Probability Class Problem Set Solutions

We will find out the number of times that certain poker hands appear.Recall that a normal hand consists of 5 cards out of a set of 52 cards. The 52 cards are composed of 4 different suits ♣(clubs),♠(spades),♡(hearts),♢(diamonds). Each suit carries 13 cards. An “ace”, cards numbered 2 through 10, a “jack”, a “queen”, and a “king”. Thus in total there are

52
5
=2598960
different hands. We will then calculate the probability of drawing certain specific hands.

1. What is the probability of drawing a pair?

An example of a hand with a pair is 3♠,3♣,4♡,6♢,A♣
For each of the following problems from 1 - 8 we take the number of hands that satisfy our requirement and divide it by the number of total hands. We have counted the hands previously at https://www.wolframcloud.com/obj/scamach2/Published/Counting%20Class%20Problems%20Solutions.nb the
In[]:=
1098240/2598960.//PercentForm
Out[]//PercentForm=
42.26%

2. What is the probability of drawing two pairs?

An example of a hand with two pairs is 3♠,3♣,4♡,4♢,A♣
In[]:=
123552
2598960.
//PercentForm
Out[]//PercentForm=
4.754%

3. What is the probability of drawing a hand with three of a kind?

An example of a hand with three of a kind is 3♠,3♣,3♡,4♢,A♣
In[]:=
54912/2598960.//PercentForm
Out[]//PercentForm=
2.113%

4. What is the probability of drawing a straight?

For example {A♣,2♢,3♣,4♡,5♢}, {10♣,J♣,Q♣,K♣,A♢}
In[]:=
10200/2598960.//PercentForm
Out[]//PercentForm=
0.3925%

5. What is the probability of drawing a flush?

For example {5♣,3♣,4♣,6♣,A♣}
In[]:=
5108/2598960.//PercentForm
Out[]//PercentForm=
0.1965%

6. What is the probability of drawing a full house?

For example{3♠,3♣,3♡,4♢,4♣}
In[]:=
3744/2598960.//PercentForm
Out[]//PercentForm=
0.1441%

7. What is the probability of drawing a poker?

For example {3♠,3♣,3♡,3♢,4♣}
In[]:=
624/2598960.//PercentForm
Out[]//PercentForm=
0.02401%

8. What is the probability of drawing a straight flush?

For example{10 ♣, J♣, Q♣, K♣, A♣}
In[]:=
40/2598960.//PercentForm
Out[]//PercentForm=
0.001539%

9. What is the probability of drawing a full house given that there are no Aces in the deck?

Choose the pair C(12,1)=12
Choose the triple number C(11,1)=11
Choose the suits for the pair C(4,2) =6
Choose the suits for the triple C(4,3)=4
In[]:=
12*11*6*4
Out[]=
3168
In[]:=
Binomial[48,5]
Out[]=
1712304
In[]:=
3168
1712304.
//PercentForm
Out[]//PercentForm=
0.185%

10. What is the probability of drawing a poker, given that at least one of the cards drawn is an Ace?

First thing is find the number of hands with poker and an Ace in the hand.
There are two cases. Case 1, we have a poker of Aces and a random number. Case 2 we have a poker of something that is not an ace, and then the extra card is an Ace
Case 1: The first four cards are already decided so all I have left to choose is the additional card C(48,1)=48Case 2: We first have to choose the number that gets the poker C(12,1)=12 Technically choose the suits C(4, 4)=1 Choose the ace C(1,1) Choose the suit of the Ace C(4,1)=4 So for case 2 we have a total of 12*1*1*4= 48So total we have 48 + 48 = 96P(Poker and having an Ace) = 96/2598960​
Then we calculate the number of hands with at least 1 Ace. TO do this we calculate the number of hands with NO aces and then subtract that from the total of hands.
So counting hands with no Ace. C(48,5)=
1712304
​
So hands with an Ace are 2598960 -
1712304
=
886656
P (Having an Ace) = 886656/2598960
P(PokerandhavinganAce)
P(HavinganAce)
=
96/2598960
886656/2598960
=
0.000108272