The probability of $|+\rangle$ to be in $|0\rangle$ or $|1\rangle$? The following image is from IBM Q Experience beginner's guide HERE.
The gate sequences are being applied to the qubit $|0\rangle$. I only do not understand why in the second row, the probabilities are 85, 15. I see that the H and then T gates take $|0\rangle$ to the plane X-Y with angle $\frac{\pi}{4}$ from both axis (in Bloch sphere), but I cannot see where the Hadamard gate before the measurement takes this. Any hint is appreciated. 
 A: The states $|0\rangle$ and $|1\rangle$ lie on the $+z$-axis and $-z$-axis of the Bloch sphere, and the $|+\rangle$ and $|-\rangle$ states lie on the $\pm x$-axis. The Hadamard gate $H$ performs a rotation that transforms $|0\rangle$ to $|+\rangle$ and $|1\rangle$ to $|-\rangle$ (i.e. it's a 180-degree rotation about an axis 45 degrees from both the $z$-axis and the $x$-axis). So if you apply $HZH$, you first turn $|0\rangle$ into $|+\rangle$, then you turn $|+\rangle$ into $|-\rangle$ (since the $Z$ operator is a 180-degree rotation about the $z$-axis), then you turn $|-\rangle$ into $|1\rangle$.
If, on the other hand, you apply $HTH$, then you first turn $|0\rangle$ into $|+\rangle$, as before, but then you rotate only 45 degrees around the $z$-axis. This leaves you midway between the $x$ and $y$ axes. When you apply the second $H$, you will end up on the opposite side of the axis of rotation (which, again, evenly splits the $z$ and $x$ axes). Doing this rotation will land you much closer to the $|0\rangle$ state (the "north pole" of the Bloch sphere) than the $|1\rangle$ state (the "south pole"); trying this on a globe (or some other sphere you have laying around) might offer some intuition as to how this works. As such, the probability of being $|0\rangle$ should be much higher than the probability of being $|1\rangle$.
