double

[Jellby]

And the fact that you saw a 1-in-280,000-times event once says.. nothing.

[Romain672]

Following this discussion, of course, and just wanted to chime in on a couple of data points. I've had several games recently where the algo has rolled multiple doubles in a row for my opponent. These almost always occur toward or at the endgame portion, but of course the algo knows nothing of the situation on the board. We're not talking two or three, but four five and recently SEVEN! Yes, I've played a lot of games of backwardsgammon here, a little over 2,000. But the odds of a fair pair of dice rolling seven doubles in a row is....wait for it....

1 in 280,000

One would expect to have to play 232,845 games to have a 50/50 chance of this occurring, and the average number of games you'd need to play to see this once would be 335,922. This assumes random dice rolls, of course...LOL

I guess I am well above average, then...

First part is right. Probability is 1 in 280,000.

But that probability can arise for: yourself, the opponent, both players.
And it can arise at the start of each roll.

There is about 26 rolls in each game for each player (correct me if I'm wrong, I'm not sure of that).

So, every game, you got 19 starting streak of 7 for yourself, 19 starting streak of 7 for the opponent, and 46 starting streak of 7 for both players.
Which give a total of 19+19+46=84 potential starting streak.

Now you played a total of 2215 games.

Which mean there was a total of potential starting streak of 7 equal to 84*2215=186 060 starting streak possible.

You gave yourself the number 232 845 to have a 50/50 chance of this occuring, I can believe yoiu, but you already tried 186 060 times to have that

We are pretty close than what you can expect.
If you do it again, that will be weird, but totally possible.
If you do it twice, that will be really weird.


But thanks for sharing it though, usually the unlikely event shared on the forum are around 1/10k or less. Sometimes we find a 20k/30k. You got a 280k which is impressive

[Romain672]

If you want veggivet, I note on this forum the number of games you have done which is 2215 games.
Now, if you find another weird event between the three i have proposed (your number of double, opponent number of double, or number of double for both players in a row), I can start my probability calculation from the 2216th game.

You should have an hard time to find a probability under 1/10k.


Or I can propose you that:
In your next (100?) games, you note the number of time (or the games, i have an easy way to extract the doubles), you, the opponent or both players did 4+ or 5+ doubles in a row. Then you share the results on the forum. If you do it, I can try to find how likely it was.

And if you do it but find nothing, feel free to share it too. That could be great for us

[veggivet]

Will do! If it happens again, I hope it happens to me and not my opponent!

[veggivet]

If you want veggivet, I note on this forum the number of games you have done which is 2215 games.
Now, if you find another weird event between the three i have proposed (your number of double, opponent number of double, or number of double for both players in a row), I can start my probability calculation from the 2216th game.

You should have an hard time to find a probability under 1/10k.


Or I can propose you that:
In your next (100?) games, you note the number of time (or the games, i have an easy way to extract the doubles), you, the opponent or both players did 4+ or 5+ doubles in a row. Then you share the results on the forum. If you do it, I can try to find how likely it was.

And if you do it but find nothing, feel free to share it too. That could be great for us

You say you have an easy way to extract doubles. Is there any way to determine if a certain double occurs far more often than chance would dictate? The specific situation I have in mind is this: It's been my experience, and that of many other players I chat with, that double sixes are seen way too frequently when you have a man out on the bar and your opponent has only the 6 point covered in his/her home board. Can your method help with this question? Thx!

[Romain672]

You say you have an easy way to extract doubles. Is there any way to determine if a certain double occurs far more often than chance would dictate? The specific situation I have in mind is this: It's been my experience, and that of many other players I chat with, that double sixes are seen way too frequently when you have a man out on the bar and your opponent has only the 6 point covered in his/her home board. Can your method help with this question? Thx!

Hi, my method is pretty archaic: https://docs.google.com/spreadsheets/d/ ... 1659177939 , that document was done to extract double of Catan's games.

I just have to open the replay page, do ctrl+a, ctrl+c, then in page "logs" I do ctrl+v.

Then in page dice rolls, I've got all the dices of the game. And I can put any formula I want on them.

So no I can't do that



But I got a similar request in the Catan forum, and he was claiming that the dice rolls were weird, there was a pattern, like if there was two X, there will be an high chance to have lots of X later on.
So what you see here in white page 'dicerolls' is when a number has been roll twice in the last three rolls.
And those are the games which was considered weird.



An easy statistical way to do that is for you, to choose to count the next roll or not before the roll occurs when you think your situation arise.

[Silene]

The specific situation I have in mind is this: It's been my experience, and that of many other players I chat with, that double sixes are seen way too frequently when you have a man out on the bar and your opponent has only the 6 point covered in his/her home board. Can your method help with this question? Thx!

A methodology to investigate this might look like this:

1. Get a large data-set. I.e. the 2245 games of backgammon you played would suffice to have a good first estimate.
2. Extract the game-logs for the whole data-set. It should work with the text-log BGA provides when visiting game-replay
3. Count_total = Analyze the logs to count all situations where it occurred that (a) a stone was on the bar AND (b) the opponnent had 6 point covered AND (c) the opponnent had 1-5 points NOT covered. This could of course happen multiple times in a game and also multiple times in a row within that game.
4. Count_double_6 = Check in how many of these situations the next roll was 6-6
5. Devide Count_double_6 by Count_total

The result should be about 1/36.

I would volunteer to do the coding for the extracting from the logs. But the problem is that I see no way to retrieve the data because (a) website-scraping (to automatically retrieve the game-logs) is not allowed on BGA and (b) the amount of replays you're allowed to load is limited per person per day.

[veggivet]

Thanks for those responses. Very helpful. I won't this idea any further. Does anyone else who reads these backgammon threads think that there are more 'boxcars' (double 6 rolls) in the situation I described than chance would account for? Would appreciate any feedback.

[veggivet]

It didn't take long for another unlikely event to occur. This time, it was only 4 doubles in a row. Compared to 7 in a row, the odds don't look outrageous, but it works out to 1296 to 1.

[Romain672]

It didn't take long for another unlikely event to occur. This time, it was only 4 doubles in a row. Compared to 7 in a row, the odds don't look outrageous, but it works out to 1296 to 1.

You did 48 games.
So you got a total of 48*84=4032 starting streak possibles.

The probability for yourself to have that unlikely event for yourself/the opponent or both players was 95.6%
So it would have been weird instead for you to not have 4 double in a row in those 48 games.