Probability question, Very hard

[zgozvrm]

I'll run 4 more when I get some time (I'm at work!). Do you have Excel? If so, I can post it in spreadsheet form.

 

[SW VandeCarr]

I'll run 4 more when I get some time (I'm at work!). Do you have Excel? If so, I can post it in spreadsheet form.

OK. Thanks. BTW I meant 11 players after excluding the originator.

 

[zgozvrm]

Here's the complete data ...

(Note that the numbered columns represent how many wins each player has per successive 100,000 games played. That is, each column 1, 2, 3 ... 10 represents 100,000 games each, the "Total" column is the sum of the previous 10 columns; all 1,000,000 games)

 

[Jamma]

It's almost as if you guys still don't trust my proof =D haha

Aradesh's other problem is really cool. The reason I liked it, is that if you were asked to solve it before seeing this problem, it would have took a pretty amazing leap of insight to think of solving this one first. I don't think that I'd have managed to solve it, at least not in a day. But when you've solved the circular problem, you can show that the probability of it falling off the n end is i/(n+1) if the game starts at the i'th player. It's a pretty cool result!

 

[SW VandeCarr]

It's almost as if you guys still don't trust my proof =D haha

Yes. I couldn't let go of this problem. Actually once I considered that it had the Markov property:

$$P(q_i|q_{i-1},..,q_1)=P(q_i|q_{i-1})$$ for $$i>1$$

it should have been clear to me that the whole problem could have been broken down to just two players plus the originator. You were correct all along; but, of course, you are a mathematician.

 

[SW VandeCarr]

Here's the complete data ...

(Note that the numbered columns represent how many wins each player has per successive 100,000 games played. That is, each column 1, 2, 3 ... 10 represents 100,000 games each, the "Total" column is the sum of the previous 10 columns; all 1,000,000 games)

Thanks zgozvrm. I don't doubt the proof, but I'm curious about the actual data variations

 

[zgozvrm]

Thanks zgozvrm. I don't doubt the proof...

I didn't give a proof; I merely presented some data.

...I'm curious about the actual data variations

What are you curious about?

 

[SW VandeCarr]

I didn't give a proof; I merely presented some data.

I was referring to Jamma's previous post about a proof which I took to be in reference to your simulation. Sorry if that wasn't clear.

What are you curious about?

I was interested in the aggregate percentage of wins for the 11 positions across all four runs or as large a sample as you may have in order to to look at convergence. All I have with me is my laptop. Since you posted your simulation results, I didn't think you'd mind sharing this.

 

[zgozvrm]

Not sure what you mean by "aggregate percentage of wins..." exactly.

 

[SW VandeCarr]

Not sure what you mean by "aggregate percentage of wins..." exactly.

Nevermind. Thanks for posting the PDF and some of your results.

 

[Jamma]

Hmm, this seems wrapped up. I wonder if there are more difficult versions, with players placed in a lattice around a sphere for example, or on a square table and the one who drops it off loses =D

Actually, let's not go there...