Expected value for blackjack In the game of blackjack, the odds of winning each hand are slightly less than 50 percent. As you play an infinite amount of hands, you would always lose money because you would win less than 50 percent of the time. By this rationale, wouldn't your highest odds of winning be if you only played one game?
 A: Definitely. Consider going to the casino with $100\$$. Your best strategy would be to bet the $100 \$ $ immediately in a single hand.
Betting a lesser amount each time does not change the odds of winning a single hand, but it does lower the variation of your return. In other words, the probability that you end up in the plus gets lower the more bets you make.
NB: The above is assuming you bet the same amount multiple times, e.g. betting $100\$$ is a better strategy than betting $10$ times $10\$$, which is a better strategy than betting $100$ times $1\$$. But there are other strategies, for instance this one, that uses Martingales. 
And ofcourse the best strategy is to bet $0\$$ and go buy a sandwich across the street.
A: You are correct in that a typical "house edge" falls just below 50 percent in most casinos.
Note that probabilities of winning fluctuate as cards are constantly being dealt, due to ratio of high cards (good) to low cards (bad) remaining in the deck fluctuates. The study of these changing probabilities is what lead to the birth of card counting. 
If one becomes a skilled counter, they can place higher bets during stages in which the player has the probabilistic advantage over the casino. If played correctly, card counters generate a positive EV using strategic betting, and actually generate a profit over a long period of time. 
A: The expected value changes on a per hand basis and depends on more than one factor.
I assume what you're referring to is the basic strategy for Blackjack having a negative EV in a typically available variant of Blackjack.
Here's the first rub, the composition of remaining cards is dependent on the cards that are already dealt.  Imagine a standard deck consisting of 52 cards with each card being played in a game of Blackjack with the exception of an Ace of spades, king of diamonds, 10 of clubs, and a jack of hearts.  Let's assume you've been wagering 10 dollars per hand up to this point and for arguments sake have played a total of 20 hands.  Unfortunately, you've lost all 20 hands thus far.  You're down 200 dollars.  You came to the casino with $1,200.
Here's the second rub, a Blackjack pays the player 1.5 times the initial wager, but costs the player 1 times the initial wager.  This means if you decide to gamble the remaining 1,000 dollars on your next hand that you're only liable to lose 1,000 dollars while you could also possibly win 1,500 dollars.  This is the equalizer in this game; hence the name Blackjack!  You have a 50/50 chance at winning or losing this hand, but there's one certainty.  The certainty is that either you or the dealer will have a Blackjack.  You can calculated your EV here simply as .5 * 1,500 + .5 * -1,000 = 250 dollars.
In the example I've outlined you can see a few of the basic dynamics of the game in action.  Number 1, the final 4 cards in this example are dependent on the previous 48 cards played.  The composition of the four remaining cards is knowable even if the order isn't.  Number 2, your EV is affected by the particular rules of the game (in this example the fact that a Blackjack results in a 1.5 times your wager win with a maximum loss of only your wager).  Number 3, your EV is affected by the initial value of the wager.
Your initial premise stated an infinite number of hands, but I'm assuming there wouldn't be an infinite number of decks since you can play an infinite amount of hands with only one deck.     
