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

At least you have a rank - in Amazons, I have a rating, but no rank! :-(

(Bug where rank is not updated if rating does not change, and my rating changed from "unrated" to XXXX - but moved "0" and hence not really "updated", hence not given a rank) :-)

MM: See http://brainking.info/bugs.html.

Actually harls, I did start to! But, I realized I was making an error since the fact we both had a pair changes the math considerably.

Fencer, just checked my rating in my profile and it says i am 3rd on BB+ , but when you click on it and goto the ratings it says 2nd.

Oh don't, Niki! He'll give us a huge mathematical formula as to exactly what the chances are! (Only kidding, GI!)

Wow ! what are the chances ??
don't worry, that was a rhetorical question ;-D

Actually, it was part of this picture where my daughter and I both had the same exact hand in poker. I was getting ready to play in a Texas Hold'Em tournament in Australia, so I was using the Adelaide Casino deck (blue diamonds and green clubs.)

Brittany likes to play 5-card draw so I was taking a break playing a few hands with her.

We both ended up with a pair of tens, and an Ace-Queen-2. It was kind of funny, because at the show down she said "Pair of tens". So I asked her what her next high card was. She said "I have an Ace, so I win." I laughed and said I had an Ace too. I told her I had a Queen, then she laughed and said "darn, you probably will win, I only have a 2."

It was such an unlikely occurence, one of the other players took a picture of it for us.

Yours was taken in your moody period then? ;oÞ

...someone could create the "Back In The Day" Fellowship where members could upload everything from their baby pictures to when they were teenages, to young adults, then old cronies :)

thank you, everything worked out perfectly:)

town tell wednesday. Sry if i did not get to our game..
:o{P

but I really do look like this :)

It should be done.

ughaibu: It will be a server-side cache problem [I've written the cache myself and it contains some minor bugs]. The application server reload should fix it, I'll ask Liquid to do it.

Yes it seems most people are using their own picutres. Mine hame is certainly an original. There are some that crack me up though like AD's. Too funny AD. I think I saw Pres Clintons out there somewhere too or I might be mistaken.