Texas Hold em Question

Question

in the game of texas hold em a player is dealt two cards (called hole cards) from a shuffled standard deck of 52 playing cards. The order in which the cards are dealt does not matter. A. Find the number of hole cards combination that is possible? b. how many combinations are there in which both cards are aces? C. Find the probability that a hand consists of two aces. D. Find the probability of getting at least one aces.

 

This is for my stats class i have no clue how to start it.

Answer 1

Hi, Javi,

 

A. If we are dealt two cards, there are ₅₂C₂ different possible combinations of "hole cards".

 

    ₅₂C₂ =     52!               =   52 * 51       =  1326 different combinations

              2! * (52-2)!                 2

 

 

B. If both cards must be aces (of which there are only 4 total), the number of "two ace" combinations would be:

 

     ₄C₂ =      4!                =        4 * 3    =    6 different combinations

              2! * (4-2)!                      2

 

 

C. The probability of receiving two aces is equal to:

 

    (# combinations of two aces)       =      6       =     1       ≈ .00452

    (# total possible combinations)           1326         221

 

 

D.  To find the probability of getting at least one ace, we need to know the number of combinations that would involve at least one ace. For a probability/statistics question, involving "at least", it is usually easiest to instead find the probability of that thing NOT happening - then subtracting that probability from 1. So, let's first find the probability of the opposite - not receiving ANY aces.

 

Of the 52 cards, there are 48 non-aces. We now want our hand to only include these cards. There would then be ₄₈C₂ possible combinations that would NOT include an ace:

 

    ₄₈C₂ =        48!        =       48 * 47         = 1128 combinations without any aces

               2! * (48-2)!                2

 

The probability of getting a combination WITHOUT any aces would be:

 

    (# combinations without aces) =     1128      =     188  ≈   .851
    (# total possible combinations)        1326             221

 

So the probability of getting ANY aces would be 

 

    1 -  188    =   221 - 188    =     33     ≈  .149

    1    221          221   221           221

 

I hope this helps!