# Probabilities of a Poker Hand

• March 8th 2008, 10:47 AM
IronMan
Probabilities of a Poker Hand
Hi,

Here is my question

I randomly draw 5 cards from a pack. (Ace is the highest card) What would be the probability of drawing a king, Queen, Jack etc as your highest card?

I think below I have worked out the probability of an ace being your highest card:

The probability of drawing an ace = 1- the probability of drawing 0
aces
The probability of drawing 0 aces is: C(48.5)/C(52,5)
=1712304/2598960
=0.65

David

So The probability I think of drawing an ace = 1-0.65 = 0.35

Any help with this question would be much appreciated,

Thank you
• March 8th 2008, 11:40 AM
Plato
Suppose we assign numerical values to the different cards.
An ace=1 for the highest; king=2 for second highest, etc until 3=12 for lowest possible.
Then $P(X = k) = \frac{{\sum\limits_{j = 1}^4 {\binom {4}{j}\binom {52-4k}{5-j}} }}{{\binom {52}{5}}}$

So the probability that the hand has the a jack as the highest card is P(X=4).
• March 8th 2008, 12:49 PM
IronMan
Great Help, Thank you so much!!

I have just 2 more questions which follow on from this that I would be grateful you could answer:

1) What is the highest card likely to be?

2) Someone offers you a "all or nothing bet", that in a randomly selected set of five cards, the highest card will be either King or Ace. Would you place the bet?
(An "all or nothing bet" just means that if you bet $1, you will get$2 back if you are right and nothing if you are wrong)
• March 9th 2008, 11:34 AM
CaptainBlack
Quote:

Originally Posted by IronMan
Hi,

Here is my question

I randomly draw 5 cards from a pack. (Ace is the highest card) What would be the probability of drawing a king, Queen, Jack etc as your highest card?

I think below I have worked out the probability of an ace being your highest card:

The probability of drawing an ace = 1- the probability of drawing 0
aces
The probability of drawing 0 aces is: C(48.5)/C(52,5)
=1712304/2598960
=0.65

David

So The probability I think of drawing an ace = 1-0.65 = 0.35

Any help with this question would be much appreciated,

Thank you

The probability of not drawing an ace is:

$\frac{48}{52}\times \frac{47}{51}\times \frac{46}{50}\times \frac{45}{49} \times \frac{44}{48} \approx 0.66$

Probability of not drawing an ace or king (which is the probabilty that the highest card is a Q or less) is:

$\frac{44}{52}\times \frac{43}{51}\times \frac{42}{50}\times \frac{41}{49}\times \frac{40}{48} \approx 0.42$

$P(highest~card~is~K)=1-P(highest~card~is~less ~than~Q)-P(highest~card~is~A)$

and so on

RonL
• March 9th 2008, 11:54 AM
Plato
Quote:

Originally Posted by CaptainBlack
The probability of not drawing an ace is:
$\frac{48}{52}\times \frac{47}{51}\times \frac{46}{50}\times \frac{45}{49} \approx 0.72$RonL

But five cards are dealt not four.
• March 9th 2008, 12:46 PM
IronMan
Thanks
Hi thanks so much for the help. I'm a little confused though, which answer is right lol? I'm finding it a little hard to comprehend the question.
• March 9th 2008, 12:55 PM
CaptainBlack
Quote:

Originally Posted by Plato
But five cards are dealt not four.

Of course just need to put in the extra terms

RonL
• March 9th 2008, 01:40 PM
IronMan
Thanks
Ok got it. I really can't thank you enough, forever indebted thanks (Wink)