cool!!! that really speeds things along!

For paying members: The dropdown box containing all games for "move and go to" action should be now sorted in the same way as your game list on Main Page is.

shudders....stop that....

Cole: It's not fully fixed yet. It can happen again.

My highlighted phantasm message bug has disappeared...which is good ...I was beginning to feel like one of those conditioned dogs who respond to lights and sounds...which of course meant I ignored the message light for the longest time today....till I couldn't stand it anymore....and there it was.... a genuine message...sigh...

....sing it brother!...jumps in with harmonica

*breaks out into song*
"I feel like a number"

Everything has an ID. You, fellowships, discussion boards, games, tournaments, etc.

LOL, I did not know the Fellowships had an ID. grin.

I think they are by ID, as in the order they were created :-)

Easier not to be in so many fellowships :-D
They will be sorted by a name, later.

easier to search by name though

Aren't they ordered by the fellowship ID?

He said will put back to alphabetical sometime in future when I asked

Hey Fencer, what order is our Fellowships listed on the right side? I seem to keep srolling up and down looking for certain ones.

You are right, thanks. It will be fixed.

http://brainking.com/game/ArchivedGame?g=387609

It said I just checkmated my opponent but, my opponent still has pieces to drop between me and the bishop I checked with.

I currently have 5 BrainKing windows, 1 DailyGammon, 1 GT, 1 IYT, 1 "start page", and my mail.

or 4 or 6 or 8 or 10(like me)

He can have two open BrainKing windows.

By someone's nic it still says 'browsing waiting games' when they are writing a message.

Kind of anomaly, I guess :-)

I know they are there now, this was yesterday. they are saying they did not receive any invite in there message box (which seems odd as i know it works), but when they went to the Main Madhouse page they just accepted.
Now, i know i kept inviting them, so how come it let me keep doing it ??

Have a look at your fellowship page if the list of invited users contains him/her.

In BK 1.0, once you had invited a person to a fellowship, you did not have the option to re-invite them unless they left or declined.
I was asked to invite someone (while playing them), which i did in the normal manner. After a while they said they had not received the invite, so i returned to there profile and it let me do it again (I did get a message say they may have already received an invite though). Same outcome 4 times. In the end i told them to just search for the fellowship and they were able to join. Bug ???

Please read http://BrainKing.info, List of changes.
And thanks for pointing out this entry in tips. I forgot to remove it.

Ever since the new changes i've had to type my name and password everytime i come to the site..i saw on the "tips" that it says i can change to auto-login on my edits page ...but it's not there to change anywhere....is that just something for members...??

BBW: No, there is no way until they finish at least 4 rated games. This is how it works since the very beginning of BrainKing.com.

Is there any way to actually see what an "unrated" players rating is?

Why I ask is for example, in Dark Battleboats - I have a 13xx rating, and I've won a couple more games against unrated players and both times my ratings did not change - If there is a way to see what a current unrated players rating is, I could do my own math and double check my rating.

update: So I won 2 game, rating did not change. Lose 1 game, rating drop almost another 100 points. (of course againts an unrated person so there is no way for me to check it for myself..) :-(

Thanks, it's okay now.

No more off-topic posts please. You have General Chat for it.

'

'
where basic algorithms are shared ... if you have access to the
Fred Fish Library of the Amiga Community, I'd recommend you
try out Mathlab from 1990 - it comes with an extra floppy full of old
Fortran sources - fancy implementations partly and, highly accurate
- fairly easy to transcode them to Pascal later ... ~*~

I am currently writing a program to go through every possible set of two hands and checking for the same pair and three other (different) cards that are the same. When i'm done, we'll have the actual answer.

Yes I have.

The odds of drawing 1 out of 12 if 1 out of 12. It does not matter is I have all 4 suits, or 3 of the 4 suits, or just 1 of the four suits.

The odds of picking 1 out of 12 is 1 out of 12.

We happened to know, AFTER THE FACT, that the other card was the same.

Put another way: Say I draw 4 cards out of the deck, and they are all tens. What are the odds the next card I draw will be a ten?

You are saying 1 out of 48 (52-4) and I am saying 0 chance (of the 12x4 remaining, none are tens.)

The "specificity" is accounted for.

As this is basic combinatorics, I think we can leave it off here.

But just so you know, if you click here I showed how to count all of the Gothic Chess positions before any one piece comes off of the board. That number is 32,099,674,107,692,140,366,789,953,222,888,490,987,180,838,400,000,000 which makes doing "card math" a piece of cake :)

Sorry Kevin I am with Ed. LOL

Harley, this is off topic for the board????

If you know you are not drawing from the group that has two, then you have not accounted for that in your math. Also, it does matter whether the group has four or the group has three. The chances of selecting each individual card is the same (1 / (# of cards)) - not the chance of getting a 9 is the same as getting a Queen, regardless of how many of each are left.

Kevin, if you have 4 groups of 13 different cards, then pull out all 4 10's, there are just 12 different ones remaining, correct?

2 2 2 2
3 3 3 3
4 4 4 4
. . . .
A A A A

OK, now I am going to remove 1, then remove another exactly like it.

Surely there are 12 sets of matching cards to start with. And, even after I draw 1 card, there are still 12 identical sets. One set will only have 3 cards, but that does not matter.

Now, there are 11 of the same sets with 4 cards, and one set of cards only has 2.

I know I am not drawing from the set that has 2, or I will have another pair.

That means I am drawing from one of the 11.

Etc.

poohy LOL

No more Stevie or I'll have to delete. Not even with a smiley face!

go play your games ;oÞ

I believe your answer is not correct either.

You say:
--------
{there are just 12 "different" cards now to start...}

4 x (12 x 12) x (11 x 11) x (10 x 10) =

6,969,600.
-----
However, after the first "x12", it is no longer 12:1 that another card of the same number will be drawn - there are only three of it remaining, while there are four of the rest. Same for the 2nd "x11" and "x10".

So then what is the answer, if my answer is wrong?

Except that the answer is wrong.

There are 13 of the same cards, disregarding suits.


The chances of 4 of the same being drawn by two people would be...

4 x (13x12x11x10/4x3x2) then divided by 2.

This is one in 5720.

Now, given that 4 cards have been removed, with 6 more to be dealt, with 3 matching pairs, the odds are:

{there are just 12 "different" cards now to start...}

4 x (12 x 12) x (11 x 11) x (10 x 10) =

6,969,600.

Multiply this by 5720 and you get:


39,866,112,000 to 1.

and it only took Kevin 44 minutes to do as well, :)

rofl - It's amazing what you can do with a calculator :-)

*faint* Amazing odds, however you worked that out!

The chances of the two players getting the same pair and the same three (different) other cards is:

(52/52 * 3/51 * 48/50 * 44/49 * 40/48) * (2/47 * 1/46 * 3/45 * 3/44 * 3/43) = 1.24 * 10 ^ -8 = 0.0000000124 = 0.000000124 % = 124 / 10,000,000,000 = 1 / 80,667,372.92

ok
I had not bothered to look as the date still says 30th June lol :-)
No wonder i keep jumping about on the ranking at the moment, mainly against unrated players.

That doesn't surprise me! And it probably wouldn't surprise you that I wouldn't have understood a number of it! :oD