Will the Resting Offer Get Hit? Overlapping Random Time Windows
On a single asset, participant A posts a new best offer (a sell quote) of 100 lots that rests in the market for 11 seconds. Participant B wants to buy 100 lots and will send a market order. Each of A and B starts their action at a moment chosen uniformly and independently at random between 8:45:00 and 8:45:30.
Assume that while A's quote is live nobody else quotes better, and that A cancels the quote when its 11 seconds expire or at 8:45:30, whichever comes first. B's market order will execute against A precisely when B's start time falls within the interval during which A's quote is resting.
What is the probability that B trades with A?
Hints
- Map both random start times to coordinates in a $30 \times 30$ square and turn the question into the area of a favorable region (geometric probability).
- B trades iff B's arrival $b$ satisfies $a \le b \le \min(a+11, 30)$ -- a diagonal strip of width 11 above the line $b = a$, clipped at the top edge.
- Split the integral at $a = 19$ (where $a + 11$ first exceeds 30) to handle the clipping, then divide the area by $900$.
Worked Solution
How to Think About It: B trades with A exactly when B's arrival time lands inside A's live quoting window. A starts at a uniform time $a \in [0,30]$ (seconds after 8:45:00) and rests until $\min(a+11, 30)$. B arrives at a uniform time $b \in [0,30]$. The event is $a \le b \le \min(a+11, 30)$. This is a geometric probability over the unit-area square $[0,30]^2$ -- compute the favorable area and divide by $900$.
Quick Estimate: The favorable region is a band just above the diagonal $b = a$ of width 11 (in the $b$ direction), clipped at $b = 30$. As a rough cut, a band of vertical width 11 across a square of side 30 covers roughly
Formal Solution: Let $a, b$ be i.i.d. Uniform$[0,30]$. We need
$P\big(a \le b \le \min(a+11,\,30)\big).$
The favorable area $S$ in the $30 \times 30$ square is the region above the line $b=a$ and below $b = a + 11$ (capped at $b=30$). Compute the complementary/strip area directly. The strip between $b=a$ and $b=a+11$, intersected with the square:
- For $a \in [0, 19]$: $b$ ranges over $[a, a+11]$, full height 11.
- For $a \in [19, 30]$: $b$ ranges over $[a, 30]$, height $30 - a$.
So
$S = \int_0^{19} 11\,da + \int_{19}^{30} (30 - a)\,da = 11\cdot 19 + \Big[30a - \tfrac{a^2}{2}\Big]_{19}^{30}.$
The second integral: at 30, $900 - 450 = 450$; at 19, $570 - 180.5 = 389.5$; difference $= 60.5$. Thus $S = 209 + 60.5 = 269.5$.
$P = \frac{269.5}{900} \approx 0.2994.$
Answer: $\dfrac{269.5}{900} \approx 0.299$, about $30\%$.
Intuition
Geometric probability -- recasting 'two random times overlap' as an area in a square -- is the right hammer for a whole family of problems (the classic 'two friends meet within 15 minutes' is the same shape). The microstructure framing matters: a market maker's resting quote has a finite lifetime, and the chance an incoming order interacts with it depends on the overlap of arrival processes, which is exactly what this models. The only real subtlety is the boundary clipping when A's window would extend past the 8:45:30 cutoff, and forgetting that clipping is the most common error -- it shaves the naive