A practical everyday puzzle: will AI solve it?

Here’s a genuine problem, as per the drawing:

• A person who shopped at a supermarket is at point A. They could go take the bus either from the bus stop S1 or from the bus stop S2. The bus goes in the direction from S1 to S2.
• The distance from A to S1 is 150 meters.
• The distance from A to S2 is 400 meters.

It’s impossible to see when the bus is coming in the distance, but we know this:

• A bus passes on average every 12 minutes.
• A bus has a speed of 30 km/h when it moves.
• A bus spends 15 seconds in S1 before moving towards S2.
• There is a 40% chance that the bus needs to stay 60 seconds at the traffic light in the crossing at A. In 60% of the cases, the traffic light is green.

For the pedestrian:

• Being heavy with purchases, their speed is 4 km/h and no more.
• There is no delay if they choose to go from A to S2.
• If they decide to go from A to S1, there is a delay required to cross the street. It takes 15 seconds with a chance of 60% and 60 seconds with a chance of 40% (when the traffic light is red for pedestrians).

Question: What’s the best strategy for the pedestrian? Should they go to S1, or to S2? The best strategy implies that the average waiting time for the bus is minimized.

⚓ Copilot | ChatGPT | Kimi (2 versions) | Claude | Grok | GLM-4.5 | Qwen (2 versions) | DeepSeek | Mistral | Gemini | Lumo | Summary | An improved puzzle

Copilot (Think Deeper)

I need to analyze this step-by-step by calculating the expected waiting time for each strategy.

Let me define the problem clearly:

• Person at A needs to catch a bus going from S1 → S2
• Bus frequency: every 12 minutes on average
• Bus speed: 30 km/h = 8.33 m/s
• Walking speed: 4 km/h = 1.11 m/s

Strategy 1: Go to S1

• Distance: 150m
• Walking time: 150m ÷ 1.11 m/s = 135 seconds
• Crossing delay: 15s (60% chance) or 60s (40% chance)
• Expected crossing delay: 0.6 × 15 + 0.4 × 60 = 33 seconds
• Total time to reach S1: 135 + 33 = 168 seconds

Strategy 2: Go to S2

• Distance: 400m
• Walking time: 400m ÷ 1.11 m/s = 360 seconds
• No crossing delay
• Total time to reach S2: 360 seconds

Now I need to consider bus timing. The key insight is that buses arrive randomly every 12 minutes (720 seconds) on average.

For Strategy 1 (S1):
When the person arrives at S1 after 168 seconds, they need to wait for the next bus. Since buses are random with 12-minute intervals, the expected waiting time at S1 is 6 minutes (360 seconds) on average.

Total expected time = Travel time + Expected wait time = 168 + 360 = 528 seconds

For Strategy 2 (S2):
The person arrives at S2 after 360 seconds. But here’s the crucial part: they need to account for when buses that started from S1 will arrive at S2.

Time for bus to travel from S1 to S2:

• Distance S1→A: 150m (from diagram geometry)
• Distance A→S2: 400m
• Total S1→S2 distance ≈ 550m (assuming roughly straight line)
• Bus time S1→S2: 550m ÷ 8.33 m/s = 66 seconds
• Plus 15 seconds stop at S1 = 81 seconds total
• Plus potential 60-second delay at traffic light (40% chance) = 81 + (0.4 × 60) = 105 seconds expected

So buses arrive at S2 approximately 105 seconds after leaving S1.

Since the person takes 360 seconds to walk to S2, and buses come every 720 seconds, the expected waiting time at S2 is also 360 seconds on average.

Total expected time = 360 + 360 = 720 seconds

Comparison:

• Strategy 1 (go to S1): 528 seconds expected total time
• Strategy 2 (go to S2): 720 seconds expected total time

Answer: The person should go to S1.

Despite the crossing delay and longer walking time to S1, the total expected time is about 3.2 minutes shorter than going to S2. This is because the person can catch the bus earlier in its route, and the time saved by catching an earlier bus outweighs the additional walking and crossing time.

Additional clarification of the model

• I assume bus arrivals are strictly periodic with headway H = 720 s, but the phase at the instant the shopper finishes shopping is unknown and uniformly distributed on [0,H).
• The pedestrian arrival at a stop is deterministic once they choose which stop to walk to (walk time + crossing delay if applicable).
• Under a uniform unknown phase the expected residual wait when arriving at a fixed clock time equals H/2 = 360 s, unless the arrival time is correlated with the bus schedule (for example because the bus must pass an upstream stop and that imposes a known offset).
• The bus moves only S1 → S2 and passes point A on the segment between them; a pedestrian walking from A toward S2 could potentially meet an on-coming bus before reaching S2, which introduces a different arrival-time correlation than a pure “arrive at a stop and wait” model.

Why the simple uniform-residual calculation was valid and what it missed

• If your arrival time at a chosen stop has no systematic correlation with the bus phase, the expected residual wait is H/2 and the expected total time is simply (walk time to chosen stop) + H/2. That was the calculation used earlier and it correctly shows S1 wins because the walk+cross to S1 (168 s) is much shorter than the walk to S2 (360 s), giving S1 a 192 s advantage.
• What this simple model misses is the possibility of intercepting a bus while walking (not waiting at the stop) and the effect of the S1 dwell + traffic-light delays which shift the bus arrival time at S2 relative to S1. Those effects can create correlations that change the expected waiting time at the two stops.

Quick analysis of interception and schedule-shift effects

1. Interception while walking toward S2

• A bus that departed S1 earlier may pass point A (and thus pass you if you are walking toward S2) before it reaches S2. If you meet the bus while walking you board immediately and waiting is zero (ignoring any safe-boarding time). The probability of that depends on where the bus is at your shopping finish time and on your walking duration to the meeting point.
• Roughly: because S2 is farther, walking to S2 increases the chance of meeting a bus in motion between A and S2 compared with standing at S2 from the start. But to offset the larger walk time you must have a nontrivial probability of catching a bus en route early enough to beat the shorter walk to S1 plus its expected wait.

1. Effect of S1 dwell and light delays on expected arrival at S2

• The bus schedule at S2 is the S1 schedule shifted by a random travel/delay amount L (here L ≈ 69 s on average). That shift does not change the uniformity of the phase if the initial phase is uniform; it only shifts the arrival epoch. Therefore, with no additional information the expected residual wait at S1 and at S2 are both H/2, even accounting for L. The shift only matters if you know something about the bus’s current position (for example you see a bus approaching or you time your walk so you arrive at S2 after the typical S1→S2 lag).

Short, rigorous conclusion and robustness checks

• Under the natural assumption that you have no information about the bus phase or its current position, the expected residual wait at either stop is H/2 = 360 s, so the correct decision reduces to minimizing walking time (including expected crossing delay). With numbers given: total expected times are 528 s (S1) vs 720 s (S2). S1 is strictly better by 192 s.
• Sensitivity: because the only difference between options in that model is the deterministic extra walking time to S2 (192 s), S1 remains the better choice unless one of these happens:
  • You can intercept the bus en route to S2 with probability large enough that the reduction in expected wait exceeds 192 s; or
  • The model for residual wait is wrong (for example bus arrivals are not uniformly phased but correlated so that arriving later systematically reduces expected residual), or
  • Walking speed, distances, or crossing delays change so that (walk2) ≤ (walk1 + crossing expectation).

Numeric threshold: for the uniform-phase model the pedestrian should choose S2 only if walk2 ≤ walk1 + expected crossing delay. Here walk2 (360 s) > walk1 + crossing (168 s), so S1 is better. If you could increase walking speed enough that walk2 ≤ 168 s, then S2 becomes preferable.

Practical rule of thumb for the shopper

• If you have no visual on the bus and no other timing cue, go to S1. It minimizes expected total time to board by a comfortable margin.
• If you can see or hear an approaching bus while you start walking toward S2, or if you can time your walk so you’ll be between A and S2 when a bus normally passes (visible evidence of a bus approaching S1 or A), then attempt to intercept en route to S2 — only then could S2 or an interception beat the deterministic advantage of S1.

ChatGPT

Short answer: Go to S1. It gives a smaller average total time to catch the bus.

Explanation + numbers (clean, minimal assumptions shown)

Assumptions I use (the result is robust to either common assumption about arrivals):

• Headway (mean time between buses) = 12 min = 720 s.
  If arrivals are periodic with random phase, the expected residual wait on arrival = ½·headway = 360 s.
  If arrivals are Poisson (memoryless), the expected residual = headway = 720 s.
  In either model the expected residual waiting after you reach a stop is independent of where you go (it’s the same constant for S1 and for S2), so the choice reduces to minimizing the time to reach the stop.

Compute walking times and crossing delay:

• Pedestrian speed = 4 km/h = 4 000 m / 3 600 s = 1.111… m/s.
• Time A → S1 walking = 150 m / 1.111… ≈ 135 s.
• Time A → S2 walking = 400 m / 1.111… ≈ 360 s.
• Extra crossing delay when going to S1: 60% chance 15 s, 40% chance 60 s ⇒ expected delay = 0.6·15 + 0.4·60 = 33 s.
• So expected total time to reach S1 = 135 + 33 = 168 s.
• Expected total time to reach S2 = 360 s.

Total expected time to catch a bus = (time to reach chosen stop) + (expected residual wait when you arrive).

Under the periodic headway model (uniform random phase):

• Expected residual = 360 s.
• Total S1 = 168 + 360 = 528 s (≈ 8.8 min).
• Total S2 = 360 + 360 = 720 s (12.0 min).

Under the Poisson / memoryless model:

• Expected residual = 720 s.
• Total S1 = 168 + 720 = 888 s (≈ 14.8 min).
• Total S2 = 360 + 720 = 1 080 s (18.0 min).

Either way S1 is better (by about 3.2 min under the periodic model, about 3.2 min under the Poisson model too — the absolute gap is the difference in reach times: 360−168 = 192 s ≈ 3.2 min).

Small practical note: you might occasionally catch a bus immediately if you hit S1 exactly when a bus is there (or still dwelling) — that possibility is already built into the residual distribution used above. The light delay on the bus between S1 and S2 changes the bus arrival time at S2 but does not change the fact that the expected residual at arrival is the same constant for both stops.

Conclusion: Walk to S1 (the nearer stop). It minimizes your average total time to board.

Kimi

Kimi K2 can only use images for text extraction, but it accepted to “think” even without grasping the graphics. K1.5 has fully visual understanding.

K2:

Let’s break the problem down step by step, clearly and logically.

🧠 Step 1: Clearly Restating the Problem

We have a pedestrian at point A who wants to catch a bus. There are two bus stops nearby:

• Bus stop S1: 150 meters away from A.
• Bus stop S2: 400 meters away from A.

The bus travels from S1 → S2. The pedestrian can choose either stop, but the bus direction is fixed (S1 to S2).

We want to minimize the average waiting time for the bus. To do this, we need to consider:

• Pedestrian walking time to each stop.
• Pedestrian crossing delays (if any).
• Bus arrival patterns and possible delays at intersections.

🚶‍♂️ Step 2: Pedestrian Walking Speed and Delays

Given pedestrian speed: 4 km/h
Convert to meters per second (m/s):

4 km/h = (4000 m) / 3600 s ​≈ 1.11 m/s

Thus, walking times:

Distance	Time (seconds)
150 m (A→S1)	150 ÷ 1.11 ≈ 135 s
400 m (A→S2)	400 ÷ 1.11 ≈ 360 s

Additionally, pedestrian delays at intersections:

• A→S1: Must cross street. Crossing delay:
• 60% chance: 15 s delay
• 40% chance: 60 s delay
• Average delay = 0.6×15 + 0.4×60 = 33 s
• A→S2: No delay mentioned.

Thus, total average walking times:

Route	Walking Time	Crossing Delay	Total Average Time
A→S1	135 s	33 s	168 s
A→S2	360 s	0 s	360 s

🚌 Step 3: Bus Movement and Delays

Bus characteristics:

• Speed: 30 km/h = 30,000 m / 3600 s ≈ 8.33 m/s
• Frequency: one bus every 12 minutes = 720 seconds (average interval)
• Bus stops at S1 for 15 seconds before moving toward S2.
• At intersection A (between S1 and S2), bus has:
• 40% chance: 60 seconds delay (red light)
• 60% chance: no delay (green light)
• Average delay at A = 0.4×60 = 24 seconds

Thus, bus travel times between stops:

Segment	Distance	Speed	Base Travel Time	Average Delay	Total Average Travel Time
S1→A	150 m	8.33 m/s	150 ÷ 8.33 ≈ 18 s	24 s (traffic light at A)	42 s
A→S2	250 m	8.33 m/s	250 ÷ 8.33 ≈ 30 s	0 s (no delay mentioned)	30 s

Thus, total average travel time from S1 to S2 is:

15 s (stop at S1) + 18 s (S1→A) + 24 s (delay at A) + 30 s (A→S2) = 87 s

⏳ Step 4: Bus Arrival Pattern and Pedestrian Waiting Strategy

Buses arrive randomly, on average every 720 seconds (12 minutes). Since we don’t know the exact schedule, we assume buses arrive randomly (Poisson process). Thus, the average waiting time for a randomly arriving passenger at a bus stop is half the average interval:

Average waiting time at stop = 720 / 2 ​=360 seconds

This is crucial: no matter which stop you choose, once you arrive at the stop, your average waiting time is 360 seconds.

⚖️ Step 5: Comparing Strategies Clearly

Now, let’s clearly compare the two strategies:

Strategy	Average Walking Time	Average Waiting Time at Stop	Total Average Time
Go to S1	168 s	360 s	528 s
Go to S2	360 s	360 s	720 s

Clearly, going to S1 results in a significantly shorter total average time (528 s vs. 720 s).

📌 Step 6: Important Clarification (Bus Direction)

You might wonder: “But the bus goes from S1 to S2. If I go to S2, won’t I see the bus coming and avoid waiting?”
This intuition is natural but incorrect here. Why?

• Buses arrive randomly every ~12 minutes. You don’t know when the last bus passed.
• Even though the bus moves from S1 to S2, you cannot see it approaching (as explicitly stated: “It’s impossible to see when the bus is coming in the distance”).
• Thus, you must treat bus arrivals as random at each stop. The direction of travel doesn’t help you predict arrivals better at S2.

Therefore, the average waiting time at each stop remains 360 seconds.

✅ Final Answer & Recommendation

Strategy	Total Average Time
Go to S1 ✅	528 seconds
Go to S2	720 seconds

Clearly, the best strategy is to go to bus stop S1.
Despite the crossing delay, the much shorter walking time makes this the optimal choice.

📝 Quick Summary

• Pedestrian speed: 4 km/h (~1.11 m/s)
• Walking times: A→S1 (~135 s), A→S2 (~360 s)
• Pedestrian crossing delay (A→S1 only): avg. 33 s
• Bus frequency: every 12 min (avg. waiting time at stop: 360 s)
• Bus delays: 15 s at S1, avg. 24 s at intersection A
• Bus direction (S1→S2) doesn’t help predict arrivals better at S2 (no visibility)

Thus, the optimal strategy is:

🚶‍♂️➡️🚏 Go to S1. Your total average time (walking + waiting) is significantly lower.

Answer clearly highlighted:

🎯 Best Strategy: Go to Bus Stop S1 ✅

K1.5:

To determine the best strategy for the pedestrian, we need to calculate the expected waiting time for the bus at each bus stop, taking into account the walking time, the bus schedule, and the traffic light delay.

Walking Speed:

• The pedestrian’s walking speed is 4 km/h, which is equivalent to 4000​73600 ≈ 1.111 meters per second.

Walking Times:

• Time to walk from A to S1: 150/1.111​ ≈ 135 seconds.
• Time to walk from A to S2: 400/1.111​ ≈ 360 seconds.

Bus Schedule:

• A bus passes every 12 minutes (720 seconds).
• The bus spends 15 seconds at S1.

Traffic Light Delay:

• 60% chance of a 15-second delay.
• 40% chance of a 60-second delay.

Expected Waiting Time at S1:

1. If the pedestrian arrives at S1 exactly when the bus is there, they wait 0 seconds.
2. If they arrive after the bus has left S1 but before it arrives at S2, they wait until the next bus.
3. If they arrive after the bus has passed S2, they wait for the next bus cycle.

The expected waiting time at S1 can be calculated by considering the time they might arrive relative to the bus schedule and the traffic light delay.

Expected Waiting Time at S2:
Since there is no delay in reaching S2, the pedestrian will always wait for the next bus cycle if they arrive after the bus has passed S2.

Calculating Expected Waiting Time:

At S1:

• The pedestrian has a 60% chance of a 15-second delay and a 40% chance of a 60-second delay.
• If they arrive at S1 within the 15 seconds the bus is there, they wait 0 seconds.
• If they arrive after the bus leaves S1 but before the next bus arrives, they wait (720 – (135 + delay)) seconds.
• If they arrive after the next bus has passed S2, they wait 720 seconds.

Expected Waiting Time Calculation:

• E(S1) = 0.6 × max(0,720 − (135 + 15)) + 0.4 × max(0,720 − (135 + 60)) + (135/720) ​× 720
• E(S1) = 0.6 × max(0,570) + 0.4 × max(0,525) + 135
• E(S1) = 0.6 × 570 + 0.4 × 525 + 135
• E(S1) = 342 + 210 + 135 = 687 seconds

At S2:

• Since the pedestrian will always miss the bus if they arrive after it has passed S2, they will wait for the next bus cycle.
• E(S2) = ​(360/720) × 720 = 360 seconds

Comparison:

• Expected waiting time at S1: 687 seconds.
• Expected waiting time at S2: 360 seconds.

Conclusion:
The best strategy for the pedestrian is to go to S2, as the expected waiting time is significantly lower (360 seconds) compared to going to S1 (687 seconds).

Claude

I need to analyze this step-by-step by calculating the expected waiting time for each strategy.

Let me define the problem clearly:

• Person at A needs to catch a bus going from S1 → S2
• Bus frequency: every 12 minutes on average
• Bus speed: 30 km/h = 8.33 m/s
• Walking speed: 4 km/h = 1.11 m/s

Strategy 1: Go to S1

• Distance: 150m
• Walking time: 150m ÷ 1.11 m/s = 135 seconds
• Crossing delay: 15s (60% chance) or 60s (40% chance)
• Expected crossing delay: 0.6 × 15 + 0.4 × 60 = 33 seconds
• Total time to reach S1: 135 + 33 = 168 seconds

Strategy 2: Go to S2

• Distance: 400m
• Walking time: 400m ÷ 1.11 m/s = 360 seconds
• No crossing delay
• Total time to reach S2: 360 seconds

Now I need to consider bus timing. The key insight is that buses arrive randomly every 12 minutes (720 seconds) on average.

For Strategy 1 (S1):
When the person arrives at S1 after 168 seconds, they need to wait for the next bus. Since buses are random with 12-minute intervals, the expected waiting time at S1 is 6 minutes (360 seconds) on average.

Total expected time = Travel time + Expected wait time = 168 + 360 = 528 seconds

For Strategy 2 (S2):
The person arrives at S2 after 360 seconds. But here’s the crucial part: they need to account for when buses that started from S1 will arrive at S2.

Time for bus to travel from S1 to S2:

• Distance S1→A: 150m (from diagram geometry)
• Distance A→S2: 400m
• Total S1→S2 distance ≈ 550m (assuming roughly straight line)
• Bus time S1→S2: 550m ÷ 8.33 m/s = 66 seconds
• Plus 15 seconds stop at S1 = 81 seconds total
• Plus potential 60-second delay at traffic light (40% chance) = 81 + (0.4 × 60) = 105 seconds expected

So buses arrive at S2 approximately 105 seconds after leaving S1.

Since the person takes 360 seconds to walk to S2, and buses come every 720 seconds, the expected waiting time at S2 is also 360 seconds on average.

Total expected time = 360 + 360 = 720 seconds

Comparison:

• Strategy 1 (go to S1): 528 seconds expected total time
• Strategy 2 (go to S2): 720 seconds expected total time

Answer: The person should go to S1.

Despite the crossing delay and longer walking time to S1, the total expected time is about 3.2 minutes shorter than going to S2. This is because the person can catch the bus earlier in its route, and the time saved by catching an earlier bus outweighs the additional walking and crossing time.

Grok

To determine the best strategy, calculate the expected waiting time at each bus stop, assuming bus headways at S1 are deterministic at 720 seconds (12 minutes) with a random phase, leading to an expected waiting time of 360 seconds at S1. At S2, the variable stop times at the traffic light introduce variability in headways, resulting in interarrivals of 660 seconds with probability 0.24, 720 seconds with probability 0.52, and 780 seconds with probability 0.24. This yields E[H] = 720 seconds and E[H²] = 520,128 seconds², so the expected waiting time at S2 is 520,128 / (2 × 720) = 361.2 seconds.

Since 360 < 361.2, the expected waiting time is lower at S1.

Explanation of the solution:

1. Convert all times to seconds for precision: bus interval = 720 s, pedestrian speed = 1.111 m/s (walking times: 135 s to S1, 360 s to S2), bus speed = 8.333 m/s (travel times: 18 s S1 to A, 48 s A to S2), dwell at S1 = 15 s, bus stop at A = 0 s (60%) or 60 s (40%), pedestrian crossing delay to S1 = 15 s (60%) or 60 s (40%).
2. At S1, deterministic headways of 720 s give E[wait] = 720 / 2 = 360 s, as the time to the next bus is uniform [0, 720] regardless of the variable arrival time to S1 (since independent of the bus schedule).
3. At S2, each bus experiences an independent stop at A, leading to headway differences of +60 s (p=0.24), 0 s (p=0.52), -60 s (p=0.24). Thus, headways H are 780 s (p=0.24), 720 s (p=0.52), 660 s (p=0.24).
4. Compute E[H] = 0.24×780 + 0.52×720 + 0.24×660 = 720 s.
5. Compute E[H²] = 0.24×780² + 0.52×720² + 0.24×660² = 0.24×608,400 + 0.52×518,400 + 0.24×435,600 = 146,016 + 269,568 + 104,544 = 520,128 s².
6. E[wait at S2] = E[H²] / (2 × E[H]) = 520,128 / 1,440 ≈ 361.2 s.
7. Compare: S1 has the lower expected waiting time.

The best strategy is to go to S1.

GLM-4.5 (Z.ai)

Analysis: Best Strategy for the Pedestrian

To determine the best strategy, I need to calculate the expected total time for both options (going to S1 or S2) and compare them.

Basic Information:

• Distance from A to S1: 150 meters
• Distance from A to S2: 400 meters
• Pedestrian speed: 4 km/h
• Bus frequency: every 12 minutes on average
• Bus speed: 30 km/h
• Bus stops at S1 for 15 seconds before moving to S2
• At traffic light A: 40% chance bus waits 60 seconds, 60% chance it passes immediately
• If pedestrian goes to S1: 60% chance of 15-second delay, 40% chance of 60-second delay
• If pedestrian goes to S2: no delay

Calculation for Going to S1:

1. Walking time:

• 150 meters = 0.15 km
• Time = Distance/Speed = 0.15/4 = 0.0375 hours = 2.25 minutes

2. Expected crossing delay:

• Expected delay = 0.6 × 15 seconds + 0.4 × 60 seconds = 9 + 24 = 33 seconds = 0.55 minutes

3. Total time to reach S1:

• Total time = 2.25 + 0.55 = 2.8 minutes

4. Expected waiting time at S1:

• For buses arriving every 12 minutes on average, the expected waiting time is half the interval, which is 6 minutes

5. Total expected time for S1 option:

• Total time = 2.8 + 6 = 8.8 minutes

Calculation for Going to S2:

1. Walking time:

• 400 meters = 0.4 km
• Time = Distance/Speed = 0.4/4 = 0.1 hours = 6 minutes

2. No delay for going to S2
3. Expected waiting time at S2:

• For buses arriving every 12 minutes on average, the expected waiting time is half the interval, which is 6 minutes

4. Total expected time for S2 option:

• Total time = 6 + 6 = 12 minutes

Comparison:

• S1 total expected time: 8.8 minutes
• S2 total expected time: 12 minutes

Conclusion:

The best strategy is to go to S1, as it minimizes the average waiting time (8.8 minutes vs 12 minutes). Although going to S1 involves crossing the street with potential delays, the shorter walking distance more than compensates for this, resulting in a lower total expected time.

Qwen

Qwen doesn’t accept images as input, but I gave it a try. Qwen3-Max-Preview doesn’t support thinking.

Qwen3-Next-80B-A3B (Thinking)

To determine the best strategy for the pedestrian, we need to minimize the total expected time from point A to boarding the bus, which includes:

• Walking time from A to the chosen stop (S1 or S2),
• Crossing delay (if applicable),
• Waiting time at the bus stop before the next bus arrives.

🔹 Step 1: Understand the Problem Setup

• Distance from A to S1: 150 meters
• Distance from A to S2: 400 meters
• Pedestrian speed: 4 km/h = 1.111 m/s
• Bus headway (average time between buses): 12 minutes = 720 seconds
• Bus speed: 30 km/h = 8.333 m/s
• Bus spends 15 seconds at S1 before moving toward S2
• Traffic light at A affects the bus:
• 40% chance of 60-second delay
• 60% chance of no delay
• Pedestrian crossing at A (for S1):
• 60% chance of 15 seconds
• 40% chance of 60 seconds
• No crossing delay for S2

🔹 Step 2: Calculate Walking Time

For S1:

• Distance: 150 m
• Time = 150/1.111 ≈ 135 seconds
• Crossing delay:
  E[Crossing] = 0.6 × 15 + 0.4 × 60 = 9 + 24 = 33 seconds
• Total time to reach S1:
  135 + 33 = 168 seconds

For S2:

• Distance: 400 m
• Time = 400/1.111 ≈ 360 seconds
• No crossing delay
• Total time to reach S2:
  360 seconds

🔹 Step 3: Expected Waiting Time at Each Stop

The bus arrives every 12 minutes on average, so the headway is 720 seconds. Assuming a regular schedule with no randomness in arrival times (i.e., deterministic headway), the expected waiting time at either stop is:

Expected waiting time = Headway / 2 = 720 / 2 = 360 seconds

This assumes that the pedestrian arrives at a random time relative to the bus schedule, and the bus spends only a short time (15 seconds) at each stop — which is negligible compared to the headway. Thus, the expected waiting time remains approximately 360 seconds for both S1 and S2.

Note: If we consider the exact distribution where the bus is present for 15 seconds, the expected waiting time would be slightly less than 360 seconds (≈345 seconds), but this difference is small and does not change the conclusion.

🔹 Step 4: Total Expected Time for Each Option

Option 1: Go to S1

• Walking + crossing time: 168 seconds
• Waiting time: 360 seconds
• Total:
  168 + 360 = 528 seconds

Option 2: Go to S2

• Walking time: 360 seconds
• Waiting time: 360 seconds
• Total:
  360 + 360 = 720 seconds

🔹 Step 5: Compare and Conclude

• S1 total time: 528 seconds
• S2 total time: 720 seconds

Clearly, going to S1 minimizes the total expected time from point A to boarding the bus.

Even though the pedestrian must cross the street at A (with an average delay of 33 seconds), the shorter walking distance to S1 (150 m vs. 400 m) more than compensates for this delay.

Additionally, the traffic light at A that affects the bus (not the pedestrian) has no impact on the pedestrian’s waiting time at either stop, since the headway is already given as an average of 12 minutes.

✅ Final Answer

Go to S1

Qwen3-235B-A22B-2507 (Thinking)

To determine the optimal strategy for the pedestrian, we analyze the total expected time for both options (S1 and S2), considering walking time, crossing delays, and waiting time for the bus.

Calculations for S1:

1. Walking Time:
   Distance = 150 m; Speed = 4 km/h ≈ 1.111 m/s.
   Time = 150 / 1.111 ≈ 135 seconds.
2. Crossing Delay at Traffic Light:
   Expected delay = (0.6 × 15 s) + (0.4 × 60 s) = 33 seconds.
3. Waiting Time for Bus at S1:
   Buses arrive every 12 minutes (720 seconds). The bus spends 15 seconds at S1.
   Probability of catching the bus immediately = 15/720 ≈ 0.0208.
   Expected waiting time = (1 – 0.0208) × 360 s (average wait for 12-minute interval) ≈ 352.5 seconds.

Total Expected Time for S1:
135 s (walking) + 33 s (crossing) + 352.5 s (waiting) ≈ 520.5 seconds.

Calculations for S2:

1. Walking Time:
   Distance = 400 m; Speed = 4 km/h ≈ 1.111 m/s.
   Time = 400 / 1.111 ≈ 360 seconds.
2. Crossing Delay:
   No delay required (direct path).
3. Waiting Time for Bus at S2:
   Assuming the bus’s arrival at S2 is uniformly distributed over the 12-minute interval (due to lack of distance data between S1 and S2), the average waiting time is 360 seconds (half of 12 minutes).

Total Expected Time for S2:
360 s (walking) + 0 s (crossing) + 360 s (waiting) = 720 seconds.

Conclusion:

The total expected time for S1 (520.5 seconds) is significantly shorter than for S2 (720 seconds). Despite the crossing delay at S1, the reduced walking time and comparable waiting time make S1 the optimal choice.

Answer: The pedestrian should go to S1 to minimize the average waiting time.

DeepSeek

DeepSeek also only uses images for text recognition, but I hit it with this puzzle nonetheless.

The best strategy for the pedestrian is to go to bus stop S1 to minimize the average waiting time for the bus.

The expected total time from point A to boarding the bus at S1 is calculated as follows:

• Walking time from A to S1: 150 meters at 4 km/h (10/9 m/s) takes 150 / (10/9) = 135 seconds.
• Expected delay at crossing: 60% chance of 15 seconds and 40% chance of 60 seconds, so expected delay = 0.6 × 15 + 0.4 × 60 = 9 + 24 = 33 seconds.
• Total expected time to reach S1: 135 + 33 = 168 seconds.
• Average waiting time at S1: Since buses arrive every 12 minutes (720 seconds) periodically, the average waiting time is half the headway, so 720 / 2 = 360 seconds.
• Expected total time at S1: 168 + 360 = 528 seconds.

The expected total time from point A to boarding the bus at S2 is calculated as follows:

• Walking time from A to S2: 400 meters at 4 km/h takes 400 / (10/9) = 360 seconds.
• Average waiting time at S2: Due to the random delay at the traffic light, the inter-arrival time at S2 has a variance, leading to an average waiting time of approximately 361.2 seconds.
• Expected total time at S2: 360 + 361.2 = 721.2 seconds.

Since 528 seconds is less than 721.2 seconds, going to S1 results in a shorter average waiting time.

S1

Mistral

The best strategy for the pedestrian is to go to bus stop S1. This minimizes the average waiting time for the bus.

Therefore, the pedestrian should go to S1.

Here’s a summary of the reasoning:

1. Time to reach S1: Approximately 168 seconds (including expected delay at the traffic light).
2. Average waiting time at S1: Approximately 360 seconds.

Total average time: ~528 seconds.

3. Time to reach S2: 360 seconds.
4. Average waiting time at S2: Approximately 360 seconds.

Total average time: 720 seconds.

Thus, going to S1 results in a shorter average total time to board the bus.

S1

Gemini 2.5 Pro

The best strategy is for the pedestrian to go to bus stop S1.

Going to S1 results in a significantly lower average waiting time for the bus compared to going to S2. Here is the detailed breakdown of the two strategies.

Common Calculations

First, let’s convert the speeds and times into consistent units (meters and seconds) for easier calculation.

• Bus Frequency: 12 minutes = 12 × 60 = 720 seconds.
• Pedestrian Speed: 4 km/h = 4 × 1000 m / 3600 s = 10/9 m/s (approx. 1.11 m/s).
• Bus Speed: 30 km/h = 30 × 1000 m / 3600 s = 25/3 m/s (approx. 8.33 m/s).

Strategy 1: Go to Bus Stop S1

In this scenario, we calculate the time it takes for the person to reach S1 and then determine their average wait for the bus.

1. Pedestrian’s Travel Time to S1:
   • Walking time = Distance / Speed = 150 m / 10/9 m/s = 135 seconds.
   • Average street crossing delay = (0.60 × 15 s) + (0.40 × 60 s) = 9 + 24 = 33 seconds.
   • Total time to reach S1 = 135 + 33 = 168 seconds (or 2 minutes and 48 seconds).
2. Average Waiting Time at S1:
   • The buses arrive on average every 720 seconds. Since the pedestrian’s arrival at the stop is random relative to the bus schedule, the average wait time for the next bus is half of the frequency.
   • Average Wait Time = 720 s / 2 = 360 seconds (or 6 minutes).

The average time the person will wait for the bus at stop S1 is 360 seconds.

Strategy 2: Go to Bus Stop S2

This strategy is more complex because it’s a “race” between the pedestrian and the bus. We need to find the average time the bus takes to get from S1 to S2 and then calculate the pedestrian’s chances of catching it.

1. Pedestrian’s Travel Time to S2:
   • There is no crossing delay for this path.
   • Time to reach S2 = Distance / Speed = 400 m / 10/9 m/s = 360 seconds (or 6 minutes).
2. Bus’s Average Travel Time from S1 to S2:
   • Distance from S1 to S2 = (Distance A to S1) + (Distance A to S2) = 150 m + 400 m = 550 m.
   • Moving time = 550 m / 25/3 m/s = 66 seconds.
   • Average traffic light delay at A = (0.40 × 60 s) + (0.60 × 0 s) = 24 seconds.
   • Time stopped at S1 = 15 seconds.
   • Total average bus time from S1 to S2 = 15 + 24 + 66 = 105 seconds.
3. Average Waiting Time at S2:
   The person arrives at S2 after 360 seconds. A bus that departs S1 will arrive at S2, on average, 105 seconds later.
   • The person misses the bus if the bus arrives at S2 before the person does (at 360s). This means the bus must have arrived at S1 less than 360 – 105 = 255 seconds after the person started walking.
   • The bus cycle is 720 seconds. The person will miss the bus if they start walking during the last 255 seconds of this cycle.
     • Probability of missing the bus = 255/720 ≈ 35.4%
   • The person catches the bus if they start walking during the first 720 – 255 = 465 seconds of the cycle.
     • Probability of catching the bus = 465/720 ≈ 64.6%
   Now we find the average wait time by combining these two outcomes:
   • If they catch the bus, their average wait is 232.5 seconds.
   • If they miss the bus, they must wait for the next one (an additional 720 seconds), making their average wait 952.5 seconds.
   Average Wait Time = (0.646 × 232.5 s) + (0.354 × 952.5 s) ≈ 150 + 337.5 = 487.5 seconds (or 8 minutes and 7.5 seconds).

Conclusion

To minimize the average waiting time for the bus, the pedestrian should choose the strategy with the lower waiting time.

• Average Wait at S1: 360 seconds
• Average Wait at S2: 487.5 seconds

Therefore, the best strategy is to go to bus stop S1. 👍

Lumo

Lumo doesn’t support images.

Answer – walk to S₁ is the better choice

Below is a compact “expected‑time” calculation for the two alternatives.
All times are expressed in seconds ( 1 min = 60 s ) and then converted back to minutes for readability.

Step	Pedestrian speed	Distance	Pure walking time
A → S₁	4 km/h = 1.111 m s⁻¹	150 m	150 ÷ 1.111 ≈ 135 s
A → S₂	4 km/h = 1.111 m s⁻¹	400 m	400 ÷ 1.111 ≈ 360 s

1️⃣ Going to S₁

• Cross‑street delay at A (only when heading for S₁)
  15 s with 60 % probability → 0.6 × 15 = 9 s
  60 s with 40 % probability → 0.4 × 60 = 24 s
  Expected delay = 9 + 24 = 33 s
• Bus‑waiting time at S₁
  Buses arrive on average every 12 min (= 720 s).
  With a uniformly random arrival instant, the expected residual waiting time is half the headway:
  E[wait] = 720 / 2 = 360 s (The 15‑second dwell that the bus spends at S₁ is already part of the 12‑minute cycle, so it does not change the average residual time.)
  Total expected time to board T_S1 = 135 walk + 33 crossing + 360 bus wait = 528 s approx 8.8 min

2️⃣ Going to S₂

• No crossing delay (the problem states “no delay if they choose to go from A to S₂”).
• Bus‑waiting time at S₂
  Same 12‑minute headway ⇒ expected residual wait = 360 s.
  However, the bus may be held at the traffic light located at A for 60 s with 40 % probability. This extra delay propagates to S₂:
  Extra expected delay = 0.4 × 60 = 24 s
  So the expected waiting component becomes 360 s + 24 s = 384 s. Total expected time to board
  T_S2 = 360 walk + 384 bus wait + traffic‑light delay = 744 s approx 12.4 min

3️⃣ Comparison

Option	Expected total time
Walk to S₁	≈ 8.8 min
Walk to S₂	≈ 12.4 min

Result: Going to S₁ yields a lower average time to catch the bus (by roughly 3½ minutes). Therefore, the pedestrian should head for the nearer stop S₁.

Summary

While all LLMs but one (Kimi K1.5) decided that going to S1 was the right solution, there’s a wild difference in the estimated waiting times in S1 and S2 (including the walking time in each case):

	Model	S1 total time	S2 total time
1	Copilot (Think Deeper)	528 seconds	720 seconds
2	ChatGPT (uniform random)	528 seconds	720 seconds
3	ChatGPT (Poisson model)	888 seconds	1080 seconds
4	Kimi K2	528 seconds	720 seconds
5	Kimi K1.5	687 seconds	360 seconds
6	Claude	528 seconds	720 seconds
7	Grok	360 seconds	361.2 seconds
8	GLM-4.5	528 seconds	720 seconds
9	Qwen3-Next-80B-A3B	528 seconds	720 seconds
10	Qwen3-235B-A22B-2507	520.5 seconds	720 seconds
11	DeepSeek (DeepThink)	528 seconds	721.2 seconds
12	Mistral	528 seconds	720 seconds
13	Gemini 2.5 Pro	528 seconds (168+360)	847.5 seconds (360+487.5)
14	Lumo	528 seconds	744 seconds

WTF?!

An improved puzzle

Let’s make the situation more realistic. Let’s replace “it’s impossible to see when the bus is coming in the distance” with this:

• From point A, looking towards S1, it can be seen if a bus is coming if the bus is at most 150 m away from S1, which is 300 m away from A.
• However, the bus number cannot be identified.
• The chance that a bus that’s seen coming is the right bus is 20%.

This way, I expect the following elements to become (more) relevant:

• The fact that the bus spends 15 seconds at S1.
• The possible waiting at the traffic light.
• The possible crossing time for the pedestrian.

While in theory this could change the outcome, with the particular values used, the choice of S1 doesn’t change, and here’s why:

• The time from seeing a bus until the bus reaches and departs S1 is 33 seconds or shorter, while the walking time to S1 is 150 seconds in the best case (no waiting at the red light, so crossing in 15 seconds). So it’s impossible to make it to S1 after seeing a bus.
• The total time from seeing a bus until the bus reaches S2 is no longer than 99 or 159 seconds, depending on the traffic lights. But the pedestrian needs 360 seconds to reach S2. So seeing the bus is non-actionable in this case, too.

Seeing a bus from A remains non-actionable for catching that same bus.

🟢 LLMs that got it right: Copilot, ChatGPT, Kimi K2, Claude, GLM-4.5, Qwen, DeepSeek, Gemini, Lumo.

🔴 Kimi K1.5 got it wrong in the first case, but here it made an even worse “reasoning”:

Given the updated scenario, we need to re-evaluate the expected waiting time at each bus stop, S1 and S2, considering the visibility of the bus and the probability that the visible bus is the correct one.

Revised Scenario:

• Visibility at S1: From point A, a bus can be seen if it is within 150 meters of S1, which is 300 meters away from A.
• Probability of Correct Bus: If a bus is seen, there is a 20% chance it is the correct bus.
• Bus Speed: 30 km/h (8.33 m/s).
• Bus Stop Time at S1: 15 seconds.
• Pedestrian Speed: 4 km/h (1.11 m/s).

Calculating Expected Waiting Time at S1:

1. Walking Time to S1: 150/1.11 ≈ 135 seconds.
2. Visibility Time: The bus can be seen 15 seconds before reaching S1 if it is moving at 8.33 m/s, covering 125 meters in 15 seconds. Thus, the bus can be seen from 165 meters away from S1.
3. Probability of Correct Bus: 20%.

Expected Waiting Time Calculation at S1:

• If the pedestrian sees a bus, they have a 20% chance it is the correct bus. If it is, they wait 0 seconds.
• If not, they wait for the next bus cycle (720 seconds).
• If they do not see a bus, they wait for the next bus cycle.

E(S1) = 0.2 × 0 + 0.8 × 720 + (1 – P(see bus) × 720
P(see bus) = 135/720​ (since they start walking 135 seconds before a new bus cycle begins)

E(S1) = 0.8 × 720 + (1 − 135/720​) × 720
E(S1) = 576 + 585/720 ​× 720
E(S1) = 576 + 585 = 1161 seconds

Calculating Expected Waiting Time at S2:

1. Walking Time to S2: 400/1.11 ​≈ 360 seconds.
2. No visibility advantage at S2.

Expected Waiting Time Calculation at S2:

• The pedestrian will always wait for the next bus cycle if they arrive after the bus has passed S2.

E(S2) = 360/720 ​× 720 = 360 seconds

Conclusion:

Given the updated scenario, the expected waiting time at S1 is significantly higher (1161 seconds) compared to S2 (360 seconds). Therefore, the best strategy for the pedestrian remains to go to S2, as the expected waiting time is minimized. The visibility of the bus towards S1 does not significantly impact the decision due to the low probability of the visible bus being the correct one and the long walking time to S1.

🔴 Grok was confused by the change:

The best strategy to minimize the average waiting time is to go to S2 if you see a bus coming, and to S1 otherwise.

Explanation of the solution:

The expected waiting time under this strategy is approximately 358 seconds. This is lower than the expected waiting time of approximately 360 seconds if always going to S1 or approximately 361 seconds if always going to S2.

To arrive at this conclusion:

1. Assume the right bus arrives at S1 with deterministic headways of 720 seconds and a uniform random phase, resulting in waiting times uniform over [0, 720] seconds upon random arrival.
2. Assume four other bus lines with identical headways and independent uniform random phases, ensuring that when a bus is seen approaching, there is a 20% chance it is the right bus.
3. The visibility zone corresponds to the next bus (any line) arriving at S1 within 18 seconds with probability approximately 0.12.
4. When no bus is seen, the distribution of time to the next right bus is shifted slightly (uniform over [18, 720] seconds), reducing the expected waiting time at S1 to approximately 356 seconds due to a lower probability of arriving just after a bus has passed.
5. When a bus is seen, the expected waiting time at S2 is approximately 381 seconds, which is higher than average but, given the low probability of seeing a bus, the overall expected waiting time is minimized by this conditional choice.
6. Simulations with 100,000 samples confirm this strategy yields the lowest expected waiting time among always S1, always S2, and the opposite conditional (go to S1 if seen, S2 otherwise).

🔴 Mistral was fooled, too!

To determine the best strategy for the pedestrian, we need to consider the following factors:

Walking Times:

Time to reach S1: 168 seconds (including expected delay at the traffic light).
Time to reach S2: 360 seconds.

Bus Arrival Rate:

Buses arrive at S1 every 720 seconds on average.
Only 20% of buses are the correct one for the pedestrian.
Therefore, correct buses arrive at S1 every 3600 seconds on average.

Expected Waiting Times:

At S1: 1800 seconds (half the interval between correct buses).
At S2: 1800 seconds (since the bus arrival times at S2 are offset from S1 by about 69 seconds).

Impact of Seeing a Bus:

• If a bus is seen arriving at S1 now:

With 20% chance, it’s the correct bus (but the pedestrian can’t catch it at S1 or S2).
With 80% chance, it’s not the correct bus.
The next correct bus is expected in about 2160 seconds at S1 and about 1869 seconds at S2.

• If no bus is seen arriving at S1 now:

The next correct bus is expected in about 1800 seconds at both S1 and S2.

Given these factors, the optimal strategy is:

If a bus is seen arriving at S1 now: go to S2.
If no bus is seen arriving at S1 now: go to S1.

Therefore, the pedestrian should go to S2 if a bus is seen arriving at S1 now, and go to S1 if no bus is seen arriving at S1 now.

S2​ if a bus is seen coming, otherwise S1​.

FFS!

⚓ Copilot | ChatGPT | Kimi (2 versions) | Claude | Grok | GLM-4.5 | Qwen (2 versions) | DeepSeek | Mistral | Gemini | Lumo | Summary | An improved puzzle