Conditional Expectation and the Tower Property

Roll two fair dice, call the first one $D_1$, and let $X$ be the sum. Before you look there are 36 equally likely worlds, and the sum balances at $E[X] = 7$. Picture 36 blocks on a beam with the fulcrum under 7.

Now someone tells you the first die shows 5. Thirty worlds vanish. The six that survive have sums $6, 7, 8, 9, 10, 11$, each now with probability $1/6$, and the beam re-balances at

That is all conditioning on an event does. You throw away the worlds that are ruled out, rescale the survivors so their probabilities add to 1, and find the new balance point. In symbols, for an event $A$ with $P(A) > 0$ and its indicator $\mathbf{1}_A$ (1 on $A$, 0 off it),

With $A = \{D_1 = 5\}$ we get $E[X \mathbf{1}_A] = 51/36$ and $P(A) = 6/36$, so the ratio is $51/6 = 8.5$ again.

If the first die had shown 2, the survivors would have sums 3 through 8 and the fulcrum would slide left to $5.5$. New information moves your expected value, and which way it moves depends on what you learn.