+0  
 
0
1185
3
avatar+4 

You are playing a poker game called "Omaha" and are dealt 4 cards from a standard 52 card deck. How many possible combinations of hands can you be dealt that contain both at least one Ace and at least one King? Please explain your answer.

 Jan 27, 2016

Best Answer 

 #2
avatar+118667 
+10

Hi Lance and Fellowes,

 

You are playing a poker game called "Omaha" and are dealt 4 cards from a standard 52 card deck. How many possible combinations of hands can you be dealt that contain both at least one Ace and at least one King? Please explain your answer.

 

It would be easier for me to check if you had given some number answers

I am not sure if my logic is the same as yours or not.

A for ace, K for king, x for any other card

 

1ace

AKxx            4*4* 44C2  =  15136

AKKx            4*4C2*44  =   1056

AKKK            4* 4C3      =      16

 

2 aces

AAKx           4C2 * 4 * 44 = 1056

AAKK            4C2 * 4C2 =       36

 

3 aces

AAAK             4C3 * 4     =     16

 

15136+1056+16+1056+36+16 = 17316

 

That is what I think anyway :)

 Jan 28, 2016
 #1
avatar+1316 
+5

To do this you have to add a lot of combinations

To have at least mean you can have more so you have to add up all the combinations where you have exact amounts. 

Start with only aces and kings

There are 4 aces and 4 kings to select from. 

There are

4C3 + 4C1 to draw 3 aces and 1 king.

4C2 + 4C2 to draw 2 aces and 2 kings

4C1 + 4C3 to draw 1 aces and 3 kings

 

Then do this with the other 46 cards

4C2 + 4C1 + 46C1 to draw 2 aces and 1 king and 1 of something else. 

4C1 + 4C2 + 46C1 to draw 1 ace and 2 kings and 1 of something else. 

4C1 + 4C1 + 46C2 to draw 1 ace and 1 king and 2 of something else. 

 

Add all these up to get the total number of ways you can draw 4 cards with at least 1 ace and 1 king.

 

On the site calc use this nCr(4, 3) for 4C3

 

One of moderators should check that I not make a mistake. Maybe there is a easier way to do this too.

 Jan 27, 2016
 #2
avatar+118667 
+10
Best Answer

Hi Lance and Fellowes,

 

You are playing a poker game called "Omaha" and are dealt 4 cards from a standard 52 card deck. How many possible combinations of hands can you be dealt that contain both at least one Ace and at least one King? Please explain your answer.

 

It would be easier for me to check if you had given some number answers

I am not sure if my logic is the same as yours or not.

A for ace, K for king, x for any other card

 

1ace

AKxx            4*4* 44C2  =  15136

AKKx            4*4C2*44  =   1056

AKKK            4* 4C3      =      16

 

2 aces

AAKx           4C2 * 4 * 44 = 1056

AAKK            4C2 * 4C2 =       36

 

3 aces

AAAK             4C3 * 4     =     16

 

15136+1056+16+1056+36+16 = 17316

 

That is what I think anyway :)

Melody Jan 28, 2016
 #3
avatar+1316 
0

OK I make 2 mistakes

I put 46 instead of 44 for the leftover cards (I left the jokers in hahaha)

I add instead of multiplying the combinations going across.

 

After fixing these mistakes I get this which is the same as your answer :)

 

 

 

NCR(4,3) * NCR(4,1)  =  16

NCR(4,2) * NCR(4,2) =  36

NCR(4,1) * NCR(4,3) =   16                           

 

NCR(4,2) * NCR(4,1) * NCR(44,1)  =   1056

NCR(4,1) * NCR(4,2) * NCR(44,1)  =   1056

NCR(4,1) * NCR(4,1) * NCR(44,2)  = 15136

 

 

16+36+16+1056+1056+15136 = 17316

 

 

Thank you Miss Melody.

 Jan 28, 2016

2 Online Users