Your intuition is right. A simple way to say it is that your mean losses are the same whether you bet $1000 in one hand or $100 in ten hands, but the variance in your losses is ten times smaller in the second case. This means that the probability that you’re up by $1000 is much smaller in the second case.
For the cases where the individual-hand bets are small compared to your total bankroll, you can use the Law of Large Numbers to argue that a Gaussian approximation to your bankroll after k hands is good. Integrate the tails of the Gaussian to estimate the probability that you’ve won or busted.
If you want to compute exact probabilities (e.g., for the cases where your bet is a significant fraction of your bankroll), you can do something like this: Let your initial bankroll be B and your individual-hand bets be b. Then your bankroll must always be B+nb dollars, where n is an integer (the number of wins minus the number of losses). If N=B/b (I will assume N is an integer), then you’re only interested in -N<=n<=N (smaller and you’re broke; larger and you’ve won, assuming you’re just trying to double your money). So you can set up a Markov transition table of 2N+1 equations relating the P(k,n) (the probability that you have B+nb dollars after k hands) to the corresponding values P(k-1,n) after k-1 hands:
P(k,n) = p P(k-1,n-1) + q P(k-1,n+1)
except that
P(k,N) = p P(k-1,N-1) + P(k-1,N)
P(k,N-1) = p P(k-1,N-2)
P(k,1-N) = q P(k-1,2-N)
P(k,-N) = q P(k-1,1-N) + P(k-1,-N)
where p=1-q=48% the one-hand win probability. That is, in each hand you either gain $b, with probability p, or lose $b, with probability q. Once you win or bust, you quit, so there is a deterministic transition from P(k-1,N) to P(k,N).
This can be expressed as a transition matrix equation, of the form P(k)=M P(k-1); for example, if N=2 (B=1000, b=500),
[ 1 0.48 0 0 0 ]
[ 0 0 0.48 0 0 ]
M = [ 0 0.52 0 0.48 0 ]
[ 0 0 0.52 0 0 ]
[ 0 0 0 0.52 1 ]
Now P(k)=M[sup]k[/sup]P(0), and P(0) is just a 1 at n=0 and 0 everywhere else, so you can calculate the win and loss probabilities for any number k of hands.
If you compute the eigenvalues and eigenvectors of M, you can also meaningfully compute M[sup]infinity[/sup] and figure out your long-term winning percentage. (Because M is a stochastic matrix, all of its eigenvalues are between -1 and 1, and it has at least one eigenvalue 1. For large k, M[sup]k[/sup] is dominated by the contributions for its 1 eigenvectors.)
For N of reasonable size, these calculations are trivial to do in something like Matlab or Mathematica. For the case N=2, for example, with p=0.48, you have
# of hands Prob{win} Prob{loss}
-------------------------------------
2 0.2304 0.2704 [the case Giles mentioned]
4 0.3454 0.4054
6 0.4028 0.4728
8 0.4315 0.5064
10 0.4458 0.5232
infinite 0.4601 0.5399