Probability calculation

[Ursula1972]

Has anyone determined the probability that a particular row or column is filled in with 0, 1, 2, etc.? So, for ABC and for JKLMN? The probability that, for example, B has six colored squares is, I believe, 33%. Has anyone figured this out? So:
A and C: 0 1 2 3 4
B, K, M: 0 1 2 3 4 5 6
J, L, N: 0, 1, 2, 3

Of course, I know that the probability that B has 0 colored areas is zero percent.

[BaiBlagoi]

I think I did.

The probability of 0,1,2,3 and 4 spaces for A, D and G is 6.67%, 26.67%, 33.33%, 26.67% and 6.67%.
The probability of 0,1,2,3, 4, 5 and 6 spaces for B, E and H is 0%, 4.44%, 4.44%, 17.78%, 26.67%, 13.33% and 33.33%.
The probability of 0,1,2,3 and 4 spaces for C, F and I is 0%, 0%, 6.67%, 46.67%, 46.67%

The probability of 0,1,2 and 3 spaces for J and O is 0.25%, 7.90%, 43.35% and 48.40%.
The probability of 3, 4, 5 and 6 spaces for K and P is 18.52%, 47.40%, 29.63% and 4.45%.
The probability of 0,1,2 and 3 spaces for L, N , Q and S is 1.48%, 18.15%, 49.25%, 31.11%.
The probability of 3, 4, 5 and 6 spaces for M and R is 31.11%, 49.25%, 18.15% and 1.48%.

Hope this helps, good luck!

[Ursula1972]

Wow, thanks so much. That really helps!

[jatubio]

@BaiBlagoi, thank you so much!

How can we use these numbers as a strategy?
- Assuming high probabilities as answer when guessing?
- Asking low probability options first?

[BaiBlagoi]

@Jatubio
well this is The question that I am trying to find myself the answer to...

[jatubio]

@BaiBlagoi, hahahaha

At least for now, I'm using these statistics when I have to guess. I'm not certain, but I think they have improved my win rate.

[Ursula1972]

You can, of course, use it in solo games. I also wanted to know it to increase the chances of getting a meaningful result when there's still zero information about a digit, or even more information. Which column or row is best to ask about to eliminate more digits?

[Wolf_4D]

I think I did.

The probability of 0,1,2,3 and 4 spaces for A, D and G is 6.67%, 26.67%, 33.33%, 26.67% and 6.67%.
The probability of 0,1,2,3, 4, 5 and 6 spaces for B, E and H is 0%, 4.44%, 4.44%, 17.78%, 26.67%, 13.33% and 33.33%.
The probability of 0,1,2,3 and 4 spaces for C, F and I is 0%, 0%, 6.67%, 46.67%, 46.67%

The probability of 0,1,2 and 3 spaces for J and O is 0.25%, 7.90%, 43.35% and 48.40%.
The probability of 3, 4, 5 and 6 spaces for K and P is 18.52%, 47.40%, 29.63% and 4.45%.
The probability of 0,1,2 and 3 spaces for L, N , Q and S is 1.48%, 18.15%, 49.25%, 31.11%.
The probability of 3, 4, 5 and 6 spaces for M and R is 31.11%, 49.25%, 18.15% and 1.48%.

Hope this helps, good luck!

useless info

[ct219]

Actually it tells you a lot, especially so in the physical game. Let me explain:

I get different probabilities when I worked it out, but you can then apply information theory to see how "much" each question tells you. This is what's actually important.

In the actual 2 player game (not BGA) this guides you heavily - ask questions with highest information at each stage for the fastest mean answer. Bonus points for tracking your opponent's remaining entropy and adapting your risk strategy accordingly. It also tells you to set codes like 232/565 as the better questions yield less information, and not to use 1s so much.

For BGA, or 3+ players, *If* my (casual and unchecked!) sums are right, the game needs 18.83 bits of information to reduce which of the 465,120 combinations we seek. So, with 10-12 "good" questions needed to win, you want to choose one that leaves the correct amount of entropy left so that you will be appropriately able to guess the solution. You can obviously pick highly informative questions (count "B" at 2.27 bits) or utterly redundant questions (segments you've deduced must be lit/empty).

I've not looked further to see how feasible this is, and how correlations on the questions affect your future knowledge of how that entropy is burnt down. At it's simplest, at the end, you can see with two player games how you might avoid asking a useful question because it won't let you guess, but the next question might let your opponent do so.

I leave working that out properly as an exercise for the reader

C.