# Math Help - Card game probabilities help needed

1. ## Card game probabilities help needed

Hey, me and my friend have been thinking about designing a program which works out the probabilities of a simple card game, however it is harder than we anticipated.

This card game works by randomly giving each of the 4 players 13 cards. The first player then has to put down an Ace, the second player a two and so on until a player has managed to loose all their cards and wins. A player may put more than one card of that number down. Since this is not possible players must lie about what cards they are placing on the table and the other players must work out if that player was lying or not. If they were lying they have to pick up the pile of cards on the table, if they weren’t the player who said they were lying has to.

The objective of our program is to predict the chance of a player having certain cards or to produce a probability of the chance that they are lying or not.

Anyway, this turns out to be very difficult and we are stuck on some of the basics of card game probabilities...

For example if we have 4 cards and 4 players where random amounts of the cards are given out to random players.
The chance of a player getting one card is 1/4 however since we have 4 cards it is (1/4)*4 = 1. But since this is random it cannot be a certain chance that every player will get 1 card.

Furthermore, we simply cannot seem to be able to work out a reasonable answer to the probability of a hand of 13 to get a pair of matching cards or a triplet of matching cards when through the practical test we usually have 2 pairs and around 0.4 triplet pairs.

Could anyone please shed some light onto this topic?

Thank you

2. Hello jamescockayne

Welcome to Math Help Forum!

Let me answer two of your specific questions, as I understand them. I shall re-phrase them, but I hope I keep your original meaning.
Originally Posted by jamescockayne
...For example if we have 4 cards and 4 players where random amounts of the cards are given out to random players.
The chance of a player getting one card is 1/4 however since we have 4 cards it is (1/4)*4 = 1. But since this is random it cannot be a certain chance that every player will get 1 card.
The problem I think you're posing is this:
Four cards are given, one at a time, to a player chosen at random from the four players in the game. What is the probability that a particular player receives at least one card?
You work this out, by working out the probability that a particular player receives no cards at all, and subtracting the answer from 1. So:
When the first card is allocated, the probability that a particular player does not receive it is $\tfrac34$.

When the second card is allocated, the probability that this player does not receive it is also $\tfrac34$.

Therefore the probability that this player does not receive either of the first two cards is $(\tfrac34)^2 = \tfrac{9}{16}$.

Similarly with the third and fourth cards. So the probability that a particular player receives no cards at all is $(\tfrac34)^4 = \tfrac{81}{256}$.

Thus the probability that a particular player receives at least one card is $1- \tfrac{81}{256}= \tfrac{175}{256} \approx 0.6836$.
...Furthermore, we simply cannot seem to be able to work out a reasonable answer to the probability of a hand of 13 to get a pair of matching cards or a triplet of matching cards when through the practical test we usually have 2 pairs and around 0.4 triplet pairs.
I shall simplify this a little and answer the following question:
When a hand of 13 cards is dealt at random from a standard pack of 52, what is the probability that this hand contains at least two matching cards? (By two matching cards we mean, for example, two 5's.)
This is a simpler question than the one you asked, but this is complicated enough for now, as I think you'll appreciate.

We work this out using a similar technique to the first question: we find the probability that it doesn't happen - in other words, all 13 cards are different - and subtract this from 1. Like this:
It doesn't matter what the first card is. Of the remaining 51 cards, 3 will match and 48 won't. So the probability that the second card doesn't match the first is $\tfrac{48}{51}$.

In the same way, when the third cards is dealt, 44 cards won't match either of the first two, out of the remaining 50 cards. So the probability that this one is different from the first two is $\tfrac{44}{50}$.

Thus the probability that none of the first three cards match is:
$\frac{48}{51}\times\frac{44}{50}$
Continuing in the same way through all 13 cards, the probability that none of them match is:
$\frac{48.44.40.36.32.28.24.20.16.12.8.4}{51.50.49. 48.47.46.45.44.43.42.41.40}= \frac{4^{12}\times12!\times39!}{51!}\approx0.00010 6$
So the probability that you get at least two matching cards $\approx 1-0.000106=0.999894$.
You may be able to use a similar technique to work out further probabilities, but as you'll see, it ain't too easy!