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#1 2011-04-04 09:33:06

micheko
Member
Registered: 2011-04-04
Posts: 3

Expected Value

Hello,

I am having A LOT OF TROUBLE with this problem. Honestly, I did check my sources before coming here and I still cannot figure it out. =(

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

I want to find the theoretical expected value for the above 2 questions. I don't know where to start.

My second question is related to the above as well. Say in this game (with the rules described above) the opponent is dealt 4 cards. and you are dealt the number of cards according to the sum of the 2 dice rolled (a sum of 4 gives 4 cards ...and so on. doubles give 7 cards). How many cards do you need to be dealt before you have an advantage to win over the opponent? Can someone do this in a theoretical approach?

THANK YOU.
Please help me, I am really stuck =(

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#2 2011-04-04 09:41:39

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Expected Value

Hi micheko;

Welcome to the forum.

You are supposed to have a lot of difficulty with those three problems they are very tough. The first one for six cards was already posed here and at the time I could only get a computer simulation answer.

The other two are naturally even harder.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2011-04-04 09:46:35

micheko
Member
Registered: 2011-04-04
Posts: 3

Re: Expected Value

Oh ...=( I knew, I saw your answer already in the 2009 thread from way back. But that was the expected value of just 3 cards. Not 3 highest cards out of 6. I actually got that question (your computer answer question) doing the same as you did. For that question, I did the expected value of a single card and multiplied it by 3 to get the total of the 3 cards.

sad this is for a school midterm project and those 3 questions are worth basically 50% of my entire project.

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#4 2011-04-04 10:58:45

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Expected Value

Hi;

Not that it is really going to matter but you incorrect. That thread in 2009:

http://www.mathisfunforum.com/viewtopic … 01#p123901

Is for 3 highest cards out of 6. your exact first question. The two posters Mofaye and froggy. I have a good memory for problems I worked on and do not solve satisfactorily. I will review my findings back then and you should see what Mofaye and froggy accepted because it got them through the same type questions.

Incidentally they should be 100% of your grade! 50% for questions that tough is ridiculous. I will post when I get something.

Also I do have an exact answer for 3 highest cards out of 6 dealt. it is


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#5 2011-04-04 12:27:50

micheko
Member
Registered: 2011-04-04
Posts: 3

Re: Expected Value

Sorry I think I didn't see that one. Thanks for bringing it up. Wow...it's looked like the same questions pop up again after 2 years. Ah.

Yes..so I looked over how you did the question, but how did you get the 2 numbers in that fraction/probability? I don't know where it came from. I know the 52 choose 6 part ...but where does that fraction came from (the numbers I mean)? Can you explain it to me?

Thanks.

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#6 2011-04-04 12:57:54

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Expected Value

Hi micheko;

That is the exact answer but not every time you see an exact answer should you assume that some method got it. I had a computer deal out every possible combination of 6 cards then I just had it pick out the 3 highest cards and add them up. Because it generated all the possibilities it was possible to get an exact answer.

This method is not possible for the 8 card problem. Why?

Almost a billion 8 card hands that is a little too much.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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