger than $$\frac{3}{32}$$.

How to mitigate this depends a bit more on the exact setup of the game. There is no way to change the fact that she is more likely to draw the value on the blue card. However, it you randomly picked 2 cards to be the "blue" ones (without looking at them before dealing), then the probability of picking any given number would be uniform.

• Thank you for your answer - since I'm doing this programatically, there are indeed ways to apply "weight" to the chance, but I think I have enough to work on from your answer. I just needed to make sure my thoughts were correct, and how the math behind it works exactly – Scherling Dec 20 '18 at 8:30