Probability: Expected distance of an accelerating car - with a twist

In summary: This happens with probability P(T=i).In summary, the problem presented is about a race car on a straight track, where a random number is generated each turn to change its velocity and move it forward. If a 60 is rolled, the car crashes and its velocity is reset to 0. The first problem is to find the average distance the car will travel before crashing once, which is 3540 units. The second problem asks for the expected distance after a given number of turns, with the assumption that the car is still "in the race" even after crashing. This can be solved using a recurrence formula and the probability of the car crashing at a given time.
  • #1
HexNash
2
0
I have an expected value probability problem. Unfortunately, I do not know enough about mathematics to determine if it is even remotely feasible to try to solve it. The best way to explain it is to disguise it as a physics problem:

Imagine a race car on a never-ending straight track. Let's call the position it starts at 0 units, and its starting velocity is 0 units per turn. Suppose we generate a random number from 1 to 60. If any number except 60 comes up, its current velocity is increased by 1, and the car is moved forward the number of units corresponding to the new velocity. However, if a 60 is rolled, it crashes: the current velocity is reset to 0, and the car is not moved (as the new velocity is 0u/t).

Now, the initial problem I worked out was how many units, on average, the car will travel before it crashes once. That was fairly easy: because the distance the car travels corresponds to a triangular number [n(n+1)/2], and the probability of the trial ending is always 1/60 each turn, the expected value corresponds to the equation: sum of (n(n+1)/2)(1/60)(59/60)^n from n=0 to n=infinity = 3540

However, I have another problem I am trying to solve that is proving much more difficult:
What is the expected distance of the car after 60 (or in general, n) turns?
In this scenario, when a car crashes, its velocity is reset to 0, but it is still "in the race" for the remaining number of turns.

I have tried all different kinds of techniques to try to figure this out, but to no avail. I'm hoping this isn't too complicated or tedious to compute, but considering there are 966,467 ways to partition 60, all with differing odds of occurring, I'm not too confident. Doing some research, I found that something called a "Markov chain" might help, but I am not familiar with how those work.

Does anyone know how I can tackle this? It is nothing really important, just a thought experiment I was curious about, but it is really bugging me.
 
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  • #2
One thing you do NOT say is how often this random number is generated. Not knowing long the car moves at a given velocity, you cannot say how far it would go.
 
  • #3
HallsofIvy said:
One thing you do NOT say is how often this random number is generated. Not knowing long the car moves at a given velocity, you cannot say how far it would go.

It is discrete: each time a number is generated it is a turn, and the velocity is changed and the car is moved forward. For example, given the sequence of generated numbers: 53 04 34 60 23 47, that is 6 turns and the car is moved 1+2+3+0+1+2 = 9 units. What I am trying to figure out is the expected value after 60 turns.
 
  • #4
Let [itex]d_n[/itex] be the expected distance that the car will travel in a race of duration [itex]n[/itex]. Let [itex]T[/itex] be the last time the car is stopped (so if there is no crash we'll put [itex]T = 0[/itex]). Now we can set up the following recurrence for [itex]d_n[/itex]:
[tex]d_n = \sum_{i=0}^n P(T=i)(d_{i-1} + \Delta(n-i))[/tex]
where [itex]\Delta(m)[/itex] is the mth triangular number, and we assume [itex]d_{-1} = d_0 = 0[/itex]. Now, [itex]P(T=i) = (1/60)(59/60)^{n-i}[/itex], so this recurrence let's you easily calculate [itex]d_n[/itex] for any n (using a computer, of course).

The explanation of this is simple: if the car crashes for the last time at time [itex]i[/itex], it traveled an average of [itex]d_{i-1}[/itex] steps in the time between the start of the race and [itex]i[/itex] and travels exactly [itex]\Delta(n-i)[/itex] steps after that time.
 
  • #5


I would suggest approaching this problem by breaking it down into smaller, more manageable parts. Instead of trying to solve for the expected distance after 60 turns, start by solving for the expected distance after 1 turn, then 2 turns, and so on. This can help you build a better understanding of the problem and potentially lead to a solution for the larger question.

Additionally, using a Markov chain may indeed be helpful in this scenario. Markov chains are a mathematical tool used to model the probability of transitioning between different states over time. In this case, the states would be the different velocities that the car could have after each turn. By constructing a Markov chain for this problem, you may be able to calculate the expected distance after a certain number of turns.

Another approach could be to use simulations to estimate the expected distance. This involves running multiple trials of the race and recording the distance traveled in each trial. By averaging these results, you can get an estimate of the expected distance. This may not give an exact answer, but it can provide a good approximation.

Finally, I would also suggest seeking help from a mathematician or statistician who may have more expertise in solving probability problems. They may be able to provide valuable insights and techniques for approaching this problem.
 

1. What is the expected distance of an accelerating car?

The expected distance of an accelerating car is the average distance that the car is expected to travel over a given period of time, taking into account the acceleration of the car.

2. How is the expected distance of an accelerating car calculated?

The expected distance of an accelerating car can be calculated by using the formula: d = (1/2)at^2, where "d" is the distance traveled, "a" is the acceleration, and "t" is the time interval.

3. What factors affect the expected distance of an accelerating car?

The expected distance of an accelerating car is affected by the initial velocity of the car, the rate of acceleration, and the time interval over which the acceleration occurs.

4. Is the expected distance of an accelerating car the same as the actual distance traveled?

No, the expected distance of an accelerating car is an estimate based on the initial conditions and acceleration rate. The actual distance traveled may vary due to external factors such as friction, air resistance, and driver behavior.

5. How can the expected distance of an accelerating car be used in real-world applications?

The expected distance of an accelerating car can be used in various real-world applications such as designing cars, calculating braking distances, and predicting the performance of vehicles in different conditions.

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