DylanN wrote: There are seven cards in a hat, and on the card $k$ there is a number $2^{k-1}$, $k=1,2,...,7$. Solarin picks the cards up at random from the hat, one card at a time, until the sum of the numbers on cards in his hand exceeds $124$. What is the most probable sum he can get? The only possible stop values are $125,126,127$ Probability of stopping at $125$ is probability to get card $2$ as last card in a seven picks game Probability of stopping at $126$ is probability to get card $1$ as last card in a seven picks game Probability of stopping at $127$ is probability to get cards $3,4,5,6,7$ as last card in a seven picks game Hence the answer : $\boxed{127}$

Let $S=\{2^0, 2^1, ..., 2^7\}$. Notice that $\sum(S) = 127$. Therefore Solarin can stop at $127, 126, 125, 124$. In each of these cases, there is a unique subset of $S$ that gives the required sum. Call a permutation $\sigma = \{i_1, .., i_k\}$ of such a subset legal if $\sum\limits_{i=1}^{k-1} \sigma(i) < 124$, while $\sum\limits_{i=1}^{k} \sigma(i) \ge 124$. Now it is easy to see that there are more legal permutations ($6!*5$) of the subset $S$ (for which $\sum(S) = 127$) than for the subsets $\{2^1, .., 2^6\}, \{2^0, 2^2, ..., 2^6\}$, etc.