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 $\displaystyle \tfrac34$.

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

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

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

Thus the probability that a particular player receives at least one card is $\displaystyle 1- \tfrac{81}{256}= \tfrac{175}{256} \approx 0.6836$.

...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 $\displaystyle \tfrac{48}{51}$.

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 $\displaystyle \tfrac{44}{50}$.

Thus the probability that none of the first three cards match is:$\displaystyle \frac{48}{51}\times\frac{44}{50}$

Continuing in the same way through all 13 cards, the probability that none of them match is:$\displaystyle \frac{48.44.40.36.32.28.24.20.16.12.8.4}{51.50.49. 48.47.46.45.44.43.42.41.40}= \frac{4^{12}\times12!\times39!}{51!}\approx0.00010 6$

So the probability that you get at least two matching cards $\displaystyle \approx 1-0.000106=0.999894$.

You may be able to use a similar technique to work out further probabilities, but as you'll see, it ain't too easy!

Grandad