To bet or not to bet Your utility from having $x$ dolars is $u(x)$. 

There is a gamble in which the winnings in dollars are a random variable, $Y$. It is known that $E[u(Y)]&gt;E[u(1)]$, so you prefer to bet than to get one dollar for sure.

But now you are offered the following options. You select a large number $T$ (which may depend on the distribution of $Y$), and then you can choose between two options:

A. Receive $T+1$ dollars.

B. Bet $T$ times, where all bets are statistically independent and distributed like $Y$.

For what utility functions $u$ would you prefer option B (for a sufficiently large $T$)?

The expected utility of option B is:

$$E\left[u\left(\sum_{t=1}^T Y_t\right)\right]$$

where $Y_t$ are i.i.d. variables distributed like $Y$. So the question is actually: for what functions is it true that, for a sufficiently large $T$:

$$E\left[u\left(\sum_{t=1}^T Y_t\right)\right] &gt; u(1 + T)\; ?$$

One obvious answer is when $u(x)=x$, since in that case: 

$$E\left[u\left(\sum_{t=1}^T Y_t\right)\right] = \sum_{t=1}^T E[Y_t] = T E[Y_t]$$

since $E[Y_t]&gt;E[u(1)]=1$, it is clear that for a sufficiently large $T$:

$$T \cdot E[Y_t] &gt; T + 1$$

What other utility functions have this property?
I think the answer depends on the space $Y$ is coming from. Consider if $Y \in \{ 1, \frac{T+1}{T}+\epsilon \}$, where $\epsilon &gt; 0$ and the probability $p$ of the better result is 

$$p &gt; \frac{T^2}{\epsilon T + 1}$$

(Which just comes from solving:)

$$1\cdot (1-p) + \left(\frac{T+1}{T} + \epsilon \right)\cdot p &gt; T+1$$

Then no matter what utility function you have, risk averse or risk loving, you will always take the gamble (option B).

Did you mean to have some sort of restriction on $Y$ for this question? Otherwise we'd have to consider both $Y$ and $u$ to determine whether the player will take option B.