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**mofaye****Member**- Registered: 2009-11-17
- Posts: 3

If each card has same points as their number (Ace is 1 point, 2 is 2 points, 3 is 3 points), the Jack, Queen, King each is 10 points...what's the expected value for the three highest cards out of six dealt cards? What's the expected value for the three highest cards out of eight dealt cards? (they're theoretical expected values)

please help~~~~ thanks!

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**soroban****Member**- Registered: 2007-03-09
- Posts: 451

If each card has same points as their number (Ace = 1, Deuce = 2, , Trey = 3, . . . Jack, Queen, King = 10),

what's the expected value for the three highest cards out of six dealt cards?

[Who assigned it anyway? . . . Professor deSade?]

.

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**gckc123****Member**- Registered: 2009-10-19
- Posts: 15

I have a feeling that we may be able to solve this without enumerating all the combination

Let y_max be the point of the largest of the six card

Let y be the point of a card drawn randomly from the deck

we can work out the probability mass function of y_max

say for example,

P(y_max = 10) = 6!/5! P(y = 10) (P(y <= 10 | a card with 10 points is drawn))^5

= 6*P(y = 10)

we can then work out the probability mass function of y_(5), which is the poit of the second largest card.

Haven't think about it carefully. But i think we might tackle this problem using the idea similar to order statistics.

Maths is fun!

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**mofaye****Member**- Registered: 2009-11-17
- Posts: 3

Thank you guys!

but i have to say this is HIGH SCHOOL question, and it's one of questions on my g12 data management project.

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**froggy21****Member**- Registered: 2009-11-16
- Posts: 8

hallo~

I'm mofaye's partner on this inane project : P What we were planning to do was just show this method with a few examples on how to get the probabilities of a few sums as a start of point. I'm having trouble though with the equation for each probability

for example, to get a sum of 30...would it be

(16C3)*(48C3)/(52C6)

or is it

(16C3)*(48C3)/52x51x50x49x48x47?

I thought at first that it was the latter one but my brother said that the hyper geometric formula took the dependency into account already. Also, if for example, we're looking for the probability for a sum of 15. If one possible way to get that is 10, 3, 2, do we need to multiply the resulting probability by 3! to account for the permutations? : ( Sorry for asking so many questions and thank you so much for all the help so far.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi mofaye;

mofaye wrote:

but i have to say this is HIGH SCHOOL question, and it's one of questions on my g12 data management project.

This is a HIGH SCHOOL question? Oh, I get it, you already have the answer. So why are you posting in Help Me? I will withdraw my previous posts and answers and work on something else.

*Last edited by bobbym (2009-11-18 20:19:40)*

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**froggy21****Member**- Registered: 2009-11-16
- Posts: 8

Aah, sorry about mofaye... she's a bit...meh. We don't actually have an answer for it, since this is part of our DM project (and before anyone says it's cheating by asking here, I already asked my teacher and she said we're allowed to ask other people for help). I think what mofaye meant was we can't present to her the answer that's achieved from a simulation (because it's not part of our curriculum) and we need to show the break down of how to get each probability....

Thank you still for all your help and sorry if we offended you : (

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**mofaye****Member**- Registered: 2009-11-17
- Posts: 3

I'm REALLY SORRY!!! It's probably a misunderstanding in my sentence, what I actually mean is the question is super hard and we are not allowed to ask teacher, so it's pretty frustrated.

Again, I really want to apologize to you if my words offended you.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi mofaye and froggy21;

I wasn't offended I just thought that you at least could get the answer from your teacher. I withdrew my stuff because if you have an analytic answer what does anybody need with the computer solution?

Here is the important part of the post:

Hi mofaye;

By enumerating all

with a computer, this is the exact answer.

as the expected value of the sum of the 3 highest cards dealt out of 6.

As a check to the above work, I did an ordinary simulation of dealing out 6 cards 25 000 000 times:

The simulation came up with an average value of 26.73757884 for the sum of the 3 highest cards dealt out of 6. This is very close to the exact answer.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi froggy21;

froggy21 wrote:

for example, to get a sum of 30...would it be

Soroban in post#2 already started that sort of analysis, take a look at what he has done .

froggy21 wrote:

I thought at first that it was the latter one but my brother said that the hyper geometric formula took the dependency into account already.

I think he is correct. For these you would use the multivariable hypergeometric distribution.

*Last edited by bobbym (2009-11-19 14:29:36)*

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**

**froggy21****Member**- Registered: 2009-11-16
- Posts: 8

Aah, thank you very much for all your help!

We ended up using a (sort of semi-guessing, ahaha) formula for our project wherein we find the E(x) for the highest card in 6 dealt and multiplying that by 3. The value's a bit off but it does give a ballpark answer and our teacher said as long as she can follow the logic of what we were trying to attempt, then it's ok. Mofaye thought of the method while looking at the methods you gave us so thank you, because you all helped a lot : D

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

Hi froggy21;

Glad to help.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,229

mofaye wrote:

If each card has same points as their number (Ace is 1 point, 2 is 2 points, 3 is 3 points), the Jack, Queen, King each is 10 points...what's the expected value for the three highest cards out of six dealt cards? What's the expected value for the three highest cards out of eight dealt cards?

Finally the long awaited answer to this. I just realized that gAr when solving one of mine over in another thread has solved this one too!

This agrees well with simulations.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**