The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Math Topics » alt.math.undergrad

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: probability of card dealing
Replies: 2   Last Post: Feb 13, 2014 4:02 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ] Topics: [ Previous | Next ]
Colin Reinhardt

Posts: 2
From: San Diego
Registered: 1/30/14
probability of card dealing
Posted: Jan 30, 2014 3:15 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Trying to compute the probability of an opponent being dealt a specified pair (two cards) from an initially well-shuffled (uniform distribution) 52-card deck, given that I know the two cards I've been dealt already, in a two-person game. To clarify, each player is dealt only two cards, face down (but I can look at mine).

For example,
What is probability that my opponent receives Ace-King (of any suit combo), given that I know I have Ace-King in my hand already?

My idea of how to do this is to use Bayes' formula
P(X+Y) = P(X|Y)P(Y)
+ here indicates the union
and the events are
X = opponent's hand is Ace-King pair
Y = my hand is Ace-King pair

Then P(Y) = 16/1326 = 0.0121
16 = # ways to get an Ace-King pair
1326 = # two card combinations (unordered) from 52-card deck

P(X|Y) = 9/1225 = 0.0073
9 = # remaining ways to get Ace-King pair, since one Ace and one King are already in my hand
1225 = # two card combos from remaining 50-card deck

P(X+Y) = 0.0073*0.0121 = 8.89e-5 or 0.89%

Is this correct? Thank you.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.