# Where will the pendulum land -- Probability Question

1. Oct 29, 2017

### Barclay

1. The problem statement, all variables and given/known data
Imagine a number line -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

A pendulum begins at -1 and then swings towards the LEFT or RIGHT. There is no knowing how fast the pendulum is travelling or how much energy it has but will only swing LEFT or RIGHT.

(I) Need to work out the probability of the pendulum landing on NUMBER -2 compared to landing on NUMBER +2. Ratio of landing on NUMBER -2 : NUMBER +2

(II) The number on which the pendulum lands corresponds to points.

If the pendulum lands on either +3, +4 or +5 then 3, 4 or 5 points is GAINED, respectively.

If the pendulum lands on either -2, -3, -4 or -5 then ONLY 2 points is LOST on any of those numbers.

There is no loss or gain if the pendulum lands on -1, 0, +1 or +2

Calculate the ratio of points lost : gained at any given time.

2. Relevant equations

3. The attempt at a solution

MY ATTEMPT at (I)

There are 11 numbers on which the pendulum can land BUT is more likely to land on the closest number.

NUMBER -1 is the starting point so the probability of it being there is 11/11

Probability of pendulum landing on NUMBER -2 is 10/11 because it is only one position away from the starting point.

Probability of pendulum landing on NUMBER +2 is 8/11 because it is three positions away from the starting point.

Ratio

Landing on NUMBER +1 : Landing on NUMBER +2

10/11 : 8/11

0.9 : 0.72

1.25 : 1

So pendulum is x1.25 more likely to land on -2 than +2

MY ATTEMPT at (II)

Probability of pendulum landing on NUMBER -2 is 10/11 and 2 points will be lost.
Probability of pendulum landing on NUMBER -3 is 9/11 and 2 points will be lost.
Probability of pendulum landing on NUMBER -4 is 8/11 and 2 points will be lost.
Probability of pendulum landing on NUMBER -5 is 7/11 and 2 points will be lost.

Probability of pendulum landing on NUMBER +3 is 7/11 and 2 points will be gained.
Probability of pendulum landing on NUMBER +4 is 6/11 and 2 points will be gained.
Probability of pendulum landing on NUMBER +5 is 5/11 and 2 points will be gained.

Probability of a loss of points is 10/11 + 9/11 + 8/11 + 7/11 = 34/11

Probability of a gain of points is 7/11 + 6/11 + 5/11 = 18/11

Loss : gain

34/11 : 18/11

1.9 : 1

The loss of points each time the pendulum lands on either -2, -3, -4 or -5 is 2. Total loss = 8 points

8 points x 1.9 = 15.2

The gain of points each time the pendulum lands on either +3, +4 or +5 is 3, 4, or 5, respectively. Total loss = 12 points

12 points x 1 = 12

Loss of points : gain of points

15.2 : 12

15 : 12 (approximately)

5 : 4

So for every 5 losses there are 4 gains.

ARE THESE CALCULATIONS CORRECT ??? Please advise. Thank you

Last edited by a moderator: Oct 29, 2017
2. Oct 29, 2017

### OmneBonum

Maybe it is just me, but I am having trouble visualizing the problem. The number line is straight, not a semi-circle, right? So the pendulum is stationary at -1? Then it is given some unknown amount of force? It can't then 'land on' any other number but -1, so do we mean something like 'appear directly above -2 at any randomly chosen moment'? Or is it a semicircle divided into arcs of equal length with -1 at the lowest point in stead of zero for some reason? Or does the pendulum start at -1, and then either receives a force or is allowed to swing. In the latter case, it will never hit 2 or -2 at all, so that can't be it. More info please! Sounds like a fun question!

3. Oct 29, 2017

### kuruman

I imagine this to be a pointer that starts at -1 already moving either to the left or to the right with an initial amount of unknown energy (fuel). It spends fuel at an unknown rate as it moves. When it runs out of fuel, it stops. It is labeled a "pendulum" which implies that it moves back and forth between the two extremes at least for a few times before it stops. That is how I visualized this picture.

4. Oct 29, 2017

### OmneBonum

Are you sure? The question uses the words 'pendulum' and 'swing.' Your visualization is more like, say, a wind up toy that automatically reverses when it hits a wall, the two walls being at -5 and 5 on an imaginary number line between them. In that scenario, given no knowledge of the amount of 'fuel' created by winding it up, isn't the toy equally likely to land anywhere no matter where you start?

5. Oct 29, 2017

### Barclay

YES THIS IS A BETTER DESCRIPTION

My question is actually based on a casino game. I've made up the question by analyzing the basic pattern that is sort of mathematical. The actual game is more complex with BONUS PRIZES and FORFEIGHTS on certain numbers. There number line cane be up to 200 either way. It is all a bit random but I'm trying to find a pattern to BREAK THE CASINO ..... then get a Hollywood movie deal.

So the POINTER moves but is hidden from view. You can stop it and see where it is and then at that point decide if you want to take your money (if you are on +3, +4, +5) or carry on in the hope you make more money (or get a BONUS .... but forget the BONUS for now because that just complicates things). You can't take your money or stop on -1 or 0 or -2 or +2

I didn't want to mention the casino element because people begin lecturing me on HOW THE CASINO ALWAYS WINS .... based on previous forum posts

Thanks

6. Oct 29, 2017

### Staff: Mentor

Are we to assume that the pendulum must land on one of these 11 numbers? The motion of a pendulum is continuous.

7. Oct 29, 2017

### Staff: Mentor

That is not relevant to the OP question. Everyone please focus (as @Barclay agreed to do in an off topic subthread which I have deleted) on the basic scenario described in the OP.

8. Oct 29, 2017

### Staff: Mentor

Let me attempt to rephrase this problem statement to make it clearer:

To start a run of this game, a motionless pendulum sitting at point -1 is given a randomly chosen impulse, with a 50-50 change of the impulse being to the left or to the right. This randomly chosen impulse causes the pendulum to travel some distance to the left or right, passing some number of points in the given direction (either -2, -3, etc. or 0, +1, etc.). At some point, the pendulum stops momentarily and then reverses direction and swings back. The farthest integer it passes before stopping and starting its swing back is considered the result of the run.

Is this a fair description?

Assuming it is, the probability you are trying to calculate is unknowable without more information. We need to know the distribution of possible impulses given to the pendulum; we know there's a 50-50 chance of being pushed to the left or right, but not, for example, how likely are pushes of 1 pound, 2 pounds, 3 pounds, etc. of force (or, equivalently, how likely are pushes resulting in an initial velocity of 1 meter/second, 2 meters/second, etc.).

9. Oct 29, 2017

### haruspex

But... that implies the impulse can get it from -1 to +5, and that the same impulse could have been applied in the other direction, which would have taken it to where?
Quite, and does it depend on the remaining distance to the last spot in the chosen direction?

10. Oct 29, 2017

### Barclay

Thank you Peter Donis for helping me continue the question.

I need to add the following:

The pendulum/pointer can swing left AND right. Basically it can be anywhere at any given time.

Basically I need to know what is the probability that the pendulum will hit -2 at which point the person loses -2 points. If the pointer was to continue travelling to the left is irrelevant really because the person has lost as soon as -2 is hit and the game stops.

What is the probability that the pendulum will hit +3 or +4 or +5 which point the person will make 3, 4 or 5 points if they stop the game. Otherwise the pointer may go back towards the negative numbers and there is risk of losing 2 points.

In summary should the person collect at +3 or wait to see if it hits + 4 or +5 when more points can be gained?

Thank you

11. Oct 29, 2017

### haruspex

To get anywhere we need to know the constraints on what the pendulum can do. So far I do not see any.
Can you at least give an example of something it will not do, or is less likely to do than something else?

12. Oct 29, 2017

### Staff: Mentor

And I've already told you that this is unknowable without more information. I explained what information is needed in post #8.

13. Oct 29, 2017

### Barclay

Hello

Constraints:
The game stops when pendulum/pointer hits -2.
Then the person loses 2 points

The game stops when pendulum/pointer hits +5 and then the person wins +5 points.

The pendulum may never reach the -2 or +5

14. Oct 29, 2017

### Staff: Mentor

This doesn't help in calculating probabilities. See my post #8.

15. Oct 29, 2017

### haruspex

Those are not constraints on the movement of the pendulum; they are consequences of the movements.
From what you have told us so far, all probability distributions for the movements are possible, so we have no information with which to answer your question.

16. Oct 30, 2017

### Barclay

OKAY you know better. There is no more information I can provide.

I thought it would have been possible to work out :
In 100 plays what is the probability of hitting NUMBER -2 (so losing -2 each time) compared with hitting either +3 +4 or +5 (and gaining 3, 4 or 5 point)

If that can't be calculated then thank you all for at least analysing my problem.

17. Oct 30, 2017

### Barclay

Constraints: it takes longer time to move from 0 to 5 than it does from say 0 to 1

Assume that the time taken to move between integers is equal .... say 10 seconds

Perhaps that helps?

18. Oct 30, 2017

### haruspex

Time is not relevant, as I understand it. All that matters is where it stops, yes? The only difference being slower makes is that your money lasts longer.

19. Oct 30, 2017

### haruspex

You can calculate something if you are prepared to guess a model for its logic.
E.g., say that having decided to move in a particular direction it picks any slot in that direction with equal likelihood. So when at -2 it will pick -3, -4 or -5 with probability 1/6 each and -1 through to +5 with probability 1/14 each. but it's only a guess.

20. Oct 30, 2017