Exact calculations of odds for a blackjack hand if you knew the remaining cards

I was wondering how to calculate the best play for a particular blackjack hand if you had perfect knowledge of the remaining cards in a shoe. The particular example I came across was the following: The player has a hard 17 against the dealer's up card of 10. The remaining cards are 2, 2, 2, 2, 3, 3, 3, 3, 10, 10, 10, Ace, Ace, Ace. The person who posed the problem claimed the player should hit, since he had 11/14 = 78.6% chance of improving his hand. But that doesn't take into account the possibilities of the dealers outcomes. To get the exact odds, I would think you would have to look at all possible outcomes. I can see how the 24 permutations of the 4 "2"s would add up to 24 loses if the player stood on the hand, but my math skills are far too weak to tackle the complications of the remaining combinations. I would much appreciate if you could take a look, and if possible show the method of calculation, or direct me on how to solve this kind of problem.
Thanks,
RDF

This is definitely a "brute force" problem - go through every possible result, then add them up.
I am assuming that one of the 14 cards is the dealer's hole card - and that it is not an Ace, as otherwise the dealer would have blackjack.

If you stand, the possible dealer results (starting with the hole card, then with any hit cards):
10 - dealer has 20 (probability 3/11 - it's 3/11 and not 3/14 as you know the first card is not an Ace)
3 10 - dealer busts (probability 4/11 x 3/13)
3 3 10 - dealer busts (probability 4/11 x 3/13 x 3/12)
3 3 3 - dealer has 19 (probability 4/11 x 3/13 x 2/12)
3 3 2 - dealer has 18
3 3 A - dealer has 17
3 2 10 - dealer busts
3 2 3 - dealer has 18
3 2 2 - dealer has 17
3 2 A 10 - dealer busts
3 2 A 3 - dealer has 19
3 2 A 2 - dealer has 18
3 2 A A - dealer has 17
2 10 - dealer busts
2 3 3 - dealer has 18
2 3 2 - dealer has 17
2 3 A 10 - dealer busts
and so on
You win with a bust, push with a 17, and lose with an 18 or 19. 20 and 21 are impossible as the highest number that hits is 16, and the highest number that doesn't bust is 3.

If you hit:
10 - you bust
3 - you stand on 20; go through every one of the dealer's combinations again, but keep in mind that there are only three 3s in the deck
2 - you stand on 19; go through every one of the dealer's combinations again, but keep in mind that there are only three 2s in the deck
A - you stand on 18; go through every one of the dealer's combinations again, but keep in mind that there are only two Aces in the deck

No, I'm not going to go through all of that manually; that's what computers are for.

(sorry I posted this at same time as above post - as you can see it's basically the same idea.)
One way is to look at all the series of cards the dealer might draw (I'm assuming the dealer doesn't have a down card - which would not be an Ace).

So you look at
2 2 2 2 (P=4/14 * 3/13 * 2/12 * 1/11) etc.
2 2 2 3
2 2 2 T
2 2 2 A
2 2 3
etc.
and add them up and derive the chances of the player winning.
Then you have the player take a 2 / 3 / A and repeat, then factor in that taking a 10 loses.
My gut feeling is you hit since the chances of the dealer busting are fairly low and if you draw a 3 you're a standoff vs dealer's 20.

Using the brute force method I get a player stand EV of -10.0% and a player hit once EV of +13.9%.