# What is the probability of drawing cards from a deck?

1. Oct 3, 2012

### Probably

It's been some time since I studied probability. So, forgive me if this seems like a very simple question to answer, but after trying to wrap my head around it for two days without any success, I'm confident that once I understand it, the rest will become very clear.

Given a well-shuffled deck, what is the probability that the first hand drawn will be a 21 (an ace and a ten-valued card [T,J,Q,K])?

I know that drawing an ace will occur (4/52) and a ten will occur (16/51). Since the ten could have been drawn first, the probability is 2*(4/52)*(16/51) = 0.4827.

First question: Technically, should the equation be written as 2! * (4/52) * (16/51) = 0.4827 (note the factorial)?

Secondly, is there a generic form that allows me to calculate the probability of drawing cards? For example, drawing three 5s? Three aces and a four?

Am I correct for three 5s to write (4/52)*(3/51)*(2/50)*3! and (4/52)*(3/51)*(2/50)*(4/49)*4! for the three aces and a four?

I'm trying to write an algorithm to gather probabilities of drawing cards. I was surprised at being able to solve a subset sum problem earlier yet these basic probability problems elude me. :-/

Thank you!

2. Oct 3, 2012

### Simon Bridge

I think you miscounted:

There is a 4/52 chance of an ace in the first draw and a 16/51 of a 10-value on the second.

There is a 48/52 chance you didn't draw an ace first time ...
in which event, there is a 16/48 chance of a 10-value first and a 4/51 chance of an ace the second time.

Spot the difference.

afaik: there s no generic form apart from the kind of reasoning you just used.

3. Oct 6, 2012

### haruspex

I would think of the first problem as drawing any one of four (4 C 1), and any one of a different 16 (16 C 1), compared with all the ways of drawing 2 from 52 (52 C 2). That gives 4*16*2/52*51 = approx .048. (You appear to have a factor of 10 error.)
But there are often many valid ways of approaching these problems.
(Simon, I think your way is not valid.)
For your more complicated example, (4 C 3)*(4 C 3)*(4 C 1)/(52 C 7). I leave you to check whether that's the same as you got.

4. Oct 6, 2012

### Simon Bridge

So see if you can fault the reasoning ;)

I just constructed the probability tree ... here's the numbers:

(4/52)*(16/51)+(16/48)*(48/52)*(4/51) = 0.048265

5. Oct 7, 2012

### haruspex

OK, I misunderstood. You said denjay had miscounted, so I assumed you had arrived at a different number. Apart from the factor of 10 error, denjay's answer was correct.

6. Oct 7, 2012

### Simon Bridge

Fair enough. denjay said: 2*(4/52)*(16/51) = 0.4827 with the "2*" in there because "the ten could have been drawn first"... i.e. two different ways of getting the result.

mine was (4/52)(16/51)+(4/51)(16/52)
as it happened ... this this is the same calculation but (possibly) arrived at by counting in a different way. I had not computed the number - so missed the factor of ten - I was commenting on the reasoning. It could have been that OP was aware of the difference though and had just missed out a step ;)

7. Oct 7, 2012

### Norwegian

Yes, that is one way the write it, and you have arrived at the result in a way that is easy to generalize.
Yes, for example the following will encompass all of your three examples:

Assume we have a deck of N cards, with ni of the cards labelled with the integer i, where i=1,2,3,...,a, and Ʃni=N. If we draw r cards at random, what is the probability of getting exactly ri labelled i, for i=1,2,...,a, and Ʃri=r?

If we simply count the number of desired possibilites, divided by all possibilities, we arrive at the following fomula involving binomial coefficients:

Probability = (n1 r1)(n2 r2)...(na ra)/(N r)
In the first case you should not multiply with 3!, and in the second case you should multiply with 4 instead of 4!. The moral is to multiply with the number of different orderings of your cards, ignoring suits. In general, this is a multinomial coefficient. Your counting strategy, including a multinomial coefficient, can also be used in full generality to arrive at the formula given above.