Assume VTES deck as follows:
LIBRARY (90 cards):
8x CardA (Master)
8x CardB (Master)
8x CardC (Master)
66x CardD (Library)
If I calculated right, I think the chances of having one copy of CardA in my
opening 7 cards is 49.12%. If not, please correct my math.
Q1: Is it correct to say that the chances of having one copy of CardA *and*
one copy of CardB *and* one copy of CardC in my opening 7 cards is 49.12%
(or whatever is the correct %)? [Yes/No]
Q2: If not, what *are* the chances of having exactly one copy of CardA *and*
one copy of CardB *and* one copy of CardC in my opening 7 cards?
BONUS
---------
Assume that, to achieve *maximum* coolness:
- CardA must be played as the 1st round Master
- CardB must be played as the 2nd round Master
- CardC must be played as the 3rd round Master
...and assume also that:
- I don't want less than 64 Computer Hacks
- I want to keep a 90-card deck
...then give me the decklist for these four cards which follows these
guidelines, and at the same time gives me the best odds of being able to
achieve the properly ordered play of the three Master cards listed. Thanks
to all you cool Math geeks that help me out, here.
Kevin M., Prince of Henderson, NV (USA)
"Know your enemy, and know yourself; in one-thousand battles
you shall never be in peril." -- Sun Tzu, *The Art of War*
"Contentment... Complacency... Catastrophe!" -- Joseph Chevalier
Kevin M. expounded:
> Assume VTES deck as follows:
>
> LIBRARY (90 cards):
> 8x CardA (Master)
> 8x CardB (Master)
> 8x CardC (Master)
> 66x CardD (Library)
>
> If I calculated right, I think the chances of having one copy of CardA in my
> opening 7 cards is 49.12%. If not, please correct my math.
Correct (sort of)
That is the probability of having *AT LEAST* one copy of card A.
> Q1: Is it correct to say that the chances of having one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards is 49.12%
> (or whatever is the correct %)? [Yes/No]
NO
The probability of having A is 49% and prob(B)=49% and prob(C)-49%.
There is a difference.
> Q2: If not, what *are* the chances of having exactly one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards?
Rough estimate: P(A and B and C)=P(A)*P(B)*P(C)= 0.49^3 = 11.85%
That assumes that the events are independent which they aren't so you
chances are actually better than that but I don't have time to work it
out properly right now.
> BONUS
> ---------
> Assume that, to achieve *maximum* coolness:
> - CardA must be played as the 1st round Master
> - CardB must be played as the 2nd round Master
> - CardC must be played as the 3rd round Master
>
> ...and assume also that:
> - I don't want less than 64 Computer Hacks
> - I want to keep a 90-card deck
>
> ...then give me the decklist for these four cards which follows these
> guidelines, and at the same time gives me the best odds of being able to
> achieve the properly ordered play of the three Master cards listed. Thanks
> to all you cool Math geeks that help me out, here.
9A 8B 7C is probably the best but that is a guess -888 might actually be
better. I presume none of them is meant to be a Parthenon?
You forgot to mention how much you want to get more of the same later in
the game -are they equally valuable after the third turn?
If you don't need more than one of each you are probably better off with
a smaller deck -you are unlikely to actually use 64 Hacks in a game. A
smaller deck would certainly improve the odds of getting the Masters you
want without jamming on them.
I'll probably work out the real figures tomorrow if someone else hasn't
by then :)
Kevin M. wrote:
> Assume VTES deck as follows:
>
> LIBRARY (90 cards):
> 8x CardA (Master)
> 8x CardB (Master)
> 8x CardC (Master)> 66x CardD (Library [Computer Hacking])>
> If I calculated right, I think the chances of having one copy of CardA in my
> opening 7 cards is 49.12%. If not, please correct my math.
1 - (82*81*80*79*78*77*76)/(90*89*88*87*86*85*84) = 49.12% of having at least
one CardA in opening hand.
> Q1: Is it correct to say that the chances of having one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards is 49.12%
> (or whatever is the correct %)? [Yes/No]
>
> Q2: If not, what *are* the chances of having exactly one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards?
Exactly one of each, no more, no less?
Odds of having CardA, CardB, CardC, and 4 CardD in that exact order:
A = (8/90)(8/89)(8/88)(66/87)(65/86)(64/85)(63/85) = 0.0232%
Since order doesn't matter, but the last four are indistinguishable,
the number of possible arrangements of this set of seven cards is:
B = 7!/4! = 210
So the odds are A*B = 4.87%
> BONUS
> ---------
> Assume that, to achieve *maximum* coolness:
> - CardA must be played as the 1st round Master
> - CardB must be played as the 2nd round Master
> - CardC must be played as the 3rd round Master
>
> ...and assume also that:
> - I don't want less than 64 Computer Hacks
> - I want to keep a 90-card deck
>
> ...then give me the decklist for these four cards which follows these
> guidelines, and at the same time gives me the best odds of being able to
> achieve the properly ordered play of the three Master cards listed. Thanks
> to all you cool Math geeks that help me out, here.
I may get back to you on this one...
--
LSJ (vte...@white-wolf.com) V:TES Net.Rep for White Wolf, Inc.
Links to V:TES news, rules, cards, utilities, and tournament calendar:
http://www.white-wolf.com/vtes/
"Kevin M." <you...@imaspammer.org> wrote in message news:<L%nsb.15598$Cj1.2685@fed1read07>...
> Assume VTES deck as follows:
>
> LIBRARY (90 cards):
> 8x CardA (Master)
> 8x CardB (Master)
> 8x CardC (Master)
> 66x CardD (Library)
>
> If I calculated right, I think the chances of having one copy of CardA in my
> opening 7 cards is 49.12%. If not, please correct my math.
Almost correct. The chance of having *at least* one copy of Card A in
the starting hand is 49,12%
>
> Q1: Is it correct to say that the chances of having one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards is 49.12%
> (or whatever is the correct %)? [Yes/No]
>
No
> Q2: If not, what *are* the chances of having exactly one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards?
>
Sorry. Too lazy for that one. Send me 50 € and i'd do it for you ;)
> BONUS
> ---------
> Assume that, to achieve *maximum* coolness:
> - CardA must be played as the 1st round Master
> - CardB must be played as the 2nd round Master
> - CardC must be played as the 3rd round Master
>
> ...and assume also that:
> - I don't want less than 64 Computer Hacks
> - I want to keep a 90-card deck
>
> ...then give me the decklist for these four cards which follows these
> guidelines, and at the same time gives me the best odds of being able to
> achieve the properly ordered play of the three Master cards listed. Thanks
> to all you cool Math geeks that help me out, here.
10xA
8xB
8xC
64xD
Reason:
Card A seems to be the most crucial to me. After playing it, you can
draw an additional card, and discard on your turn too if need be. That
makes Card B significantly easier to achieve than Card A. Thus i'd
prefer 10-8-8 to 9-9-8
On Wed, 12 Nov 2003 08:15:48 -0500, LSJ <vte...@white-wolf.com>
scrawled:
>> ...then give me the decklist for these four cards which follows these
>> guidelines, and at the same time gives me the best odds of being able to
>> achieve the properly ordered play of the three Master cards listed. Thanks
>> to all you cool Math geeks that help me out, here.>
>I may get back to you on this one...
should be easy enough to do in excel. jump into excel and type in the
formulas lsj used, except instead of '8', reference some cells for the
count of cardA, cardB, etc. then you can use excel's solvy thing to
find the maximum value the output cell can take by changing the
variables in the cardslot cells. then you'd probably have to round
your cards to whole numbers. :)
if it wasn't 2am here i'd have a stab at it myself...
well, that way would lie the max prob of having all three in your
opening hand, which should make a good approximation. due to the
shifting probs on non-replaced cards, finding the true prob will be
tricky.
now that i think of it, the max prob of having all three in your
opening hand would probably (heh) arise from having a library of
30/30/30 of them, with no computer hacks....which doesn't help.
then, there is also the moral issue of helping someone work on a deck
with 60+ computer hacks. it sounds awfully sleazy to me. :)
salem
domain:canberra http://www.geocities.com/salem_christ.geo/vtes.htm
On Wed, 12 Nov 2003 02:32:48 -0800, "Kevin M."
<you...@imaspammer.org> scrawled:
>BONUS
>---------
>Assume that, to achieve *maximum* coolness:
>- CardA must be played as the 1st round Master
>- CardB must be played as the 2nd round Master
>- CardC must be played as the 3rd round Master
>
>...and assume also that:
>- I don't want less than 64 Computer Hacks
>- I want to keep a 90-card deck
>
>...then give me the decklist for these four cards which follows these
>guidelines, and at the same time gives me the best odds of being able to
>achieve the properly ordered play of the three Master cards listed. Thanks
>to all you cool Math geeks that help me out, here.
if the masters are the only card you're playing each turn, it
shouldn't be too hard to do, or at least approximate. i love
approximating.
the problem then becomes:
maximise prob of getting at least 1 A in opening hand
max prob of getting at least 1 B in opening hand _or next card_
max prob of getting at least 1 C in opening hand or next card or next.
you want to do all these things simultaneously. which involves
trickyness, as the events are not independant.
hmmmm....
[ quoted text not captured ]
Kevin M. wrote:
> Assume VTES deck as follows:
>
> LIBRARY (90 cards):
> 8x CardA (Master)
> 8x CardB (Master)
> 8x CardC (Master)
> 66x CardD (Library)
>
> If I calculated right, I think the chances of having one copy of CardA in my
> opening 7 cards is 49.12%. If not, please correct my math.
49.12% is your chance of having at least one CardA in your hand. Your
chance of having exactly one copy is 37.49%.
(That is, (8/90) (82/89) (81/88) (80/87) (79/86) (78/85) (77/84) * (7!/6!))
> Q1: Is it correct to say that the chances of having one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards is 49.12%
> (or whatever is the correct %)? [Yes/No]
No. Your chance of having one of each would be much, much lower,
because the probabilities are going to tend to multiplicative (eg,
rolling a 1 on a ten sided die is 1/10. Rolling a 1 twice is 1/100).
> Q2: If not, what *are* the chances of having exactly one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards?
LSJ's answer looks right to me (I get 4.94% rather than 4.87%, but close
enough for government work. Could just be precision on my calculator.)
> BONUS
> ---------
> Assume that, to achieve *maximum* coolness:
> - CardA must be played as the 1st round Master
> - CardB must be played as the 2nd round Master
> - CardC must be played as the 3rd round Master
>> ....and assume also that:> - I don't want less than 64 Computer Hacks
> - I want to keep a 90-card deck
>> ....then give me the decklist for these four cards which follows these> guidelines, and at the same time gives me the best odds of being able to
> achieve the properly ordered play of the three Master cards listed. Thanks
> to all you cool Math geeks that help me out, here.
Just to theorize here...I don't think it matters, much. You can't put
in any more of the cards (to increase the chances of drawing at least
one), and from your description, it's impossible to say if any of them
are more or less important, so the assumption would need to be that you
want to draw into all of them equally.
My non-exhaustive playing around indicates that you should switch two
CardCs for two CardAs; this is because you must get a CardA in your
initial hand, whereas you have up to 4 + number of minions you influence
out first turn draws to get a CardC (two master phases, two discards,
and the number of times you can computer hack on turn 2).
(I played around with lowering the number of CardBs, but if you don't
get one in your opening hand, you only get two draws to get them -- one
master and one discard -- and that's a low enough chance already).
So I'd recommend 10 A, 8 B, 6 C.
Your chance of having at least one in your initial draw is:
A 57.48%
B 49.12%
C 39.38%
(These percentages are independant of eachother -- if you draw an A,
your chance of getting a B or C declines slightly. But I'm too lazy to
work it out.)
Your chance of getting at least one B in the next two draws, assuming
that you did not get one in your opening hand is: 18.45%
Your chance of getting at least one C in the next eight draws (assuming
four minions brought out on turn 1 that hack on turn 2) is: 46.65%.
Further manipulation of the A, B, and C cards modify the chances, but
it's pretty close to even money at this point that you'll have any given
card, which is about the best I could make it. I still see the opening
A as the weak point, but changing a B to an A makes B the weak point,
and changing a C to an A gives you a 61.20% of an opening A (+4%), but
drops your chances of getting a C in eight draws to 40.56% (-6%).
--Colin McGuigan
Colin McGuigan wrote:
> So I'd recommend 10 A, 8 B, 6 C.
Didn't see that the limit on computer hacks was 64 (assumed 66). In
that case, to ensure maximum A, B, and Cness, you should definitely add
two more. I would do 11 A, 9 B, and 6 C.
--Colin McGuigan
salem wrote:
> should be easy enough to do in excel. jump into excel and type in the
> formulas lsj used, except instead of '8', reference some cells for the
> count of cardA, cardB, etc. then you can use excel's solvy thing to
> find the maximum value the output cell can take by changing the
> variables in the cardslot cells. then you'd probably have to round
> your cards to whole numbers. :)
> if it wasn't 2am here i'd have a stab at it myself...
> well, that way would lie the max prob of having all three in your
> opening hand, which should make a good approximation. due to the
> shifting probs on non-replaced cards, finding the true prob will be
> tricky.
Because I can always use a brief mental exercise, I put together a page
to calculate the probabilities of getting at least one card in an
opening hand, or after a specified number of draws, given various
parameters.
http://64.81.143.195/deckprob.php
--Colin McGuigan
"Kevin M." <you...@imaspammer.org> wrote in message news:<L%nsb.15598$Cj1.2685@fed1read07>...
> Assume VTES deck as follows:
>
> LIBRARY (90 cards):
> 8x CardA (Master)
> 8x CardB (Master)
> 8x CardC (Master)
> 66x CardD (Library)
>
> If I calculated right, I think the chances of having one copy of CardA in my
> opening 7 cards is 49.12%. If not, please correct my math.
that's the probability of having at least 1 copy of card A
the probability of having exactly one copy is 37.5%
> Q1: Is it correct to say that the chances of having one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards is 49.12%
> (or whatever is the correct %)? [Yes/No]
no, i would say without thinking really is around 5% (37,5%^3) but i
could be totally wrong,
>
> Q2: If not, what *are* the chances of having exactly one copy of CardA *and*
> one copy of CardB *and* one copy of CardC in my opening 7 cards?
>
> BONUS
> ---------
> Assume that, to achieve *maximum* coolness:
> - CardA must be played as the 1st round Master
> - CardB must be played as the 2nd round Master
> - CardC must be played as the 3rd round Master
>
> ...and assume also that:
> - I don't want less than 64 Computer Hacks
> - I want to keep a 90-card deck
>
> ...then give me the decklist for these four cards which follows these
> guidelines, and at the same time gives me the best odds of being able to
> achieve the properly ordered play of the three Master cards listed. Thanks> to all you cool Math geeks that help me out, h.............
Andrea
Kevin M. wrote:
> BONUS
> ---------
> Assume that, to achieve *maximum* coolness:
> - CardA must be played as the 1st round Master
> - CardB must be played as the 2nd round Master
> - CardC must be played as the 3rd round Master
>
> ...and assume also that:
> - I don't want less than 64 Computer Hacks
> - I want to keep a 90-card deck
>
> ...then give me the decklist for these four cards which follows these
> guidelines, and at the same time gives me the best odds of being able to
> achieve the properly ordered play of the three Master cards listed. Thanks
> to all you cool Math geeks that help me out, here.
9x CardA
9x CardB
8x CardC
64x CardD (Computer Hacking)
Gives you the best chance at having A in your first 7, B in your first 9 and
C in your first 11 (this takes into account cards played as well as discarded
if necessary - you must discard on your first turn if you are missing either
B or C (or both) and must discard again on your second turn if you are still
missing C). (That chance is 22.37%, BTW).
If you'd like to assume that you'll be able to play CardD on your second turn
(with whatever weenie you brought out), then the conditions on C relax to
"first 12 cards". The best deck is still the same deck, but the odds are now
23.51%
With 2 weenies each able to play CardD on your second turn (say, if CardA is
Information Highway or you just want to depend on not going first), switch to
10x CardA
9x CardB
7x CardC
and you get a 24.58% chance of pulling it off, as opposed to 24.56% with the
9,9,8 set up. (But if you only get 1 weenie out, 10,9,7 drops your odds to
23.43% from the 9,9,8 set up.)
[ quoted text not captured ]
LSJ <vte...@white-wolf.com> wrote in message news:<3FB23284...@white-wolf.com>...
[ quoted text not captured ]
Given that the odds of getting A,B,C,D,D,D,D in the first hand is 4%,
the odds of your getting them in time to do a properly ordered play is
going to be hard to get above 10%, I'd imagine.
Even allowing for three turns of discarding duplicates of A-C, your
odds still aren't going to rise far enough to make it likely to happen
that you'll play A, B, C in order without swamping your hand and deck
further with Master cards.
And since I can estimate that your odds are poor, I'm not going to
bother to do the actual math and decklist, since it won't work often
enough to make it worth it. :)
-John Flournoy
LSJ wrote:
> If you'd like to assume that you'll be able to play CardD on your second turn
[...]
> With 2 weenies each able to play CardD on your second turn
[...]
Note that this doesn't take into account the (relatively small) chance that
you won't have any CardD in hand on your second turn.
[ quoted text not captured ]
"Kevin M." <you...@imaspammer.org> wrote in message news:<L%nsb.15598$Cj1.2685@fed1read07>...
> Assume VTES deck as follows:
>
> LIBRARY (90 cards):
> 8x CardA (Master)
> 8x CardB (Master)
> 8x CardC (Master)
> 66x CardD (Library)
>
> If I calculated right, I think the chances of having one copy of CardA in my
> opening 7 cards is 49.12%. If not, please correct my math.
Wouldn't this be better (cross)posted to sci.math.num-analysis or some
other group? Nobody here claims to be an expert in probability, and I
believe you are looking for the CORRECT statistics. Just a thought.
> Wouldn't this be better (cross)posted to sci.math.num-analysis or some
> other group? Nobody here claims to be an expert in probability, and I
I am an expert in probability,
David Oros
[ quoted text not captured ]
LSJ wrote:
> Kevin M. wrote:
>>> BONUS
>> ---------
>> Assume that, to achieve *maximum* coolness:
>> - CardA must be played as the 1st round Master
>> - CardB must be played as the 2nd round Master
>> - CardC must be played as the 3rd round Master
>>
>> ...and assume also that:
>> - I don't want less than 64 Computer Hacks
>> - I want to keep a 90-card deck
>>
>> ...then give me the decklist for these four cards which follows these
>> guidelines, and at the same time gives me the best odds of being able to
>> achieve the properly ordered play of the three Master cards listed.
>> Thanks
>> to all you cool Math geeks that help me out, here.>
>
> 9x CardA
> 9x CardB
> 8x CardC
> 64x CardD (Computer Hacking)
Mmm. My figures are incorrect. I was calculating the chances of
getting any of A in 7, B in 9, C in 11 (rather than getting all).
So the "optimal ratio" is incorrect as well.
Correct numbers to follow (sometime tomorrow, probably).
[ quoted text not captured ]
On Wed, 12 Nov 2003 17:36:46 -0500, "Metropolis" <do...@umich.edu>
scrawled:
>
>>> Wouldn't this be better (cross)posted to sci.math.num-analysis or some
>> other group? Nobody here claims to be an expert in probability, and I>I am an expert in probability,
>
>David Oros
i think lots of people here are. :) i did some degree thingy at
university that involved a lot of probability. mostly probability of
people dying or getting sick, but probability none the less.
and all these things we're doing here are really just extensions of
the 'what are my chances of a royal flush' type problem you get in any
simple statistics course.
[ quoted text not captured ]
"Kevin M." wrote:
> BONUS
> ---------
> Assume that, to achieve *maximum* coolness:
> - CardA must be played as the 1st round Master
> - CardB must be played as the 2nd round Master
> - CardC must be played as the 3rd round Master
>
> ...and assume also that:
> - I don't want less than 64 Computer Hacks
> - I want to keep a 90-card deck
>
> ...then give me the decklist for these four cards which follows these
> guidelines, and at the same time gives me the best odds of being able to
> achieve the properly ordered play of the three Master cards listed.
>
Does it matter if Card A is a Dreams or Capuchin or Storage Annex or Barrens or
Fragment? I especially like Dreams here as Card A improving your chances a
lot. Play it, tap it, gain 2 cards. Next turn during your untap, tap it again
and gain 2 more cards. You will have gone through 21 cards by the time turn 3's
master phase comes up (I think - assuming you are playing 1 cap weenies and they
all computer hack each turn, and none die). While you are laying out a cut and
dried math problem, if you add in the cards that cause you to draw like crazy,
you change the probabilities. Is that part of the math problem? You say "Give
me the decklist" I would definitely put Dreams in as Card A. Capuchino and
Fraggy are best bets for 2 and 3, I think.
chris
--
chris shorb
<www.vtesinla.org> (A V:TES site)
prince of torrance, california
***
Into the abyss I'll fall - the eye of Horus
Into the eyes of the night - watching me go
Green is the cat's eye that glows - in this temple
Enter the risen Osiris - risen again
- Dickinson
Chris Shorb <chr...@vtesinla.org> wrote:
> "Kevin M." wrote:
>>> BONUS
>> ---------
>> Assume that, to achieve *maximum* coolness:
>> - CardA must be played as the 1st round Master
>> - CardB must be played as the 2nd round Master
>> - CardC must be played as the 3rd round Master
>>
>> ...and assume also that:
>> - I don't want less than 64 Computer Hacks
>> - I want to keep a 90-card deck
>>
>> ...then give me the decklist for these four cards which follows these
>> guidelines, and at the same time gives me the best odds of being
>> able to achieve the properly ordered play of the three Master cards
>> listed.
>>> Does it matter if Card A is a Dreams or Capuchin or Storage Annex or
> Barrens or Fragment?
Yes. CardC is a "drawing"-type card, but is only useful in my example if it
is drawn *after* CardA and CardB.
Both CardA or CardB are just plain-old Master cards (and must remain so).
> chris shorb
> <www.vtesinla.org> (A V:TES site)
> prince of torrance, california
Kevin M., Prince of Henderson, NV (USA)
[ quoted text not captured ]
>
> Yes. CardC is a "drawing"-type card, but is only useful in my example if it
> is drawn *after* CardA and CardB.
>
> Both CardA or CardB are just plain-old Master cards (and must remain so).
>
CardA = Information Highway?
CardB = Pentex(tm) Loves You?
CardC = Dreams of the Sphinx?
LSJ <vte...@white-wolf.com> wrote in message news:<3FB2CF7F...@white-wolf.com>...
> LSJ wrote:
> > Kevin M. wrote:
> >
> >> BONUS
> >> ---------
> >> Assume that, to achieve *maximum* coolness:
> >> - CardA must be played as the 1st round Master
> >> - CardB must be played as the 2nd round Master
> >> - CardC must be played as the 3rd round Master
> >>
> >> ...and assume also that:
> >> - I don't want less than 64 Computer Hacks
> >> - I want to keep a 90-card deck
> >>
> >> ...then give me the decklist for these four cards which follows these
> >> guidelines, and at the same time gives me the best odds of being able to
> >> achieve the properly ordered play of the three Master cards listed.
> >> Thanks
> >> to all you cool Math geeks that help me out, here.
> >
> >
> > 9x CardA
> > 9x CardB
> > 8x CardC
> > 64x CardD (Computer Hacking)
The above are of course the right numbers.
The chance of success are :
53.46 % of being able to play card A
32.17 % of dropping card B on the table when you need to
19.84 % of going all the way (without playing card D)
21.08 % of scaring your oponents with card C (if one card D played on turn 2)
22.22 % of making enemies with card C (card D played twice)
(chances of not getting any D have been neglected)
There is a 11.05 % chance of getting at least on of each (ABC) in the opening hand.
Graaf Tel
VTES2004 expounded:
> Wouldn't this be better (cross)posted to sci.math.num-analysis or some
> other group? Nobody here claims to be an expert in probability, and I
> believe you are looking for the CORRECT statistics. Just a thought.
I'm certainly expert enough for this little question -it would have been
rather a hassle though and it seems to have been answered well enough
now.
blacksh...@hotmail.com (Orestes) wrote in message news:<c25ded38.03111...@posting.google.com>...
[ quoted text not captured ]
I am guessing CardA is
Jake Washington
for those Turn 1 bleeds
But that still leaves me clueless as to the next 2 masters.
-johnmeier1
"johnmeier1" <johnm...@catholic.org> wrote in message
news:2b593401.0311...@posting.google.com...
> blacksh...@hotmail.com (Orestes) wrote in message
news:<c25ded38.03111...@posting.google.com>...
> > >
> > > Yes. CardC is a "drawing"-type card, but is only useful in my example
if it
> > > is drawn *after* CardA and CardB.
> > >
> > > Both CardA or CardB are just plain-old Master cards (and must remain
so).
> > >
> >
> > CardA = Information Highway?
> > CardB = Pentex(tm) Loves You?
> > CardC = Dreams of the Sphinx?
>
> I am guessing CardA is
>
> Jake Washington
>
> for those Turn 1 bleeds
NOt only that, but because he would have 8 of them he could keep playing
them when they got killed/exploded themselves. Extra minion who can computer
hack.
>
> But that still leaves me clueless as to the next 2 masters.
>
Yeah I have no idea either.
Colin Riggs
[ quoted text not captured ]
Thanks.
But I get 19% for having at least 1 each in opening hand with 9/9/8 in the deck.
Is there a problem in my formulation?
(7!/4!) * (9/90)*(9/89)*(8/88)
[ quoted text not captured ]
[ quoted text not captured ]
CardA = a non-"drawing", non-"Master phase-duplicating" Master card. This
card must be played before CardB.
CardB = a non-"drawing", non-"Master phase-duplicating" Master card. This
card cannot be played more than once a game, and must be played after CardA
and before CardC
CardC = Info Highway
For maximum results, I'd prefer to play CardA on turn 1, CardB on turn 2,
and CardC on turn 3. For slightly less than maximum results, you could just
ignore CardC (and turn 3) entirely, but that lessens the power of the deck
by about 25%.
[ quoted text not captured ]
On Thu, 13 Nov 2003 13:59:28 -0500, LSJ <vte...@white-wolf.com>
scrawled:
>> There is a 11.05 % chance of getting at least on of each (ABC) in the opening hand.>
>Thanks.
>But I get 19% for having at least 1 each in opening hand with 9/9/8 in the deck.
>
>Is there a problem in my formulation?
>
>(7!/4!) * (9/90)*(9/89)*(8/88)
maybe i don't remember my classes correctly, but it looks like when
there are different numbers of each card, the above formula is not
exactly right.
if you get C as the first draw, it will be 8/90, which is different to
getting A or B on first draw, which are each 9/90. so just multiplying
(9/90)*(9/89)*(8/88) by the number of different ways you could get it
in 7 cards would me a little off.
[ quoted text not captured ]
salem wrote:
> On Thu, 13 Nov 2003 13:59:28 -0500, LSJ <vte...@white-wolf.com>
> scrawled:
>
>>>>There is a 11.05 % chance of getting at least on of each (ABC) in the opening hand.>>
>>Thanks.
>>But I get 19% for having at least 1 each in opening hand with 9/9/8 in the deck.
>>
>>Is there a problem in my formulation?
>>
>>(7!/4!) * (9/90)*(9/89)*(8/88)>
> maybe i don't remember my classes correctly, but it looks like when
> there are different numbers of each card, the above formula is not
> exactly right.
The formula takes into account the different numbers of each card.
a=9
b=9
c=8
hand=7
deck=90
(hand!/(hand-3)!)*(a/deck)*(b/(deck-1))*(c/(deck-2))
> if you get C as the first draw, it will be 8/90, which is different to
> getting A or B on first draw, which are each 9/90. so just multiplying
> (9/90)*(9/89)*(8/88) by the number of different ways you could get it
> in 7 cards would me a little off.
The formula computes the odds of getting A as the first draw, B as the
second, and C as the third, and then any four as the next four (which
is a probability of 1), and then accounts for the fact that the order of
those seven doesn't matter by multiplying by the number of distinct
orderings of (A,B,C,X,X,X,X) (the number of which is 7!/4!).
[ quoted text not captured ]
LSJ <vte...@white-wolf.com> wrote in message news:<3FB3D490...@white-wolf.com>...
[ quoted text not captured ]
Intuition tels me:
Lets say in your formulation you have ABCXXXX (where X can be ABCD).
You would count ABCXXXX and XBCAXXX as two separate occurrances.
When the first X is A however these would be the same.
I'll give reason a chance while i take a bath (hey, it worked for Archimedes)
Graaf Tel
"Eureka" greek word meaning "this water is too hot"
Graaf Tel wrote:
> LSJ <vte...@white-wolf.com> wrote in message news:<3FB3D490...@white-wolf.com>...>>Graaf Tel wrote:>>>There is a 11.05 % chance of getting at least on of each (ABC) in the opening hand.>>
>>Thanks.
>>But I get 19% for having at least 1 each in opening hand with 9/9/8 in the deck.
>>
>>Is there a problem in my formulation?
>>
>>(7!/4!) * (9/90)*(9/89)*(8/88)>
> Intuition tels me:
> Lets say in your formulation you have ABCXXXX (where X can be ABCD).
> You would count ABCXXXX and XBCAXXX as two separate occurrances.
> When the first X is A however these would be the same.
Rats. So what's the correct formulation that leads to 11.05%?
I thought about taking (1-na)*(1-nb)*(1-nc), but dismissed that as
counting too high as well (allowing the odds of 7 A's to coincide
with the odds of 7 C's - the events are not independent).
(Where na is the probability of drawing no A's, and so on).
[ quoted text not captured ]
Graaf Tel wrote:
>>>9x CardA
>>>9x CardB
>>>8x CardC
>>>64x CardD (Computer Hacking)>
> The above are of course the right numbers.
>
> The chance of success are :
> 53.46 % of being able to play card A
> 32.17 % of dropping card B on the table when you need to
> 19.84 % of going all the way (without playing card D)
> 21.08 % of scaring your oponents with card C (if one card D played on turn 2)
> 22.22 % of making enemies with card C (card D played twice)
> (chances of not getting any D have been neglected)
> There is a 11.05 % chance of getting at least on of each (ABC) in the opening hand.
Ah. Found the right formulation. At least, it matches your results above.
(Thanks to Hunter's enlightening email solution for the opening hand problem).
pseudo C-source:
a = number of A
b = number of B
c = number of C
nd = number of D played on round 2
ff(x,y) = falling factorial = x * (x-1) * ... * (x-y+1)
pa = ff(90-a, 7)/ff(90,7); /* prob of no A in 7 */
pb = ff(90-b, 9)/ff(90,9); /* prob of no B in 9 */
pc = ff(90-c, 11+nd)/ff(90,11+nd);
/* prob of no A and no B in 7 and of no B in the next 2 */
pab = (ff(90-a-b, 7)/ff(90,7))*(ff(90-7-b,2)/ff(90-7,2));
/* prob of no A and no C in 7 and of no C in the next 4+ */
pac = (ff(90-a-c, 7)/ff(90,7))*(ff(90-7-c,4+nd)/ff(90-7,4+nd));
/* prob of no B and no C in 9 and of no C in the next 2+ */
pbc = (ff(90-b-c, 9)/ff(90,9))*(ff(90-9-c,2+nd)/ff(90-9,2+nd));
/* prob of no A and no B and no C in 7, no B and no C in next 2, and
* still no C in next 2+
*/
pabc = (ff(90-a-b-c, 7)/ff(90,7))*(ff(90-7-b-c,2)/ff(90-7,2))*
(ff(90-9-c,2+nd)/ff(90-9,2+nd));
/* inclusion-exclusion to get odds of any one event (any one missing
* card), subtracted from 1 to get no card missing
*/
p = 1.0 - ((pa+pb+pc) - (pab+pac+pbc) + pabc);
[ quoted text not captured ]
LSJ <vte...@white-wolf.com> wrote in message news:<3FB4DF56...@white-wolf.com>...
> Graaf Tel wrote:
> > LSJ <vte...@white-wolf.com> wrote in message news:<3FB3D490...@white-wolf.com>...
> >>Graaf Tel wrote:
> >>>There is a 11.05 % chance of getting at least on of each (ABC) in the opening hand.
> >>
> >>Thanks.
> >>But I get 19% for having at least 1 each in opening hand with 9/9/8 in the deck.
> >>
> >>Is there a problem in my formulation?
> >>
> >>(7!/4!) * (9/90)*(9/89)*(8/88)
> >
> > Intuition tels me:
> > Lets say in your formulation you have ABCXXXX (where X can be ABCD).
> > You would count ABCXXXX and XBCAXXX as two separate occurrances.
> > When the first X is A however these would be the same.
>
> Rats. So what's the correct formulation that leads to 11.05%?
The formulation i used is a recursive one.
P(XYZ,n) = the chance of having at least one of each of XYZ after n draws.
P(XY,n) = the chance of having at least one of XY but none of Z after n draws.
P(X,n) = the chance of having at least one of X but none of YZ after n draws.
P(-,n) = the chance of having none of XYZ after n draws.
a,b,c = number of card A,B,C in the deck
P(ABC,n+1) = P(ABC,n) + P(BC,n)*a/(90-n)
+ P(AC,n)*b/(90-n)
+ P(AB,n)*c/(90-n)
P(AB,n+1) = P(AB,n)*(90-n-c)/(90-n) + P(A,n)*b/(90-n)
+ P(B,n)*a/(90-n)
etc.
P(A,n+1) = P(A,n)*(90-n-b-c)/(90-n) + P(-,n)*a/(90-n)
etc
P(-,n+1) = P(-,n)*(90-n-a-b-c)/(90-n)
P(A,1)=a/90
P(B,1)=b/90
P(C,1)=c/90
P(-,1)=(90-a-b-c)/90
P(BC,1)=P(AC,1)=P(AB,1)=P(ABC,1)=0
As mentioned before in this thread, it is quite easily done in Excel.
Graaf Tel