probability that subsets are successively contained The set A has $n$ elements, and so has $2^n$ subsets.  The subsets are placed into an urn, and $m$ subsets $B_1, \dots, B_m$ are drawn in order at random with replacement from the urn.  (Each subset has probability $\frac{1}{2^n}$.)  What is the probability that $$B_1 \subseteq B_2 \subseteq \dots \subseteq B_m?$$
I'm not really sure how to approach this problem; the only thing I can think of is a counting argument of some sort, but it seems like that would involve a lot of casework about the sizes of the sets.
 A: (The other answer appeared while I was typing this - the two are really the same argument, expressed differently.)
If you identify $B$ with its indicator function $\chi_B$ it's not hard to see that the probability that $B_1\subset\dots\subset B_m$ is exactly $(p_m)^n$, where $p_m$ is the probability that $X_1\le\dots\le X_m$ for iid  random variables $X_1,\dots, X_m$, uniformly distributed  on $\{0,1\}$.
Now there are $2^m$ possibilities for the sequence $(X_1,\dots,X_m)$, among which exactly $m+1$ are non-decreasing (the non-decreasing ones are just $(1,1,\dots,1), (0,1,1,\dots,1),(0,0,1,1,\dots,1),\dots,(0,0,\dots,0)$.)
So your probability is just $((m+1)/2^m)^n$.
A: First let us assume $m=2$.  For each element you must have it in $B_2$ only, neither, or both, so the chance is $\frac 34$.  The chance that $B_1 \subseteq B_2$ is therefore $\left(\frac 34\right)^n$  
For more general $m$ each element must not stop appearing in the subsets once it has started.  There are $2^m$ arrangements for the element among the $B$s of which $m+1$ are successful.  The chance that a given element does not spoil the set of inclusions is then $\frac {m+1}{2^m}$ which agrees with $\frac 34$ when $m=2$.  The chance no element spoils our inclusion is then $$\left(\frac {m+1}{2^m}\right)^n$$
