What is the probability of drawing a full house with three aces and two kings?

[YODA0311]

1. So the question is "what is the probability of drawing a full house consisting of three aces and two kings?"

2. Useful knowledge to help solve(all answers are based on order matters)a)in how many ways can five cards be drawn from the deck?= 311875200 ways
b)in how many ways can 3 aces be drawn?= 24 ways
c)in how many ways can 2 kings be drawn?= 12 ways
d)in how many ways can 3 aces and 2 kings be drawn? 24x12=288 ways

3. In order to solve question #1, I just put: 288/311875200= 9.234463, does this equation make sense? or am I computing this wrong?

 

[Dick]

1. So the question is "what is the probability of drawing a full house consisting of three aces and two kings?"

2. Useful knowledge to help solve(all answers are based on order matters)a)in how many ways can five cards be drawn from the deck?= 311875200 ways
b)in how many ways can 3 aces be drawn?= 24 ways
c)in how many ways can 2 kings be drawn?= 12 ways
d)in how many ways can 3 aces and 2 kings be drawn? 24x12=288 ways

3. In order to solve question #1, I just put: 288/311875200= 9.234463, does this equation make sense? or am I computing this wrong?

First, 288/311875200=9.234463 isn't right. There's an exponent on that, isn't there? And if you are going to do it in an 'order manners' way, you need to account for the number of ways the aces can be mixed up with the kings.

 

[YODA0311]

First, 288/311875200=9.234463 isn't right. There's an exponent on that, isn't there? And if you are going to do it in an 'order manners' way, you need to account for the number of ways the aces can be mixed up with the kings.

1.You are correct 288/311875200=9.23446 E-7.
2.You also stated "account for the number of ways the aces can be mixed up with the kings." I calculated 1680 ways aces can be mixed up with the kings.
Calculations= 4 kings and 4 aces=8 total cards, then out of the 8 is 4 aces. Therefore 8!/(8-4)!=8!/4!= 1680 number of ways aces can be mixed to acquire aces. I am sooo confused. Plz help

 

[Dick]

1.You are correct 288/311875200=9.23446 E-7.
2.You also stated "account for the number of ways the aces can be mixed up with the kings." I calculated 1680 ways aces can be mixed up with the kings.
Calculations= 4 kings and 4 aces=8 total cards, then out of the 8 is 4 aces. Therefore 8!/(8-4)!=8!/4!= 1680 number of ways aces can be mixed to acquire aces. I am sooo confused. Plz help

You only have to count the number of ways to mix up the three aces and the two kings that you drew. You've counted AAA and KK, now you count the ways to mix them, like AAAKK, AAKAK, AKAAK, etc. Notice there is no order in this count. You've already ordered AAA and KK. BTW, it's easier to handle this problem if you use unordered counts.

 

[YODA0311]

You only have to count the number of ways to mix up the three aces and the two kings that you drew. You've counted AAA and KK, now you count the ways to mix them, like AAAKK, AAKAK, AKAAK, etc. Notice there is no order in this count. You've already ordered AAA and KK. BTW, it's easier to handle this problem if you use unordered counts.

Is there a simple formula to use in this? I was taught the combo and permutation rule

 

[YODA0311]

Is there a simple formula to use in this? I was taught the combo and permutation rule

Actually since I figured out the answer to- 1.how many ways can three aces be drawn=24
2. how many ways can two kings be drawn=12

To answer the question of how many ways can three aces and two kings be drawn just add= 24+12= 36 ways in which three aces and two kings can be drawn

 

[Dick]

Is there a simple formula to use in this? I was taught the combo and permutation rule

Yes, there is. Of the five places in the hand you have to pick two places to put the aces. It's a combo problem. What's the answer? You could also do the whole thing as a combo problem.

 

[Dick]

Actually since I figured out the answer to- 1.how many ways can three aces be drawn=24
2. how many ways can two kings be drawn=12

To answer the question of how many ways can three aces and two kings be drawn just add= 24+12= 36 ways in which three aces and two kings can be drawn

You don't add them. Each of the 24 ways to draw three aces can be paired with each of the 12 ways to draw two kings. It's a multiplication. Like you did before.

 

[YODA0311]

You don't add them. Each of the 24 ways to draw three aces can be paired with each of the 12 ways to draw two kings. It's a multiplication. Like you did before.

Oh well don't I feel stupid, so it would be 24x12? which=288 ways three aces and two kings can be drawn

 

[Dick]

Oh well don't I feel stupid, so it would be 24x12? which=288 ways three aces and two kings can be drawn

Yes, but you have to multiply by another factor that counts how many ways you can assemble them into a hand of five cards. There's more than one way, right?

 

[YODA0311]

Yes, but you have to multiply by another factor that counts how many ways you can assemble them into a hand of five cards. There's more than one way, right?

okay so what I am getting from you is, there is 288 ways to draw them, but once i have 5 cards in my hand there are more ways to order them? IF i understand right here is my calculations= n!/(n-r)!= 288!/(288-8)!=288!/280!=4.2908712297265 E19

 

[YODA0311]

okay so what I am getting from you is, there is 288 ways to draw them, but once i have 5 cards in my hand there are more ways to order them? IF i understand right here is my calculations= n!/(n-r)!= 288!/(288-8)!=288!/280!=4.2908712297265 E19

actually it would be n!/(n-r)!= 288!/(288-5)!=288!/283!=1913390519040

 

[Dick]

actually it would be n!/(n-r)!= 288!/(288-5)!=288!/283!=1913390519040

Ack! You aren't ordering the 288 ways! You want to put the three ordered kings and the two ordered aces into a hand of five. You just need to pick three places out of the five to put the kings in! How many ways can you pick three places out of five. It's a combo problem!

 

[YODA0311]

Ack! You aren't ordering the 288 ways! You want to put the three ordered kings and the two ordered aces into a hand of five. You just need to pick three places out of the five to put the kings in! How many ways can you pick three places out of five. It's a combo problem!

I understand, but my professor told us that order counts. So I used the permutation formula when order counts... He acknowledges that (ace of spade, ace of heart, ace of diamond, king of heart, and king of diamonds) is not the same as (ace of spade, ace of heart, ace of diamond, king of diamonds, and king of heart). He views those two examples as separate counts. Therefore I have to take "order" into account

 

[Dick]

I understand, but my professor told us that order counts. So I used the permutation formula when order counts... He acknowledges that (ace of spade, ace of heart, ace of diamond, king of heart, and king of diamonds) is not the same as (ace of spade, ace of heart, ace of diamond, king of diamonds, and king of heart). He views those two examples as separate counts. Therefore I have to take "order" into account

Order counts if it's a problem where order counts. You can also do this problem without worrying about order at all, which is what most people would do. Then we wouldn't have to fuss about this extra factor. The three kings and the two aces are already in a specific order. Choosing the three places to put the three kings in your five card hand is not a part of the problem where order counts.

 

[YODA0311]

Order counts if it's a problem where order counts. You can also do this problem without worrying about order at all, which is what most people would do. Then we wouldn't have to fuss about this extra factor. The three kings and the two aces are already in a specific order. Choosing the three places to put the three kings in your five card hand is not a part of the problem where order counts.

I thank you for your assistance, but unfortunately I am not absorbing your concept. :(

 

[Dick]

I thank you for your assistance, but unfortunately I am not absorbing your concept. :(

No, you aren't. You are insisting on 'order counts' in a problem where it really doesn't, and that's making it harder. Ordinarily, I'd say don't open two threads on the same problem, but maybe a fresh start with somebody else will help.

 

[vela]

I think a specific example would help illustrate what Dick is saying. To keep the numbers manageable, let's consider a deck of cards with 3 aces, which I'll denote by the letters a, b, and c, and three kings, which I'll call 1, 2, and 3, and let's count the hands consisting of two aces and one king.

Using your counting method, you have 3x2 = 6 ways to draw two aces and 3 ways to draw the king, so there are 18 ways to draw two aces and one king, as shown in the table below:

    ab  ac  bc  ba  ca  cb

1   1ab 1ac 1bc 1ba 1ca 1cb
2   2ab 2ac 2bc 2ba 2ca 2cb
3   3ab 3ac 3bc 3ba 3ca 3cb

But to come up with all of the hands, you still have to account for how the kings and aces are arranged in the hand. For example, from 1ab, you can get the hands 1ab, a1b, and ab1. So the total number of ordered hands is 18x3 = 54.

Note there are 3 hands for each entry in the table. It's pretty clear there's three if you just consider where the king can go — it can go in one of three spots — but you should be able to come up with the same number by considering where the aces go. Dick's method says you want to choose 2 out of 3 slots without caring about the order because the order is already accounted for by the way you chose the cards. So he says there are 3c2 ways to do this, which is the correct answer.

Because you insist the order of the aces matter when counting the possible arrangements, you say there are 6 ways: K A1 A2, K A2 A1, A1 K A2, A2 K A1, A1 A2 K, A2 A1 K. Hence, there would be 18x6 = 108 hands.

But let's look at the hands the two entries from the table 1ab and 1ba would generate. From 1ab, you'd get:

1ab 1ba a1b b1a ab1 ba1

while from 1ba you'd get:

1ba 1ab b1a a1b ba1 ab1

They're exactly the same hands! You're counting each hand twice, so there's really only 108/2 = 54 hands. Do you see why that happens?

 

[YODA0311]

I think a specific example would help illustrate what Dick is saying. To keep the numbers manageable, let's consider a deck of cards with 3 aces, which I'll denote by the letters a, b, and c, and three kings, which I'll call 1, 2, and 3, and let's count the hands consisting of two aces and one king.

Using your counting method, you have 3x2 = 6 ways to draw two aces and 3 ways to draw the king, so there are 18 ways to draw two aces and one king, as shown in the table below:

    ab  ac  bc  ba  ca  cb

1   1ab 1ac 1bc 1ba 1ca 1cb
2   2ab 2ac 2bc 2ba 2ca 2cb
3   3ab 3ac 3bc 3ba 3ca 3cb

But to come up with all of the hands, you still have to account for how the kings and aces are arranged in the hand. For example, from 1ab, you can get the hands 1ab, a1b, and ab1. So the total number of ordered hands is 18x3 = 54.

Note there are 3 hands for each entry in the table. It's pretty clear there's three if you just consider where the king can go — it can go in one of three spots — but you should be able to come up with the same number by considering where the aces go. Dick's method says you want to choose 2 out of 3 slots without caring about the order because the order is already accounted for by the way you chose the cards. So he says there are 3c2 ways to do this, which is the correct answer.

Because you insist the order of the aces matter when counting the possible arrangements, you say there are 6 ways: K A1 A2, K A2 A1, A1 K A2, A2 K A1, A1 A2 K, A2 A1 K. Hence, there would be 18x6 = 108 hands.

But let's look at the hands the two entries from the table 1ab and 1ba would generate. From 1ab, you'd get:

1ab 1ba a1b b1a ab1 ba1

while from 1ba you'd get:

1ba 1ab b1a a1b ba1 ab1

They're exactly the same hands! You're counting each hand twice, so there's really only 108/2 = 54 hands. Do you see why that happens?

Okay if i was using the combination method(order does not count) I am having a hard time incorporating the factors. the equation is = n!/(n-r)!r!, n=different items available, we select "r" of the "n" items(without replacement)
I do not understand how you got n=3, r=2?

 

[vela]

Three (n=3) spots and two (r=2) aces. Out of three positions in the hand, you're choosing two to place the aces in. You could also consider just the king, in which case you have n=3 and r=1. Either way, you get 3 hands from each entry in the table.

 

[YODA0311]

Three (n=3) spots and two (r=2) aces. Out of three positions in the hand, you're choosing two to place the aces in. You could also consider just the king, in which case you have n=3 and r=1. Either way, you get 3 hands from each entry in the table.

so to answer the question "in how many ways can three aces and two kings be drawn?
is 3!/(3-2)!2!=3!/1!2!=6/2=3= there are 3 ways, three aces and two kings can be drawn

 

[vela]

No. What problem are you trying to solve now? It seems like you've changed gears. Are you asking about drawing cards from the deck or arranging the already drawn cards?

 

[YODA0311]

No. What problem are you trying to solve now? It seems like you've changed gears. Are you asking about drawing cards from the deck or arranging the already drawn cards?

Vela I appreciate your help and I will start fresh so you can better understand the problem:
a. in how many ways can five cards be drawn from the deck? I answered= 311875200
b. in how many ways can three aces be drawn? i answered= 24
c. in how many ways can two kings be drawn? i answered= 12
d. in how many ways can three aces and two kings be drawn=? (this is the current question that i am trying to figure out)
My professor told us that order counts, that is why i have been using the permutation formula= n!/(n-r)!

 

[vela]

I've already read the thread, and you seemed to have figured out how to answer a-c.

You also found there are 288 ways to get 3 aces and 2 kings, but this is for one particular type of hand, e.g. the first ace drawn in the first spot, the second ace drawn in the second spot, the third ace drawn in the third spot, the first king drawn in the fourth spot, and the second king drawn in the last spot. If you were to make a table like I did in my example, you'd have 288 entries in the table.

What you need to do now is figure out how many hands you can generate from each entry in the table.

In my example, I used n=3. Why did I do that? What value should you use for your problem?

 

[YODA0311]

I've already read the thread, and you seemed to have figured out how to answer a-c.

You also found there are 288 ways to get 3 aces and 2 kings, but this is for one particular type of hand, e.g. the first ace drawn in the first spot, the second ace drawn in the second spot, the third ace drawn in the third spot, the first king drawn in the fourth spot, and the second king drawn in the last spot. If you were to make a table like I did in my example, you'd have 288 entries in the table.

What you need to do now is figure out how many hands you can generate from each entry in the table.

In my example, I used n=3. Why did I do that? What value should you use for your problem?

in my problem, n=288, r=5. I came up with that based on there are 288 ways, three aces can be drawn out of four aces, and two kings out of four kings. Then i used 5, because we want to know within a hand how many different arrangements can we get

 

[vela]

No, that's completely wrong. Please answer the question I asked earlier. Why did I use n=3 in my example?

 

[YODA0311]

No, that's completely wrong. Please answer the question I asked earlier. Why did I use n=3 in my example?

you chose 3 because that is how many spots the aces are taking up

 

[vela]

In the example in post 18?

Rather than trying to tease out what I'm looking for by asking leading questions, let me just ask that you explain in your own words why I multiplied 18 by 3 to get the total number of hands and how to get that factor of 3.

 

[YODA0311]

In the example in post 18?

Rather than trying to tease out what I'm looking for by asking leading questions, let me just ask that you explain in your own words why I multiplied 18 by 3 to get the total number of hands and how to get that factor of 3.

in that example you used 3 kings and 3 aces, 3 of each

 

[vela]

That's not an explanation.

 

[YODA0311]

That's not an explanation.

I think i figured out problem "in how many ways can three aces and two kings be drawn?"
I used the permutations rule(when some items are identical to others). 3 aces and 2 kings are different, but are all considered cards.

In the book they used an example of how many was can 11 girls and 3 boys be arranged? there children, but different because of gender. similar to the cards.

To answer the card question I used there formula. n!/ n1!*n2!, 5!/3!*2!=120/12=10. Thus I came up with 10 different ways 3 aces and 2 kings can be drawn. does this make sense? thank you

 

[YODA0311]

I think i figured out problem "in how many ways can three aces and two kings be drawn?"
I used the permutations rule(when some items are identical to others). 3 aces and 2 kings are different, but are all considered cards.

In the book they used an example of how many was can 11 girls and 3 boys be arranged? there children, but different because of gender. similar to the cards.

To answer the card question I used there formula. n!/ n1!*n2!, 5!/3!*2!=120/12=10. Thus I came up with 10 different ways 3 aces and 2 kings can be drawn. does this make sense? thank you

Actually that answer only applies to 1 hand of 5 cards. Therefore I must multiply it by 288.
in how many ways can three aces be drawn=24, in how many ways can two kings be drawn=12...12x24=288

 

[vela]

The 10 is correct, but I want to point out a few things:

1. That's the number of ways the 3 aces and 2 kings can be arranged, not drawn. You keep saying it's the number drawn, and that's wrong.
2. Note that you are ignoring the order of the aces and kings when you calculate the number of arrangements. This is the point Dick and I were trying to make.
3. 5!/(3!*2!) is exactly what you'd get if you calculate 5 choose 3 or 5 choose 2.

 

[YODA0311]

The 10 is correct, but I want to point out a few things:

1. That's the number of ways the 3 aces and 2 kings can be arranged, not drawn. You keep saying it's the number drawn, and that's wrong.
2. Note that you are ignoring the order of the aces and kings when you calculate the number of arrangements. This is the point Dick and I were trying to make.
3. 5!/(3!*2!) is exactly what you'd get if you calculate 5 choose 3 or 5 choose 2.

I understand where both you are coming from now. I guess it was easier to read from a textbook and follow the examples. However the actual question is "in how many ways can three aces and two kings be drawn?"
So i figured out how 1 hand of 3 aces and 2 kings can be arranged=10
I found out how many ways 3 aces can be drawn=24
I found out how many ways 2 kings can be drawn=12
If I were to multiply both the 24 and 12= 288, then I should multiply 288 by the 10(arrangements per hand)
That would equal 2880, thus there are 2880 ways 3 aces and 2 kings can be drawn. Is that sound about right?

 

[vela]

Yup, that's the answer!

 

[Dick]

I understand where both you are coming from now. I guess it was easier to read from a textbook and follow the examples. However the actual question is "in how many ways can three aces and two kings be drawn?"
So i figured out how 1 hand of 3 aces and 2 kings can be arranged=10
I found out how many ways 3 aces can be drawn=24
I found out how many ways 2 kings can be drawn=12
If I were to multiply both the 24 and 12= 288, then I should multiply 288 by the 10(arrangements per hand)
That would equal 2880, thus there are 2880 ways 3 aces and 2 kings can be drawn. Is that sound about right?

And you can also get the same probability answer without considering ordering. The number of unordered ways to pick the cards is C(52,5). The number of unordered ways to pick three kings is C(4,3) and two aces is C(4,2). So C(4,3)*C(4,2)/C(52,4) is the same as your 2880/311875200. That would be the more popular way to solve the probability problem.

 

[Bacle]

Are you playing with a full deck :) ?

But seriously; I don't recognize the 311875200 as an nC5 (i.e., "n choose 5" ).

Are you using replacement?

If you have x cards with 4 suits of 13 cards each ; with 1,2,...,j,q,k in each suit, and

you're dealing without replacement, then the answer would be 4C3*4C2 , i.e.,

select any 3 aces from the four you have in 4C3 ways ,choose 2 kings in 4C2.

If you deal with replacement, you can choose in 4!/(4-3)! , and 4!(4-2)! ways

respectively.

If you give us the original numbers, we may be better able to help.

 

[YODA0311]

Are you playing with a full deck :) ?

But seriously; I don't recognize the 311875200 as an nC5 (i.e., "n choose 5" ).

Are you using replacement?

If you have x cards with 4 suits of 13 cards each ; with 1,2,...,j,q,k in each suit, and

you're dealing without replacement, then the answer would be 4C3*4C2 , i.e.,

select any 3 aces from the four you have in 4C3 ways ,choose 2 kings in 4C2.

If you deal with replacement, you can choose in 4!/(4-3)! , and 4!(4-2)! ways

respectively.

If you give us the original numbers, we may be better able to help.

Yes part (a) is from a whole deck of cards, however I already checked the answers with previous tutors. However maybe you see something wrong.
I was using the permutation formula= n!/(n-r)!

 

[YODA0311]

Are you playing with a full deck :) ?

But seriously; I don't recognize the 311875200 as an nC5 (i.e., "n choose 5" ).

Are you using replacement?

If you have x cards with 4 suits of 13 cards each ; with 1,2,...,j,q,k in each suit, and

you're dealing without replacement, then the answer would be 4C3*4C2 , i.e.,

select any 3 aces from the four you have in 4C3 ways ,choose 2 kings in 4C2.

If you deal with replacement, you can choose in 4!/(4-3)! , and 4!(4-2)! ways

respectively.

If you give us the original numbers, we may be better able to help.

To make it more clearer for you:
(a) in how many ways can 5 cards be drawn from a whole deck?= 52!/(52-5)!=311875200
(b)how many ways can 3 aces be drawn from the total 4 possible aces?=4!/(4-3)!=24
(c)how many ways can 2 kings be drawn from the total 4 possible kings?=4!/(4-2)!=12
(d)how many ways can 3 aces and 2 kings be drawn?=they can be drawn 24x12=288 ways. Then I used n!/n!1*n!2=5!/3!2!=120/12=10, which equals the amount of ways 1 five card hand can be arranged. Then I took 10x288=2880.
Now I need to answer what is the probability of drawing a full house consisting of 3 aces and 2 kings?