Expected Value for # of Cards Dealt

[micheko1]

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 =(

 

[bpet]

I don't know if there's a simple way to calculate it, so try finding an approximation by Monte Carlo simulation.

 

[jz92wjaz]

Is it a full deck of cards?

I was really good at probability when I took it, but I took that class years ago. I would probably write a computer program for the answer to the first question. I know how to solve it, but not without about 1000 different terms in my equation.

What do you mean by theoretical approach for the second problem? The answer is obviously 5 and I really don't see any explanation needed for it.

 

[Stephen Tashi]

This is not the kind of problem that I like to think about, but if I were forced to, I'd begin with a simpler example.

Say a deck has only 5 cards: 2 aces, 1 two and 2 jacks. Say we draw 3 cards. How do I enumerate all possible draws if I don't care about the order in which the cards are drawn?

I'd look at the product:

(A + A + T + J + J) (A + A + T + J + J) (A + A + T + J + J)

I'd write a program (or use a computer algebra package) to multiply that out and "combine like terms". Then the coefficient of (A^2 T) would be the number of 3 card hands that have 2 aces and one two, etc.

One name for this technique is "combinatorial generating functions".

Then the program would have to go through the list of terms and figure out the value of it's associated hand based on the points assigned to each card, and how many of the highest cards are counted.

 

[jz92wjaz]

That algorithm implies that the card is put back with each draw. There should only be one Ace Ace Two hand possible, if I understood the original question correctly (you draw and keep the card, not draw it and put it back).

Edit: The problem is quite tricky, even for programming. It'll take 6 loops for the 6 card hand and 8 for the 8 card hand, as the variables are not independent.

 

[Stephen Tashi]

That algorithm implies that the card is put back with each draw. There should only be one Ace Ace Two hand possible, if I understood the original question correctly (you draw and keep the card, not draw it and put it back).

Good point. What about looking at all terms with an x^3 in them in the product

(1 + Ax)(1 + Ax)(1 + Tx) (1 + Jx)(1 + Jx)

?

 

[jz92wjaz]

Micheko, is this a homework problem? If so, for what class? If not, where does the problem come from?

Stephen, that looks like it would work, though I would just do 4 choose 3 to find all of the 10 point card combinations, then 1*(4 choose 2) for the 2 point and two 10 point card combos.

(30*4!/3! + 22*1*4!/(2!*2!))/(5!/(3!*2!)) = (30*4 + 22*6)/10 = 25.2

Edit: In this case, the expectation of 3 cards is 3 times the expectation of one (which is 8.4). If we took the best two of three, the expectation is 20. I may play around with a different set of numbers and find an easier way to do the original problem.

 

[Stephen Tashi]

I would just do 4 choose 3 to find all of the 10 point card combinations, then 1*(4 choose 2) for the 2 point and two 10 point card combos.

That''s fine for counting the combinations. I'm thinking in terms of a program that would evaluate each possible hand, even if the method of assigning points to it was complicated. So I want to generate a symbolic representation of what's in the hand as well as count its occurrences.

 

[jz92wjaz]

Here we go, there are five unknown values, a, b, c, d, and e, listed in reverse numerical order (a is the largest).
The combinations are as follows for drawing 3:
abc,abd,abe,acd,ace,ade,bcd,bce,bde,cde

If you take the best two, you get the following hand values:
a+b, a+b, a+b, a+c, a+c, a+d, b+c, b+c, b+d, c+d

That is 6*a + 6*b + 5*c + 3*d all over 10

If you break up the problem to look at how many hands each card is in, you might have better luck with the problem. Try starting with the first 10 point card in the 6 card problem. It is part of 51 choose 5 solutions. Card #2 is the highest in 50 choose 5 solutions, but part of 50 choose 4 solutions where card a is also included, and so on.

Using that same idea, the abcde example would look like the following equation:

a*(4 choose 2) or (a is the highest, and paired with a lower number)
+ b*(1 choose 1 * 3 choose 1 + 3 choose 2) or (the higher value paired with b 3 times, then b paired with a lower value)
+ c*(2 choose 1 * 2 choose 1 + 2 choose 2) or (c paired with a higher value and a lower value, and c paired with two lower values)
+ d*(3 choose 1 * 1 choose 1) or (d paired with a higher value and a lower value)

all divided by 5 choose 3 (which is 10)

The problem will still be messy, but this may simplify it. Also, I suspect that you'd find a summation if you did enough terms, and that might make it easier.

Edit:

That''s fine for counting the combinations. I'm thinking in terms of a program that would evaluate each possible hand, even if the method of assigning points to it was complicated. So I want to generate a symbolic representation of what's in the hand as well as count its occurrences.

Your program idea would work great as far as I can tell, and it is a more clever way of doing it than I had originally thought of. I just like finding a combination solution.