# A bouncing ball is giving me fits.

#### Tethers

##### New Member

My name's Philip Ripper. I'm not a student. I'm a curious person with a fairly poor math education. This is not a homework question, but that aside, I will try to show the effort I've put forth to solve it.

The problem comes from a board game called Blood Bowl, a fantasy wargame simulation of american football (it's great fun).

There is an action in the game called a Hail Mary Pass, HMP for short. When you do this, you throw the ball at a location on a grid. The ball is never accurate, and Bounces three times before coming to rest (at which point someone can try to catch it).

Bouncing: label the 8 squares surrounding the starting square from 1-8, and roll an 8 sided die to determine where to move the ball. The ball Bounces three times. (So think of it as a 3x3 grid, with the starting square in the center, the ball bounces to a random square not including where it started for that bounce - 8 possibilities per bounce).

I'm trying to figure out the odds that the ball will land on each possible grid square after a HMP (three bounces). I figured this creates a 7x7 grid of possible locations.

I know 8x8x8 = 512, so there are 512 possible dice combinations. And I know that the four extreme corners of the grid (3,3; 3,-3; -3,3; -3,-3) are each a 1 in 512 chance.

I know to figure the odds you are supposed to determine the number of events (512) that create each outcome. I also know the problem of bouncing a ball three time sounds simple, but that once anyone I know tries to solve it they give up.

I think it might be considered a markov chain? From trying to google for a formula that I could understand to find the solution without just counting out each of 512 events / 49 states manually.

I've worked in a spreadsheet (I'm not good with those) for about two days (as in 48 hours of work) trying to solve this problem. I've come close several times, but always end up with a probability total of more than 1, so I know I'm always wrong.

I charted out the options of the d8 roll as:

1: +1x +1y
2: +1x +0y
3: +1x -1y
4: +0x +1y
5: +0x -1y
6: -1x +1y
7: -1x +0y
8: -1x -1y

And considered the starting point as 0,0. I think this would allow me to brute force the problem, but I really don't want to count out 512 variations (though I would have been done by now if I had, I guess).

Is there a way to solve this that a high school drop out could understand? I'm afraid I don't understand the formula for markov chains, if that is what this problem is.

If there is no conceivable way for a guy with a spreadsheet to solve this without understanding complex mathematics, could anyone solve it for me?

I'd prefer to understand how to solve it. But I would still be happy with an answer instead, because I fear I may just not be able to understand it. Though I am eager to find out if I can. I've tried very hard to figure this out with the tools available to me. Please assist!

Sincerely,
Philip

#### Dason

So you have an equal chance of going to any of the 8 spots that aren't the location you're currently on?

#### Prometheus

##### New Member
My mate has this game, though we've never played it. My first thought was markov chain too. Just to clarify further - what happens if the ball 'bounces' at the edge of the game board? Does it bounce back, or go out of play, or go to the other side of the board?

#### Dason

A markov chain is the way to model this. Figuring out the probabilities at the end of three rounds is quite simple - you just need to set up a transition matrix and then do some matrix multiplication. It's easy enough to find the answer and I already calculated the probabilities but I'm not sure if what I'm thinking the transition probabilities are are the same as what you're thinking they are so I'm just waiting for clarification on my previous question.

#### Tethers

##### New Member
Thank-you all for the replies! It was supposed to e-mail me if I got any, but it didn't, so I didn't notice until just now. Sorry for the delay. I think I've solved it. This is what I came up with:

0.20% 0.59% 1.17% 1.37% 1.17% 0.59% 0.20%
0.59% 1.17% 2.34% 2.34% 2.34% 1.17% 0.59%
1.17% 2.34% 5.27% 5.27% 5.27% 2.34% 1.17%
1.37% 2.34% 5.27% 4.69% 5.27% 2.34% 1.37%
1.17% 2.34% 5.27% 5.27% 5.27% 2.34% 1.17%
0.59% 1.17% 2.34% 2.34% 2.34% 1.17% 0.59%
0.20% 0.59% 1.17% 1.37% 1.17% 0.59% 0.20%

---

Equal chance of each 8 possibilites per step.

For the purpose of this, imagine it cannot reach the edges of the board (say they are 50 squares away or something).

To figure it out with what happens when it goes off-table, it would be so much more complicated and I'm super proud I've come close to an answer on this as is, so I'll leave that modeling for a future project.

The game is silly fun, btw. But probably just confusing for two novices. Better to play with one who knows how to do it.

#### Dason

So are you saying that the board size is larger than 3x3 and that you have an equal chance of going to any of the adjacent 8 spots

#### Tethers

##### New Member
Correct.

You start at the theoretical center. You randomly move to any of the surrounding squares (8 of them). This means the ball cannot stay in the same place.

Where ever the ball is moved after the first bounce, you consider that the 'new' starting point, and scatter to any of the squares that surround that 'new' starting point. then you do this a third time.

From this, I tried to figure out the odds of the ball ending up on any particular square after the three bounces (scatters). I posted those chances above. If I understand, it should be a 7x7 area that the ball can land after those three bounces. I've included my guess at the odds for those above.

edit: The center square of the 7x7 grid being the location of the ball before the three bounces.