% won vs Diff. opp. #

[chapcrap]

"was andy a bonus monkey? "

Lol.

Congratulations Chap. Does anyone know the odds of only 1 pair from 7 cards?

It may make Denise feel better as she approaches 0-50.

I'm not sure the odds. They're pretty good at getting 2 sets from 7 cards.

[sherkaner]

"was andy a bonus monkey? "

Lol.

Congratulations Chap. Does anyone know the odds of only 1 pair from 7 cards?

It may make Denise feel better as she approaches 0-50.

Hmm, I think it would be 6*6*7/(2*3^7) (6 options of 7 positions divided in 5/2/0, total 3^7 possibilities, there's no bias towards certain cards at that point). 5-2-0 is the only possible division.
So around 5.7 %.

[Chariot of Fire]

The 7 cards have to comprise 5 of one colour and 2 of another colour.

1st card - can be any. 1:1
2nd card - can be any. 1:1
3rd card - must be one of 1st or 2nd. 2:3
4th card - must be one of 1st or 2nd. 2:3
5th card - must be one of 1st or 2nd. 2:3
6th card - must be the same as whatever one holds 3 of. 1:3
7th card - must be the same as whatever one holds 4 of. 1:3

T/fore 1 x 1 x .67 x .67 x .67 x .33 x .33 = .0328 (a 3.28% chance of getting landed with this combo).

So Denise had a 3% chance of getting stung like that, followed by red having a 22% chance of a 3 card set. So the combination of these two events is a very improbable 0.7% (or roughly 1 chance in 140).

Poor Denise. It's written in the cards, as the saying goes.

[Crazyirishman]

I learned my lesson, talk trash to denise and she'll kill you.

[Denise]

I learned my lesson, talk trash to denise and she'll kill you.

At least there was that small pleasure. After getting a handful of red cards, I wish I could have killed you twice.

"Congratulations Chap. Does anyone know the odds of only 1 pair from 7 cards?

It may make Denise feel better as she approaches 0-50.

So Denise had a 3% chance of getting stung like that, followed by red having a 22% chance of a 3 card set. So the combination of these two events is a very improbable 0.7% (or roughly 1 chance in 140).

For some strange reason, that doesn't make me feel at all better. The dice gods are going to keep me humble by never letting me win an 8 man, I guess.

[sherkaner]

The 7 cards have to comprise 5 of one colour and 2 of another colour.

1st card - can be any. 1:1
2nd card - can be any. 1:1
3rd card - must be one of 1st or 2nd. 2:3
4th card - must be one of 1st or 2nd. 2:3
5th card - must be one of 1st or 2nd. 2:3
6th card - must be the same as whatever one holds 3 of. 1:3
7th card - must be the same as whatever one holds 4 of. 1:3

T/fore 1 x 1 x .67 x .67 x .67 x .33 x .33 = .0328 (a 3.28% chance of getting landed with this combo).

So Denise had a 3% chance of getting stung like that, followed by red having a 22% chance of a 3 card set. So the combination of these two events is a very improbable 0.7% (or roughly 1 chance in 140).

Poor Denise. It's written in the cards, as the saying goes.

You know that's the wrong calculation, right? You're not checking whether the second card is the same as the first, and whether 6-1 doesn't show up as a possibility, and other sequences.

For example, rrrrrrr and rbrrrrr would be valid according to your calculation, but rrbbrrr or rrrrrbb wouldn't be.

Mine was just: 6 possibilities of card colour combinations (5r2b, 5r2g etc.) times the amount of divisions possible (7 over 2, or 7!/(2!*5!), shortened 7*6/2) divided by the total number of possibilities (3^7, 3 color possibilities for each card). And I think that should be accurate.

[AAFitz]

#P - %/ % World2.1 / % Expected
8p - 34% 49%(12.5%)
7p - 10% 0% (14.3%) 5 games
6p - 53% 69% (16.66%)
5p - 19% 24% (20%)
4p - 66% 82% (25%)
3p - 41% 60% (33.33%)
2p - 59% 74% (50%)

dubs - 57% 82% (50%)
trips - 53% 69% (50%)
quads - 34% 49% (50%)
My team rates are pretty low, but I really do play a lot of random games with random partners.

I seem to be much better with an even number of people for whatever reason. Odd.

[Denise]

No, even.

[AndyDufresne]

"was andy a bonus monkey? "

Lol.

Congratulations Chap. Does anyone know the odds of only 1 pair from 7 cards?

It may make Denise feel better as she approaches 0-50.

Congrats on the win, Chapcrap. I think overall, I played pretty alright and didn't make many mistakes... As for Bonus Monkeying, only in my dreams!


--Andy

[Robinette]

The 7 cards have to comprise 5 of one colour and 2 of another colour.

1st card - can be any. 1:1
2nd card - can be any. 1:1
3rd card - must be one of 1st or 2nd. 2:3
4th card - must be one of 1st or 2nd. 2:3
5th card - must be one of 1st or 2nd. 2:3
6th card - must be the same as whatever one holds 3 of. 1:3
7th card - must be the same as whatever one holds 4 of. 1:3

T/fore 1 x 1 x .67 x .67 x .67 x .33 x .33 = .0328 (a 3.28% chance of getting landed with this combo).

So Denise had a 3% chance of getting stung like that, followed by red having a 22% chance of a 3 card set. So the combination of these two events is a very improbable 0.7% (or roughly 1 chance in 140).

Poor Denise. It's written in the cards, as the saying goes.

So you're saying a 7 card fail happens 1 in 33 times,
I always just "ballpark estimated" that it was the same as a runner-runner flush, which is what? about 1 in 25? (4%)

The key to it happening is missing one color completely, and not having more than 2 of a color... all other combos work.
So what i don't like about your formula is this (let's assume the missing color is Yellow)
1st 2 cards can RB, RR, BB
so 3rd card, if not yellow, makes a set.

So how 'bout this:
1st card doesn't matter
2nd card has 1 in 3 chance to be missing color
3rd through 7 same 1 in 3 chance... we are now at 9%
Then add back in the chance of 2 sets with just 2 colors
The 1st 5 cards don't matter...
From there, 1 in 4 chance of not hitting the 2nd set
_____________________________________
So that would be 2 1/4%
Yikes... that's 1 in 44
Worse odds than either of us thought...

Oh, the other number that looks off... wouldn't RED have a 33% chance of a 3 card set (1 in 3) (not 22%)

[IcePack]

This series should keep getting played adding an 8th "winless" person each time you get a winner

[ljex]

The 7 cards have to comprise 5 of one colour and 2 of another colour.

1st card - can be any. 1:1
2nd card - can be any. 1:1
3rd card - must be one of 1st or 2nd. 2:3
4th card - must be one of 1st or 2nd. 2:3
5th card - must be one of 1st or 2nd. 2:3
6th card - must be the same as whatever one holds 3 of. 1:3
7th card - must be the same as whatever one holds 4 of. 1:3

T/fore 1 x 1 x .67 x .67 x .67 x .33 x .33 = .0328 (a 3.28% chance of getting landed with this combo).

So Denise had a 3% chance of getting stung like that, followed by red having a 22% chance of a 3 card set. So the combination of these two events is a very improbable 0.7% (or roughly 1 chance in 140).

Poor Denise. It's written in the cards, as the saying goes.

Did you mean a 33% chance for red having a 3 card set? Im assuming just a mishitting of the keys

[Chariot of Fire]

Nah, just a mishitting of my neurons too early in the day. Of course it's 33%

[Geger]

The 7 cards have to comprise 5 of one colour and 2 of another colour.

1st card - can be any. 1:1
2nd card - can be any. 1:1
3rd card - must be one of 1st or 2nd. 2:3
4th card - must be one of 1st or 2nd. 2:3
5th card - must be one of 1st or 2nd. 2:3
6th card - must be the same as whatever one holds 3 of. 1:3
7th card - must be the same as whatever one holds 4 of. 1:3

T/fore 1 x 1 x .67 x .67 x .67 x .33 x .33 = .0328 (a 3.28% chance of getting landed with this combo).

So Denise had a 3% chance of getting stung like that, followed by red having a 22% chance of a 3 card set. So the combination of these two events is a very improbable 0.7% (or roughly 1 chance in 140).

Poor Denise. It's written in the cards, as the saying goes.

So you're saying a 7 card fail happens 1 in 33 times,
I always just "ballpark estimated" that it was the same as a runner-runner flush, which is what? about 1 in 25? (4%)

The key to it happening is missing one color completely, and not having more than 2 of a color... all other combos work.
So what i don't like about your formula is this (let's assume the missing color is Yellow)
1st 2 cards can RB, RR, BB
so 3rd card, if not yellow, makes a set.

So how 'bout this:
1st card doesn't matter
2nd card has 1 in 3 chance to be missing color
3rd through 7 same 1 in 3 chance... we are now at 9%
Then add back in the chance of 2 sets with just 2 colors
The 1st 5 cards don't matter...
From there, 1 in 4 chance of not hitting the 2nd set
_____________________________________
So that would be 2 1/4%
Yikes... that's 1 in 44
Worse odds than either of us thought...

Oh, the other number that looks off... wouldn't RED have a 33% chance of a 3 card set (1 in 3) (not 22%)

It's too complicated.

With 7 cards we don't get 2 sets, if we have 5&2 combos (example 5R-2G, total 6 combo-types). Also the odd must be : 6*(7!/5!2!)/3^7 = 14/3^5 = 5,76%

[AndyDufresne]

This series should keep getting played adding an 8th "winless" person each time you get a winner

Hm, my points would take such an impact. Haha


--Andy

[IcePack]

This series should keep getting played adding an 8th "winless" person each time you get a winner

Hm, my points would take such an impact. Haha


--Andy

It would be interesting to see if after 8 games, if the original 8 all earned a win or how many left in the original group there would remain winless.

IcePack

[AndyDufresne]

Would remain winless. -------->


--Andy

[chapcrap]

This series should keep getting played adding an 8th "winless" person each time you get a winner

Well, then you guys need to find someone besides me...

[Leehar]

http://www.conquerclub.com/forum/viewto ... 2#p2444088
(Game-type neutral percentages)

Courtesy of Fruit's nifty spreadsheet:

Total average weighted win rate % 44.21%
Your average weighted win rate % 49.80%

5.6% difference

approx

[Rodion]

3/\7 = 2187

5 reds, 2 greens - > 7! divided by (5! x 2!) -> 21 possibilities
5 reds, 2 blues
5 greens, 2 reds
5 greens, 2 blues
5 blues, 2 greens
5 blues, 2 reds

21 x 6 = 126 / 2187 = 14 / 243 = slightly more than 5,76% of not having 2 sets with 7 cards

[Rodion]

Nevermind, Geger had it earlier.

With 7 cards we don't get 2 sets, if we have 5&2 combos (example 5R-2G, total 6 combo-types). Also the odd must be : 6*(7!/5!2!)/3^7 = 14/3^5 = 5,76%

[DiM]

Courtesy of Fruit's nifty spreadsheet:

where's the spreadsheet?

[nippersean]

There are 10 types of people in the world. Those that understand binary, and those who don't.

[IcePack]

Lol!

[DiM]

http://www.conquerclub.com/forum/viewtopic.php?f=6&t=108862#p2444088
(Game-type neutral percentages)

Courtesy of Fruit's nifty spreadsheet:

Total average weighted win rate % 44.21%
Your average weighted win rate % 49.80%

5.6% difference

approx

me: