Expectation for game choosing uniformly number in $[0,1]$ until it decreases We are playing a game where we keep on choosing a number from the uniform distribution U(0,1). The game goes on until we have the current number less than the previously picked number, i.e. the game will stop if we get a number less than the previous number and will continue if we get a number greater than equal to the previous number. What is the expected value of the final number?
 A: Let
\begin{equation*}
    N:=\inf\{n\ge2\colon X_{n-1}>X_n\}, 
\end{equation*}
where $X_1,X_2,\dots$ are independent random variables uniformly distributed on $[0,1]$.
We want to find
\begin{equation*}
    EX_N=\sum_{n=2}^\infty EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n). \tag{1}
\end{equation*}
We have
\begin{equation*}
\begin{aligned}
    &EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n) \\
    &=EX_n\,1(X_1\le\cdots\le X_{n-1}) \\ 
    &-EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n), 
\end{aligned}
\tag{2}
\end{equation*}
\begin{equation*}
\begin{aligned}
    &EX_n\,1(X_1\le\cdots\le X_{n-1}) \\
    &=EX_n\,P(X_1\le\cdots\le X_{n-1})=\frac12\,\frac1{(n-1)!}. 
\end{aligned}
\tag{3}
\end{equation*}
The calculation of $EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n)$ is more involved than that of $EX_n\,1(X_1\le\cdots\le X_{n-1})$, because $X_n$ and $1(X_1\le\cdots\le X_{n-1}\le X_n)$ are not independent -- in contrast with $X_n$ and $1(X_1\le\cdots\le X_{n-1})$.
The main idea in the calculation of $EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n)$ is to express $X_n$ in terms of indicators, to allow a better blending with the indicator $1(X_1\le\cdots\le X_{n-1}\le X_n)$.
Toward that end, note that $X_n=\int_0^{X_n} dx=\int_0^1 dx\,1(X_n>x)$ and
$$1(X_n>x)1(X_1\le\cdots\le X_{n-1}\le X_n) \\ 
=1(X_1\le\cdots\le X_{n-1}\le X_n>x),$$
so that
$$X_n\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\
=\int_0^1 dx\,1(X_1\le\cdots\le X_{n-1}\le X_n>x),$$
the latter expression being indeed in terms of the indicators $1(X_1\le\cdots\le X_{n-1}\le X_n>x)$.
Hence,
\begin{equation*}
\begin{aligned}
    &EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\ 
    &=E\int_0^1 dx\,1(X_1\le\cdots\le X_{n-1}\le X_n>x) \\ 
    &=\int_0^1 dx\,P(X_1\le\cdots\le X_{n-1}\le X_n>x) \\ 
    &=\int_0^1 dx\,[P(X_1\le\cdots\le X_{n-1}\le X_n) \\ 
    &\qquad\qquad-P(X_1\le\cdots\le X_{n-1}\le X_n\le x)] \\ 
    &=P(X_1\le\cdots\le X_{n-1}\le X_n) \\
    &-\int_0^1 dx\,P(X_1\le\cdots\le X_{n-1}\le X_n\le x) \\ 
    &=\frac1{n!}-\int_0^1 dx\,x^n\frac1{n!} = \frac1{n!}-\frac1{(n+1)!}. 
\end{aligned}
\tag{4}
\end{equation*}
So, by (1), (2), (3), (4),
\begin{equation*}
\begin{aligned}
    EX_N&=\sum_{n=2}^\infty \Big(\frac12\,\frac1{(n-1)!}-\frac1{n!}+\frac1{(n+1)!}\Big) \\ 
    &=\frac e2-1\approx0.359.
\end{aligned}   
\end{equation*}

One may also note that
\begin{equation*}
\begin{aligned}
    &EN=E\sum_{n=0}^\infty1(N>n)=\sum_{n=0}^\infty P(N>n) \\ 
    &=\sum_{n=0}^\infty P(X_1\le\cdots\le X_n) =\sum_{n=0}^\infty \frac1{n!}=e\approx2.72. 
\end{aligned}
\end{equation*}

Simulation with
Mathematica appears to confirm these results (click on the image below to enlarge it):

