Expected value of 3 cards dealt

[froggy21]

If each card on a regular deck has points that corresponds to their number (like 2 of hearts is 2 points, 7 of clubs is 7 points), the Jack, Queen, King each being 10 points...what's the expected value of your opponent's hand if you deal them 3 cards?

I know the empirical expected value...but I'd like to know -how- to get the theoretical expected value, please : )

Help please D: I've been stewing over this question for days now. The only way I can think of doing this is by doing a tree diagram to get each probability but that'll have like 1000 end branches -headdesk-

 

[CRGreathouse]

What's an ace worth? 1, 10, 11, 13?

The expected value wouldn't be too far off from three times the expected value of one card.

 

[CRGreathouse]

I'll assume the ace is valued at something other than 10.

So there are 52 * 51 * 50 ways to choose three cards from a deck. Of those, the breakdown is

216 ways to draw 3 of the same non-10
10368 ways to draw 2 of the same and one different, none 10
32256 ways to draw 3 different non-10s
5184 ways to draw a 10 and 2 of the same non-10s
55296 ways to draw a 10 and 2 different non-10s
25920 ways to draw 2 10s and a non-10
3360 ways to draw 3 10s

So calculate the average value for each, multiply, add, and divide.

 

[froggy21]

Ah, yes, Ace is worth 1 points : )

Uhm, sorry, would you happen to know the breakdown for 4 cards? I misread the question and apparently it's the expected value of 4 cards dealt. I tried doing the breakdown myself but I always seem a few hundred thousand short of the total ways.

Thank you very much for all the help!

 

[bpet]

Note that E(X+Y)=E(X)+E(Y) holds regardless of the dependence between X and Y - so you won't need to work out all 13^4 combinations.

 

[awkward]

To expand on bpet's remark, the expected value of one card is 85/13, assuming an ace is 1 and Jack, Queen, King are 10 each. Let's say the value of the ith card is $$X_i$$. Then the expected value of 3 cards is

$$E(X_1 + X_2 + X_3) = E(X_1) + E(X_2) + E(X_3) = 3 \times 85/13$$.

That's all it takes.