CARO'S $100 CHALLENGE #1
Mike Caro
Everyone -
I'm offering $100 reward for the first person who posts the right
answer to RGP by October 31, 1997 (might take much less time, though,
knowing you guys). To qualify, key members of this group must agree
that it is the first right answer and instruct me to whom and where to
send the check. In the event of a dispute, I am the sole decider about
who (or whether anyone) should be paid, but if I stiff you for $100,
you'll have a story to tell.
The problem under discussion (see previous posts by Barbara Yoon, Erik
Reuter, and me): How many discrete starting hand matchups are there in
a heads-up hold 'em game? In order to understand the problem better,
you need to read the itemized list in the last part of this message.
There are many good reasons for players to participate in this forum.
Wars of expertise are one reason, and we can all learn from those.
However, just so you know, my personal reason for coming here is
simply to spread affection and knowledge and to share affection and
knowledge. We have a common interest and we're on the same team
(unless we're at the same table).
I will seldom be combative, so the following is provided only to
further the exploration into a particular statistical aspect of poker,
for those few who are interested. The truth usually comes out and
benefits everyone, and those who get answers right and those who get
answers wrong are often the same people at different stages of
investigation. It is the process of posing the wrong solution and then
attacking it ourselves that sometimes makes us succeed. Everyone who
contributes intelligently to the quest, shares in the reward. Cryptic
words, I know. Sorry. Anyway, I appreciate the intelligent
contributions of those on RGP, and here are some starting ways of
thinking about the problem we're discussing.
(BTW, my original post was about setting up a matrix of 169 x 169 with
extra categories for various ways in which the ranks and suits
interacted. In that array, Hand A vs. Hand B was different from Hand B
vs. Hand A, even if the hands were simply reversed, and I said that
there were more than 50,000 distinctly different categories of
matchups. Then I said that you really didn't need to calculate the
results of duplicate matchups, even though the table we were building
would list the same answer twice. (I'm losing you, I know. Read the
earlier posts, if you're interested.) The debate now seems to have
turned to whether the number of matchups exceeds 50,000 IF you
consider, for example, Kc Qc vs. 6c 5c to be the same as 6c 5c vs. Kc
Qc. I know the answer, but I'm not saying yet.)
I'm not going to take the time to carefully edit the list below, or
even to check the numbers, so feel free to point out anything I may
have omitted or miscalculated. This is not to be considered a proofed
and published work. In fact, I've deliberately left out at least one
category and deliberately miscalculated at least one figure, so you'll
have to work for the $100…
Pair vs. same Pair (example: Ac Ad vs. Ah As) (always 0.5 win rate) =
13 (13 total).
Pair vs. other pair, both suits matched (example: Kc Kd vs. 6c 6d)
((13 x 12) / 2 orders of consideration) = 78 (91 total).
Pair vs. other pair, one suit matched (example: Kc Kd vs. 6d 6h) = 78
(169 total).
Pair vs. other pair, no suits matched (example: Kc Kd vs. 6h 6s) = 78
(247 total).
Pair vs. two suited cards that include one rank matching the pair
(example: Kc Kd vs. Kh 5h) (13 pairs x 12 other ranks; we do NOT
divide by 2, in THIS specific case) = 156 (403 total).
Pair vs. two suited cards that match a suit contained in the pair
(example: Kc Kd vs. 6c 5c) (13 x (12 x 11) / 2)) = 858 (1,261 total).
Pair vs. two suited cards of different ranks from the pair that do not
match a suit contained in the pair. (example: Kc Kd vs. 6h 5h) = 858
(2,119 total).
Pair vs. two unsuited cards that include one rank matching the pair
and one suit contained in the pair (example: Kc Kd vs. Kh 5c) = 156
(2,275 total).
Pair vs. two unsuited cards of different ranks from the pair that
match both suits contained in the pair (example: Kc Kd vs. 6c 5d) =
858 (3,133 total).
Pair vs. two unsuited cards of different ranks from the pair that
match one suit contained in the pair (example: Kc Kd vs. 6c 5h) = 858
(3,991 total).
Pair vs. two unsuited cards of different ranks from the pair that
match no suit contained in the pair (example: Kc Kd vs. 6h 5s) = 858
(4,849 total).
That's the easy part. We've so far taken care of all matchups that
include a pair. Total so far is less than 5,000. Or is it? HINT: Could
there have been a 156-matchup category left out? Could other
categories have been omitted? Are some numbers or additions wrong?
I'll leave it open to y'all to finish the itemizing. If no one posts
the answer in a few weeks, I will. HINT: It WILL often matter WHICH
rank matches. Therefore, Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs.
6d 5c). Here's to get you started…
Two suited cards vs. two suited cards of the same suit (example: Kc Qc
vs. 6c 5c) (((13 x 12) / 2) x ((11 x 10) / 2)) = (4,290 / 2) = 2,145
(7,050 total).
Two suited cards vs. two suited cards of a different suit where both
ranks are the same (example: ((12 x 11) / 2) (example: Kc Qc vs. Kd
Qd) (always 0.5 win rate) = 55.
Two suited cards vs. two suited cards of a different suit where the
high rank…
Again, there are errors in math and omissions in the list above. You
have to do your own calculations and list the categories (similar to
the way I have) to get the $100. It won't be worth your time, in a
money sense, but you'll probably get some obscure spiritual reward.
Maybe not.
Straight Flushes,
Mike Caro
Barbara Yoon
Lee Jones wrote:
> Mike Caro, *he* say...$100 reward for the first person
> who posts the right answer...Heh...how fond r.g.p. is of
> side-bets, and in the spirit of the recent Nobel award to
> the guys that figured out how to price financial derivative
> instruments, I hereby lay the o/u line on the first correct
> answer at 0000 hours, Monday 10/20/97. But no, I'm
> not accepting any action...
Lee -- Good thing for you, "not accepting any action" -- my onion farm
would be on the "under"...
> ...haven't been following this discussion...but this disturbs me...
> ...seems that you're in a semantic argument here...results of
> Player #1 holding 6c-5c and Player #2 holding Kc-Qc are
> *identical* if the hands are swapped...results are swapped as
> well...maybe that's one calculation, maybe it's two...tickles me
> no end that one of our real luminaries is willing to put his money
> where his mouth is...
Lee -- The gist of it is that if you "know" (wins, losses, ties) for 6c 5c
v. Kc Qc, then you also "know" for Kc Qc v. 6c 5c (and for Kh Qh v. 6h 5h,
etc.) -- so the question is how many (the minimum number of) effectively
"different" match-ups do you have to "know" so that you "know" ALL possible
match-ups...OK?
Todd Mummert
my quick guess
54158
hmm..want to see my other 100k guesses?
Barbara Yoon
Mike Caro wrote:
> Barbara --
> I enjoy both your spirit AND your expertise. Please see my
> separate post under new thread, "Caro's $100 Challenge #1."
> Straight Flushes,
> Mike Caro
Thank you for the kind words... I've been a fan of yours through your
writing for many years... "Caro's $100 Challenge" -- and you sure do know
how to stimulate a discussion! However, let me declare myself "ineligible"
in this, though offering services as a "judge," if you want...
Those attempting this challenge might want to see earlier "heads-up
match-ups" thread (15/Oct) for further clarification...
Erik Reuter
I haven't seen a correct answer yet, so here's my (not carefully checked
entry) entry. Actually, I hope I've left out a few or have a mistake in
the math, because otherwise, Barbara wins bragging rights! (46,618)
numerals 1,2,3,4 stand for rank patterns
letters a,b,c,d stand for suit patterns
any 4 term series is assumed to be two sequential two card hands
C(x,y) = number of combinations of x things taken y at a time
[1111]
4 suits: abcd
#suitcombs = 1
subtotal = #rankcombs * #suitcombs = 13 * 1 = 13
3 suits: abac,abca,abbc,abcb,aabc,abcc
[1112]
4 suits: abcd
3 suits: abca,abcc
#suitcombs = 1 + 2 = 3
subtotal = 13 * 12 * 3 = 468
[1122]
4 suits: abcd
3 suits: abac
2 suits: abba
#suitcombs = 1 + 1 + 1 = 3
subtotal = C(13,2) * 3 = 234
[1213]
4 suits: abcd
3 suits: abca,abbc,abcb,aabc,abcc
2 suits: abba,aabb,aaba,abbb
#suitcombs = 1 + 5 + 4 = 10
subtotal = 13 * C(12,2) * 10 = 8,580
[1123]
4 suits: abcd
3 suits: abac,abca,abcc
2 suits: abab,abaa
#suitcombs = 1 + 3 + 2 = 6
subtotal = 13 * C(12,2) * 6 = 5,148
[1234]
4 suits: abcd
3 suits: abac,abca,abbc,abcb,aabc,abcc
2 suits: abba,abab,aabb,aaab,aaba,abaa,abbb
1 suit : aaaa
#suitcombs = 1 + 6 + 7 + 1 = 15
subtotal = ( C(13,2) * C(11,2) / 2 ) * 15 = 32,175
TOTAL = 13 + 468 + 234 + 6006 + 5148 + 32175
= 46,618
--
Erik Reuter, e-re...@uiuc.edu
Dylan Salisbury
Well, I'll go for speed if not accuracy... This was done in a hurry but
I kept realizing I forgot something and revised it. If I did a good job
I'll explain it later :)
By my math, I get interesting 35230 combinations.
I broke it down into 7 numeric categories, then broke each of those down
by suit options:
I. Pair vs same Pair
13
II. Pair vs other Pair
Numerically, 78 interesting combinations of ABCD.
(12 + 11 + 10 + ... + 1).
Suit: For each combination:
1 Rainbow
1 One suit matches between hands
1 Both suits match between hands
-
3 * 78 = 234
III. Non-pair vs same non-pair
Combinations of ABCD:
13 * 12 = 156.
Ways to suit it:
1 Rainbow
2 One suit matches between hands
1 Both suits match between hands
-
4 * 156 = 624
IV. Non-pair vs different non-pair, no cards match
Combinations of ABCD:
(13 * 12 / 2) * (11 * 10 / 2) / 2 = 2145
Ways to suit it:
1 Rainbow
1 All same suit
6 Ways for any two cards to be suited, the other two unsuited
3 Ways for any two cards to be suited, the other two suited
-
11 * 2145 = 23595
V. Non-pair vs different non-pair, one card matches
Combinations of ABCD:
13 * 12 * 11 / 2 = 858
Ways to suit it:
1 Rainbow
2 One hand suited
1 Both hands suited
1 cross-suited
2 half-cross-suited
-
7 * 858 = 6006
VI. Pair vs non-Pair that includes the pair
Interesting combinations of ABCD:
13 * 12 = 156
Ways to suit it:
1 Rainbow
1 Non-pair hand is suited
1 Odd card matches one of pair
-
3 * 156 = 468
VII. Pair vs non-Pair that doesn't include the pair
Interesting combinations of ABCD:
13 * 12 * 11 = 1716
Ways to suit it:
1 Rainbow
1 Non-pair suited, matches none of pair
1 Non-pair suited, matches one of pair
2 Non-pair unsuited, one non-pair suit matches one of pair suit
1 Non-pair unsuited, suits match both pair suits
-
6 * 1716 = 10296
--
Dylan Salisbury "What part of n=n++
San Francisco, California don't you understand?
PGP Fingerprint: 2CE5 AACC DF7F 694F 0147 BFD5 ECAE 4257
Lee Jones
In article <34452d2a...@nntp.ix.netcom.com>,
Mike Caro, *he* say...
>I'm offering $100 reward for the first person who posts the right
>answer to RGP by October 31, 1997 (might take much less time, though,
>knowing you guys).
Heh. Knowing out fond r.g.p. is of side-bets, and in the spirit of the
recent Nobel award to the guys that figured out how to price financial
derivative instruments, I hereby lay the o/u line on the first correct answer
at 0000 hours, Monday 10/20/97. But no, I'm not accepting any action on
that line; Quick has to do that, and he probably wants to lay his own line,
anyway.
>The debate now seems to have
>turned to whether the number of matchups exceeds 50,000 IF you
>consider, for example, Kc Qc vs. 6c 5c to be the same as 6c 5c vs. Kc
>Qc. I know the answer, but I'm not saying yet.)
I haven't been following this discussion (I let other people come up with
fast hold'em evaluators), but this disturbs me. Mike, it seems that you're
in a semantic argument here. We all know (I hope) that the results of
Player #1 holding 6c-5c and Player #2 holding Kc-Qc are *identical* if the
hands are swapped between players, though, of course, the results are swapped
as well. So if there's a big matrix, you can enter p(P1 wins)=x in the square
corresponding to ((P1: 6c-5c) && (P2: Kc-Qc)) *then* you can enter
p(P1 wins)=1-x in the ((P1: Kc-Qc) && (P2: 6c-5c)). So, maybe that's one
calculation, maybe it's two. As I said, this seems to be a semantic argument.
However, it just tickles me no end that one of our real luminaries is willing
to put his money where his mouth is; as I said, the correct answer is
less than a week away.
Regards, Lee
--
Lee Jones | "They come running just as fast they can -
le...@sgi.com | 'cause every girl's crazy for a sharp dressed man."
650-933-3356 | -Z.Z. Top
Erik Reuter
(An typo-line crept into my previous post, here's a repost to prevent confusion)
I haven't seen a correct answer yet, so here's my (not carefully checked
entry) entry. Actually, I hope I've left out a few or have a mistake in
the math, because otherwise, Barbara wins bragging rights! (46,618)
numerals 1,2,3,4 stand for rank patterns
letters a,b,c,d stand for suit patterns
any 4 term series is assumed to be two sequential two card hands
C(x,y) = number of combinations of x things taken y at a time
[1111]
4 suits: abcd
#suitcombs = 1
subtotal = #rankcombs * #suitcombs = 13 * 1 = 13
[1112]
Tony Oresteen
Mike,
My ISP is having trouble posting to news so it's very likly that this response
will not mate to the usenet.
I assume no Joker, bug, or wild cards. Just a 52 CARD deck.
The answer to your challenge is simple: 169*169 or 28,561. Here's why. In
Hold 'Em there are 169 different two card hands. Note that 4s7s is exacly the
same as 4d7d for a *STARTING* hand. You used the term *discrete* which is NOT
*unique*. Also, 2d2c is exactly the same as 2h2s. Thus player 1 can have any
one of 169 possible two card starting hands and the loss of two cards from the
deck does NOT reduce the number of TYPES of hand player 2 can get. If player 1
can get AA player 2 can still get AA.
The key here is that we are trying to count the number of CLASSIFICATIONS a
Hold'em hand will fall into, NOT the number PERMUTATIONS. The perutations
possible would be:
52*51*50*49 or 6,497,400 starting hands if order matters. Since order really
doesn't matter in Hold 'Em then you must divide by 4*3*2*1 or 24, yeilding
270,725.
But all that aside, what does it matter? Hold 'Em is played with 7 cards not
2. 27o can beat AA if the flop comes 777 with a turn card of 6 & river card of
J. Not likely but possible.
But let's say you want to know how week is 27o against AA. The only way to
really evaluate that is to run a Monte Carlo analysis, track the number of
times AA comes up against 27o and then play the cards to compleation. I would
guess you would to to run AT LEAST 80,000,000 RANDOM hands to get a large
enough population of 27o vs AA in order to get a statisticly valid sample size.
Great question though.
--
Tony
Remove the "z" in my address to send me e-mail.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For you spammers with automated email address catchers, here are the addresses
of the current FCC board members:
Chairman Reed Hundt: rhu...@fcc.gov
Commissioner James Quello: jqu...@fcc.gov
Commissioner Susan Ness: sn...@fcc.gov
Commissioner Rachelle Chong: rch...@fcc.gov
Others:
bi...@microsoft.com
root@localhost
[Idea shamelessly stolen from David Whitwell & Bob Sweeney]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Mike Caro <ca...@caro.com> wrote in article
<34452d2a...@NNTP.IX.netcom.com>...
> Everyone -
>
> I'm offering $100 reward for the first person who posts the right
> answer to RGP by October 31, 1997 (might take much less time, though,
> knowing you guys). To qualify, key members of this group must agree
> that it is the first right answer and instruct me to whom and where to
> send the check. In the event of a dispute, I am the sole decider about
> who (or whether anyone) should be paid, but if I stiff you for $100,
> you'll have a story to tell.
>
> The problem under discussion (see previous posts by Barbara Yoon, Erik
> Reuter, and me): How many discrete starting hand matchups are there in
^^^^^^^^
Mike Caro
Tony --
By the way, the reason your calculation of combinations didn't quite
work is that you calculated for a four-card hand out of a 52-card
deck. That means cards ABCD in any order of arrival. What we need is
ABCD or CDAB, but NOT ACBD, BDAC, ADBC, or BCAD. Your answer was
exactly one-third of what we were looking for.
Straight Flushes,
Mike Caro
On 16 Oct 97 00:51:37 GMT, "Tony Oresteen" <aore...@avana.net>
wrote:
Barbara Yoon
Erik Reuter wrote:
> Actually, I hope I've left out a few or have
> a mistake in the math, because otherwise,
> Barbara wins bragging rights! (46,618)
HAH!! Chauvinist! I hope you left out a few too,
but not TOO many, not more than 2,381...
Mike Caro
The word "discrete" may be confusing, especially since it has a
mathematical meaning of not being contiguous. Maybe "distinct" would
be more appropriate. Your "unique" also makes sense to me, if it is
taken to exclude "mirror-hand" matchups. What we're looking for is
every possible category of matchup that either has its own
A-wins-to-B-wins result OR has the same result as another matchup, but
for purely coincidental reasons. In that latter case, the matchup
would be deemed conceptually distinct, anyway.
The actual combinations of hand-versus-hand, by the way, are (((52
cards x 51 remaining cards) / 2 orders of arrival) x ((50 cards
available to second hand x 49 then remaining cards available to second
hand) / 2 orders of arrival) / 2 orders of hand consideration) =
812,175.
Thanks for responding. Sorry if I didn't make the object clear, and --
you're right -- the answer doesn't have much practical meaning in a
poker game. It may have meaning for software development, which was
the original context in which the subject was introduced (may have
scrolled off, though).
Erik Reuter
"Barbara Yoon" <by...@erols.com> wrote:
> HAH!! Chauvinist! I hope you left out a few too,
> but not TOO many, not more than 2,381...
What an impartial judge we have here. You inspired me to check my work and
I did find a few more, but I still have a couple thousand to go! (47,008)
numerals 1,2,3,4 stand for rank patterns
letters a,b,c,d stand for suit patterns
any 4 term series is assumed to be two sequential two card hands
C(x,y) = number of combinations of x things taken y at a time
[1111]
4 suits: abcd
#suitcombs = 1
subtotal = #rankcombs * #suitcombs = 13 * 1 = 13
[1112]
4 suits: abcd
3 suits: abca,abcc
#suitcombs = 1 + 2 = 3
subtotal = 13 * 12 * 3 = 468
[1122]
4 suits: abcd
3 suits: abac
2 suits: abba
#suitcombs = 1 + 1 + 1 = 3
subtotal = C(13,2) * 3 = 234
[1212]
4 suits: abcd
3 suits: abca,aabc
2 suits: abba,aabb
#suitcombs = 1 + 2 + 2 = 5
subtotal = C(13,2) * 5 = 390
[1213]
4 suits: abcd
3 suits: abca,abbc,abcb,aabc,abcc
2 suits: abba,aabb,aaba,abbb
#suitcombs = 1 + 5 + 4 = 10
subtotal = 13 * C(12,2) * 10 = 8,580
[1123]
4 suits: abcd
3 suits: abac,abca,abcc
2 suits: abab,abaa
#suitcombs = 1 + 3 + 2 = 6
subtotal = 13 * C(12,2) * 6 = 5,148
[1234]
4 suits: abcd
3 suits: abac,abca,abbc,abcb,aabc,abcc
2 suits: abba,abab,aabb,aaab,aaba,abaa,abbb
1 suit : aaaa
#suitcombs = 1 + 6 + 7 + 1 = 15
subtotal = ( C(13,2) * C(11,2) / 2 ) * 15 = 32,175
TOTAL = 13 + 468 + 234 + 390 + 8580 + 5148 + 32175
= 47,008
--
Erik Reuter, e-re...@uiuc.edu
Mike Caro
Lee --
You wrote:
>>
However, it just tickles me no end that one of our real luminaries is
willing to put his money where his mouth is; as I said, the correct
answer is less than a week away.
<<
Glad you're tickled. And it's good to be back aboard RGP and to see
you still active. However, I am NOT putting my money where my mouth is
on this. I am just giving a small prize for the right answer,
'cause... well, 'cause I'm bored. There has been no bet made. We need
an RGP historian to trace this conversation (which hopped threads)
back to its roots.
The only thing I ever said was to Barbara Yoon, something like (as a
closing tease at the end of a friendly and complimentary message),
"Are you sure you want to bet under 50,000?" In the original message
-- the first conditions I defined -- there was no question that the
number was beyond 50,000, but Barbara said the number would be "almost
50,000," if I remember correctly. I now believe she had misinterpreted
what I meant, so when I subsequently pointed out that the number was
much greater, we may have been addressing two different things. Others
have taken her bet. I have not. As I said in my challenge, I know the
answer.
As far as I can remember, I have made no definite assertion that the
number is beyond 50,000 under the new definition -- but the messages
are all here for anyone to review. However, it MIGHT be beyond 50,000,
and I MIGHT still want to bet. I ain't sayin' nothin'. It's just that
no bet has been made. It is a challenge, and more challenges may be
forthcoming.
BTW, have you been back to HPC? You were there about eight months ago,
right?
Straight Flushes,
Mike Caro
On 15 Oct 1997 21:32:07 GMT, le...@diver.engr.sgi.com (Lee Jones)
wrote:
>In article <34452d2a...@nntp.ix.netcom.com>,
>Mike Caro, *he* say...
>
>>I'm offering $100 reward for the first person who posts the right
>>answer to RGP by October 31, 1997 (might take much less time, though,
>>knowing you guys).
>
Scot Allen (SYS)
Mike Caro (ca...@caro.com) wrote:
: Everyone -
:
: I'm offering $100 reward for the first person who posts the right
: answer to RGP by October 31, 1997 (might take much less time, though,
: knowing you guys). To qualify, key members of this group must agree
: that it is the first right answer and instruct me to whom and where to
: send the check. In the event of a dispute, I am the sole decider about
: who (or whether anyone) should be paid, but if I stiff you for $100,
: you'll have a story to tell.
I came up with 14066 using the following program:
DEFINT A-Z
DIM suit(12, 3)
DIM c$(51)
f$(0) = "T": f$(1) = "J": f$(2) = "Q": f$(3) = "K": f$(4) = "A"
s$(0) = "s": s$(1) = "h": s$(2) = "d": s$(3) = "c"
count = 0
FOR s = 0 TO 12
FOR cd = 0 TO 3
READ d
suit(s, cd) = d
NEXT
NEXT
FOR cd = 0 TO 51
t = INT(cd / 4)
IF t < 8 THEN d$ = CHR$(50 + t) ELSE d$ = f$(t - 8)
d$ = d$ + s$(cd - t * 4)
c$(cd) = d$
NEXT
FOR x1 = 0 TO 12
FOR x2 = x1 TO 12
FOR x3 = x2 TO 12
FOR x4 = x3 TO 12
FOR s = 0 TO 12
c1 = x1 * 4 + suit(s, 0)
c2 = x2 * 4 + suit(s, 1)
c3 = x3 * 4 + suit(s, 2)
c4 = x4 * 4 + suit(s, 3)
IF c1 < c2 AND c2 < c3 AND c3 < c4 THEN hit = 1 ELSE hit = 0
IF (s = 6 OR s = 8 OR s = 9 OR s = 10) AND c2 = c1 + 1 THEN hit = 0
IF hit = 1 THEN count = count + 1
IF hit = 1 THEN PRINT c$(c1); c$(c2); " vs "; c$(c3); c$(c4); " "; count
NEXT
NEXT
NEXT
NEXT
NEXT
DATA 0,0,0,0,0,0,0,1,0,0,1,0,0,0,1,1,0,0,1,2,0,1,0,0,0,1,0,1
DATA 0,1,0,2,0,1,1,0,0,1,1,1,0,1,1,2,0,1,2,2,0,1,2,3
The program works by generating a combination of cards, then
applies suit values. The suit table was created to take into
account the RELATIONSHIPS between the suits of the cards
and NOT specific suit values. Any relationship between the suits
of the four cards is taken into account by one of the 13 sets
of values. Also taken into account is the special
case when the first two cards have the same value.
The case when the second pair of cards has the same value is
handled automatically by the "If C1<C2" etc.
Some examples:
2s2h vs KsAs is the same as 2c2d vs KcAc
KsKh vs AsAc is the same as KdKc vs AdAh
and so on. Email me if you have any questions about it.
(gawd, I hope I haven't screwed up somewhere! I'm sure
you'll all let me know about it :))
-Scot
Tony Oresteen
Mike-
If my answer calulated ABCD in any order then I must have counted too many
hands.
What I think you are saying is that I counted the "same" hands more than once.
This implies that I have too many thus I should REDUCE my answer from 28,561 to
some lower number.
If 28,561 is exactly ONE-THIRD of the answer, then the answer is 85,683 (28,561
* 3). However, I think that is wrong.
I did calculate ABCD and CDAB as different hands but AB and BA are allready
accounted for as the same hand by using 169 as the starting value. My
calculation does *not* consider order of the cards. 27o is the same as 72o
when you use 169.
If by *discrete* you meant from *EITHER* player's view, then that is a
different question. I interpeted *discrete* to mean "from a single player's
view*. Thus ABCD *IS* different from CDAB. For example, if it doesn't matter
then the next time you get pocket Aces I'll swap my 27o for them!
I like these tough questions!
Tony
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Mike Caro wrote: {FROM DEJANEWS as my ISP newsfeed doesn't get all the posts!}
Tony --
By the way, the reason your calculation of combinations didn't quite
work is that you calculated for a four-card hand out of a 52-card
deck. That means cards ABCD in any order of arrival. What we need is
ABCD or CDAB, but NOT ACBD, BDAC, ADBC, or BCAD. Your answer was
exactly one-third of what we were looking for.
^^^^^^^^^
Straight Flushes,
Mike Caro
On 16 Oct 97 00:51:37 GMT, "Tony Oresteen" <aore...@avana.net>
wrote:
>Mike,
>
>My ISP is having trouble posting to news so it's very likly that this response
>will not mate to the usenet.
>
>I assume no Joker, bug, or wild cards. Just a 52 CARD deck.
>
>The answer to your challenge is simple: 169*169 or 28,561. Here's why. In
>Hold 'Em there are 169 different two card hands. Note that 4s7s is exacly the
>same as 4d7d for a *STARTING* hand. You used the term *discrete* which is NOT
>*unique*. Also, 2d2c is exactly the same as 2h2s. Thus player 1 can have any
>one of 169 possible two card starting hands and the loss of two cards from the
>deck does NOT reduce the number of TYPES of hand player 2 can get. If player
1
>can get AA player 2 can still get AA.
>SNIP
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For you spammers with automated email address catchers, here are the addresses
of the current FCC board members:
Chairman Reed Hundt: rhu...@fcc.gov
Commissioner James Quello: jqu...@fcc.gov
Commissioner Susan Ness: sn...@fcc.gov
Commissioner Rachelle Chong: rch...@fcc.gov
Others:
bi...@microsoft.com
root@localhost
dgg...@worldnet.att.net, dgra...@aol.com, dgte...@aol.com,
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tboli...@aol.com
[Idea shamelessly stolen from David Whitwell & Bob Sweeney & others]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Jeff Woods
Mike Caro wrote:
> I'll leave it open to y'all to finish the itemizing. If no one posts
> the answer in a few weeks, I will. HINT: It WILL often matter WHICH
> rank matches. Therefore, Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs.
> 6d 5c). Here's to get you started=85
Mike, I don't understand your supposition above, at least with regard to
that specific hand. Why SHOULD these two instances be different, when
the outcome ofthe two doesn't appear to be affected? This is simply
KQs vs. 65o, with one suit matched. Since that suit is essentially
DRAWING DEAD (if a club flush comes for the 65o, it will lose,
regardless of if the 5 or 6 is the club -- likewise, if the offsuit
flush comes four on the board, the KQs will lose, same regardless). =
Since suits only matter in flushes and straight flushes, I can't see
how it really matters here whether the 5 or the 6 is suited with the
KQ.... I would have to say that for the purpose of this drill (incited
by Barbara Yoon, who made the same supposition that I do here in
counting discrete starting hands), KxQx vs. 5x6y is utterly equivalent
to KxQx vs. 5y6x. =
Now, KQs vs. A6o, that's another story, and its situations like these
that make counting such hands VERY VERY difficult.
I look forward to your explanation of why the two cited hands above must
be considered unique.
> =
> Two suited cards vs. two suited cards of the same suit (example: Kc Qc
> vs. 6c 5c) (((13 x 12) / 2) x ((11 x 10) / 2)) =3D (4,290 / 2) =3D 2,14=
5
> (7,050 total).
> =
> Two suited cards vs. two suited cards of a different suit where both
> ranks are the same (example: ((12 x 11) / 2) (example: Kc Qc vs. Kd
> Qd) (always 0.5 win rate) =3D 55.
> =
> Two suited cards vs. two suited cards of a different suit where the
> high rank=85
> =
> Again, there are errors in math and omissions in the list above. You
> have to do your own calculations and list the categories (similar to
> the way I have) to get the $100. It won't be worth your time, in a
> money sense, but you'll probably get some obscure spiritual reward.
> Maybe not.
> =
> Straight Flushes,
> Mike Caro
Randy Buckland
"Tony Oresteen" <aore...@avana.net> writes:
>52*51*50*49 or 6,497,400 starting hands if order matters. Since order really
>doesn't matter in Hold 'Em then you must divide by 4*3*2*1 or 24, yeilding
>270,725.
BZZZT! Sorry. Order does matter. It's true that AcKc vs. 5c4c is that same
as KcAc vs. 4c5c, but Ac5c vs. Kc4c is not! You calculated "pick four from
52 with order not important" when you really want "pick 2 from 52 (order
not important) THEN pick 2 from 50 (order not important)"
--
Randy Buckland "It's difficult to work
Independent computer consultant in a group when you're
r...@nospam.ipass.net omnipotent" -- Q
(Remove the "nospam." from the above address. Just say no to spam!)
James Layland
I was getting all ready to collect $100 by correcting Erik Reuter's analysis,
but he seems to have done it first.
I now agree with his number (47008) and have appended one checkpoint.
Double check #1:
Ignoring suitedness of hands, there are 91 unique 2-card hands
C(13,2) + 13 = 91 (There are 169 unique hands considering suitedness.)
Then, if I ignore the suitedness in all of Erik's categories, I should get
a total of
C(169,2) + 169 = 4186
different matchups. (The added 169 is for same hand-type matchups, like
AK vs AK).
[1111]
subtotal = 13 1 = 13
[1112]
subtotal = 13 * 12 = 156
[1122]
subtotal = C(13,2) = 78
[1212]
subtotal = C(13,2) = 78
[1213]
subtotal = 13 * C(12,2) = 858
[1123]
subtotal = 13 * C(12,2) = 858
[1234]
subtotal = ( C(13,2) * C(11,2) / 2 ) = 2145
TOTAL = 13 + 156 + 78 + 78 + 858 + 858 + 2145
= 4168 (ignoring suits)
So this checks out.
Of course the hard part is counting unique suit combinations. But at least
this shows that Erik's analysis is not ignoring any hand matchup categories
(as it was before his correction).
--
James Layland
jlay...@grissom.jpl.nasa.gov
dy...@alumni.cse.ucsc.edu
Eric, you're the man!
Since you took the same approach as I did, and your second post contained
the same seven categories as me, I compared our results
category-by-category and I think you are correct on every count. I think
you're 47,008 number will win the prize unless we both misinterpreted the
question.
In article <e-reuter-161...@news.cso.uiuc.edu>,
e-re...@uiuc.edu (Erik Reuter) wrote:
> [1111]
> 4 suits: abcd
> #suitcombs = 1
> subtotal = #rankcombs * #suitcombs = 13 * 1 = 13
Agreed!
>
> [1112]
> 4 suits: abcd
> 3 suits: abca,abcc
> #suitcombs = 1 + 2 = 3
> subtotal = 13 * 12 * 3 = 468
Agreed!
> [1122]
> 4 suits: abcd
> 3 suits: abac
> 2 suits: abba
> #suitcombs = 1 + 1 + 1 = 3
> subtotal = C(13,2) * 3 = 234
Agreed!
> [1212]
> 4 suits: abcd
> 3 suits: abca,aabc
> 2 suits: abba,aabb
> #suitcombs = 1 + 2 + 2 = 5
> subtotal = C(13,2) * 5 = 390
You're right! I left off the aabb suiting and
Used 13 * 12 instead of C(13,2) = 13 * 12 / 2.
> [1213]
> 4 suits: abcd
> 3 suits: abca,abbc,abcb,aabc,abcc
> 2 suits: abba,aabb,aaba,abbb
> #suitcombs = 1 + 5 + 4 = 10
> subtotal = 13 * C(12,2) * 10 = 8,580
I got 7 interesting suit combinations here and you got 10.
I left out abcb, aaba, abbb! You're right!
>
> [1123]
> 4 suits: abcd
> 3 suits: abac,abca,abcc
> 2 suits: abab,abaa
> #suitcombs = 1 + 3 + 2 = 6
> subtotal = 13 * C(12,2) * 6 = 5,148
I got twice as many combos, and same 6 suitings, my error.
> [1234]
> 4 suits: abcd
> 3 suits: abac,abca,abbc,abcb,aabc,abcc
> 2 suits: abba,abab,aabb,aaab,aaba,abaa,abbb
> 1 suit : aaaa
> #suitcombs = 1 + 6 + 7 + 1 = 15
> subtotal = ( C(13,2) * C(11,2) / 2 ) * 15 = 32,175
I left out all the suitings where 3 cards were suited!
>
> TOTAL = 13 + 468 + 234 + 390 + 8580 + 5148 + 32175
> = 47,008
>
> --
> Erik Reuter, e-re...@uiuc.edu
-------------------==== Posted via Deja News ====-----------------------
http://www.dejanews.com/ Search, Read, Post to Usenet
Charles Friedman
Mike Caro wrote:
> Everyone -
>
> I'm offering $100 reward for the first person who posts the right
> ...
I suppose this isn't what Mr. Caro wanted, and it depends somewhat on
how one interprets his
question, but it seems to ma that the most useful way of interpreting the
question leads to the following
calculation:
Player A can have (52)(51)/2 possible starting hands, and then Player
2 can have (50)(49)/2 possible
hands. The product of these is 1624350.
Of course, this counts some "equivalent" matchups several times, but
different pairs of equivalent
matchups have to be counted different numbers of times, so if we want a
sample space in which all
pairs of matchups have the same probability, this seems the way to go.
(Almost always, probabilistic
calculations are easier when the sample space contains equally likely
outcomes. )
If one wants to make an actual table to store for computational purposes,
then some other kind of counting
or enumeration of possibilities might take up less space; however, then
the calculations using this
table might be much more complicated.
Comments?
chas friedman
>
Barbara Yoon
Mike Caro:
> ...Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs. 6d 5c...
Jeff Woods:
...I don't understand your supposition...with regard to that
specific hand. Why SHOULD these two instances be
different, when the outcome ofthe two doesn't appear to be
affected? This is simply KQs vs. 65o, with one suit matched.
Since that suit is essentially DRAWING DEAD (if a club flush
comes for the 65o, it will lose, regardless of if the 5 or 6 is the
club -- likewise, if the offsuit flush comes four on the board,
the KQs will lose, same regardless). Since suits only matter in
flushes and straight flushes, I can't see how it really matters
here whether the 5 or the 6 is suited with the KQ...
Mike Caro:
> Although the matchups we're talking about may look identical,
> they don't yield the same answer. Think in terms of playing the
> board [mainly when all five upcards are diamonds -- the 6d
> wins more than the 5d]... I have specified that matchups that
> coincidentally yield the same result cannot be lumped into a
> category. Therefore, AA vs. AA and 22 vs 22 are different
> matchups, as are Ac Kc vs. Ah Kh and 7c 2c vs. 7d 2d.
> We KNOW these must be even-money matchups, but they
> are still separate data points on the table.
If the information we are interested in is "wins/losses/ties" (always
adding up to 1,712,304) in a match-up, then it becomes clearer that AA v.
AA, and 22 v. 22, etc. are indeed "different," as Mike says -- the wins are
equal to the losses in each example, of course, but there are differences
in the ties...
Mike Caro
Barbara -- Excellent point. What I meant was that you could cut
corners and lump all hands that obviously are identically matched into
a single even-money category. But the definition we're using for this
problem doesn't allow this.
Your point about pair vs. same pair matchups arriving at their 50/50
split differently is yet another powerful reason why these categories
are separate. Thanks.
Straight Flushes,
Mike Caro
Erik Reuter
jlay...@kilroy.jpl.nasa.gov (James Layland) wrote:
> I was getting all ready to collect $100 by correcting Erik Reuter's analysis,
> but he seems to have done it first.
I was thinking of doing that myself, but no one had worked it out yet even
close to correct when I started!
> Double check #1:
> Ignoring suitedness of hands, there are 91 unique 2-card hands
> C(13,2) + 13 = 91 (There are 169 unique hands considering suitedness.)
>
> Then, if I ignore the suitedness in all of Erik's categories, I should get
> a total of
>
> C(169,2) + 169 = 4186
You got the right answer, but I think you meant to say
C(91,2) + 91 = 4186
> TOTAL = 13 + 156 + 78 + 78 + 858 + 858 + 2145
> = 4168 (ignoring suits)
Actually, the total is 4186
>
> So this checks out.
Yup. Thanks for the checkpoint.
Erik Reuter
> Player A can have (52)(51)/2 possible starting hands, and then Player
> 2 can have (50)(49)/2 possible
> hands. The product of these is 1624350.
>
...
> If one wants to make an actual table to store for computational purposes,
> then some other kind of counting
> or enumeration of possibilities might take up less space; however, then
> the calculations using this
> table might be much more complicated.
>
> Comments?
I agree with your premise of simplicity. Unless my calculating program was
capable of enumerating the cases distinctly, I would probably make the
full table memory resident(1,624,350) if I were trying to calculate
something that needed this data as quickly as possbile, since I have
plenty of megabytes of RAM.
However, consider the problem of calculating the table in the first place.
On a top performance desktop computer, it takes about 1 sec to calculate
each matchup of wins/losse/ties where,
wins + losses + ties = C(48,5) = 1,712,304
so that to calculate all 1624350 would take almost 19 days. If my figure
of 47,008 is right, then we only need 13 hours to calculate the table. We
could always expand it back to full form after we calculate it.
Tony Oresteen
Randy Buckland <r...@nospam.ipass.net> wrote in article
<625h1j$e64$1...@news.ipass.net>...
SNIP
> BZZZT! Sorry. Order does matter. It's true that AcKc vs. 5c4c is that same
> as KcAc vs. 4c5c, but Ac5c vs. Kc4c is not! You calculated "pick four from
> 52 with order not important" when you really want "pick 2 from 52 (order
> not important) THEN pick 2 from 50 (order not important)"
SNIP
Oops! Guess I was asleep at the keyboard. Thanks for pointing it out Randy.
--
Tony
Remove the "z" in my address to send me e-mail.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Mike Caro
No guessing allowed. To be eligible, you need to itemize by category,
in a manner similar to the example I gave. Otherwise, the judges won't
be able to validate you claim. This applies to your other 100,000
entries as well. :-)
Thanks for responding. I happen to know that, 54,917 was the official
paid attendance announced for an NFL game 15 years ago. That's only
759 away from your entry! That seems too close to be a pure
coincidence. Were you at that game?
Straight Flushes,
Mike Caro
On 15 Oct 1997 18:46:44 -0400, mum...@hendrix.nectar.cs.cmu.edu (Todd
Barry Paul
On Wed, 15 Oct 1997 20:59:08 GMT, ca...@caro.com (Mike Caro) wrote:
>Everyone -
>
>I'm offering $100 reward for the first person who posts the right
>answer to RGP by October 31, 1997 (might take much less time, though,
>knowing you guys). To qualify, key members of this group must agree
>that it is the first right answer and instruct me to whom and where to
>send the check. In the event of a dispute, I am the sole decider about
>who (or whether anyone) should be paid, but if I stiff you for $100,
>you'll have a story to tell.
POSSIBLE SPOILER
.
.
.
.
.
.
I make it 45,370
Describe a matchup as "C1,C2 vs C3,C4" where the C's are cards.
Looking
at just the the ranks of the cards we can write the hand as
"R1,R2 vs R3,R4" where R1 is the rank of C1, R2 the rank of C2, etc.
Similarly we can write S1,S2 vs S3,S4 for the ranks.
A matchup is in "Standard Rank Form" if all the following hold
a) R1 >= R2 and R3 >= R4
e.g. J8 vs 94 not 8J vs 49
b) If both hands are paired then R1 >= R3
e.g. JJ vs 77 or 99 vs 99
c) If only one hand is paired then it is C1,C2
e.g. 55 vs AK
d) If neither hand is paired then R1 >= R3 and if R1 = R3
then R2 >= R4.
e.g. K2 vs QJ or K9 vs K8
Recall the standard order of the suits: S > H > D > C
A matchup is in "Standard Form" if it is in Standard Rank Form and
suits are assigned to C1,C2,C3,C4 in order according to
RULE1: If the card does not match a previous card in rank then give
it the highest available rank.
e.g. Ks5h vs Jd8s or Ks5h vs Jd8h or Ks5h vs Jd8c
but not Ks5h vs Jc8s or Ks5d vs Js8s
Additionally
RULE2: If neither hand is paired and they have the same two ranks
and
exactly one hand is suited then it is C1,C2 that is suited.
Then, in an obvious way, every matchup is equivalent to exactly one of
our Standard Forms. e.g. Ks5h vs Js8s would be the standard form for
itself and Kc5h vs Jc8c and 8dJd vs 5sKd and 93 other matchups.
So we could do the job by having a table entry (encoding percentages
for C1,C2 wins, C3,C4 wins, and ties) for each Standard Form. It is
unlikely that any programmer would want to look any deeper into the
matter than that. It would certainly cost Mike a lot more than $100
to get me to do it. {:-) It may happen that some of our entries are
duplicated but so what? We could take advantage of that to reduce our
table by a few entries but the extra code we would have to write to
produce and then use the table could hardly be worth it.
Here is my count of the standard forms:
We write C(n,k) for "combinations of n things taken k at a time"
1) Pair against same pair: 13
2) Pair against different pair. C(13,2) = 78 pair combos. For each
combo two, one, or no suits may match. : 3 * 78 = 234
3) Pair against no pair, no pair contains pair's rank.
e.g. KK vs K9 Pick pair rank, then odd card rank.
13 * 12 = 156 Rank Forms. For a given Rank Form odd card's suit
may match its companion, or one of the pair, or be of the fourth
suit. 3 * 156 = 468
4) Pair against no pair, no pair has two different suits. e.g.
KK vs J9. Pick two suits for the no pair then one of remaining
ranks for the pair. C(13,2) * 11 = 858. We can have the
following
patterns for S1,S2 vs S3,S4 : SH vs SH, SH vs SD, SH vs DS,
SH vs DC. 4 * 858 = 3432
5) No pairs, two matching ranks. e.g. JT vs JT. C(13,2) = 78 Rank
Forms. Patterns for suits may be SH vs HS, SH vs HD, SH vs DS,
SH vs DC, SS vs HH, SS vs HD. 6 * 78 = 468.
6) No pairs, one matching rank. e.g. JT vs J9. Pick the two
unmatched ranks then pick the matched rank from the remaining
ranks. C(13,2) * 11 = 858. Patterns for suits may be SH vs HS,
SH vs HH, SH vs HD, SH vs DS, SH vs DH, SH vs DD, SH vs DC,
SS vs HS, SS vs HH, SS vs HD. 10 * 858 = 8580.
7) No pairs, four different ranks. e.g. A7 vs Q9. Pick the four
ranks
then pick R2 from the three lower ranks. (Recall R1 must be the
highest rank.) C(13,4) * 3 = 2145. Patterns for suits may be
SH vs SS, SH vs SH, SH vs SD, SH vs HS, SH vs HH, SH vs HD,
SH vs DS, SH vs DH, SH vs DD, SH vs DC, SS vs SS, SS vs SH,
SS vs HS, SS vs HH, SS vs HD. 15 * 2145 = 32,175
GRAND TOTAL: 13 + 234 + 468 + 3432 + 468 + 8580 + 32,175
= 45,370
--------------------------------------------------
Barry Paul
mailto:bp...@hoflink.com
http://hoflink.com/~bpaul
Great Neck, NY
--------------------------------------------------
Barbara Yoon
Mike Caro wrote:
> ...Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs. 6d 5c...
Jeff Woods wrote:
Mike, I don't understand your supposition above, at least with regard to
that specific hand. Why SHOULD these two instances be different, when
the outcome ofthe two doesn't appear to be affected? This is simply
KQs vs. 65o, with one suit matched. Since that suit is essentially
DRAWING DEAD (if a club flush comes for the 65o, it will lose,
regardless of if the 5 or 6 is the club -- likewise, if the offsuit
flush comes four on the board, the KQs will lose, same regardless).
Since suits only matter in flushes and straight flushes, I can't see
how it really matters here whether the 5 or the 6 is suited with the
KQ.... I would have to say that for the purpose of this drill (incited
by Barbara Yoon, who made the same supposition that I do here in
counting discrete starting hands), KxQx vs. 5x6y is utterly equivalent
to KxQx vs. 5y6x. I look forward to your explanation of why the two
cited hands above must be considered unique.
Jeff -- Mike is right, but the differences there are very subtle, mainly in
when FIVE DIAMONDS come up on the board, the 6d can more often improve the
board and win, whereas the 5d more often has to settle for a
playing-the-board tie... A complicated question, I know, but re-read my
post carefully, and you will see that I did NOT make your "same
supposition"...OK?
Mike Caro
Jeff --
Thanks for your comments. Maybe the way I defined the problem is
confusing you. The original premise was that I envisioned a research
program that would automatically devise a table of all possible
head-to-head hold 'em outcomes from which a second program could
simply look up the answer, rather than calculate. For this purpose, a
PRECISE answer (at least, precise to, say, 17.123 percent) is needed.
Although the matchups we're talking about may look identical, they
don't yield the same answer. Think in terms of playing the board and
it might become clearer. I have specified that matchups that
coincidentally yield the same result cannot be lumped into a category.
Therefore, AA vs. AA and 22 vs 22 are different matchups, as are Ac Kc
vs. Ah Kh and 7c 2c vs. 7d 2d. We KNOW these must be even-money
matchups, but they are still separate data points on the table.
Straight Flushes,
Mike Caro
On Thu, 16 Oct 1997 11:41:08 -0400, Jeff Woods <je...@delta.com> wrote:
>Mike Caro wrote:
>
>> I'll leave it open to y'all to finish the itemizing. If no one posts
>> the answer in a few weeks, I will. HINT: It WILL often matter WHICH
>> rank matches. Therefore, Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs.
>> 6d 5c). Here's to get you started…
>
>Mike, I don't understand your supposition above, at least with regard to
>that specific hand. Why SHOULD these two instances be different, when
>the outcome ofthe two doesn't appear to be affected? This is simply
>KQs vs. 65o, with one suit matched. Since that suit is essentially
>DRAWING DEAD (if a club flush comes for the 65o, it will lose,
>regardless of if the 5 or 6 is the club -- likewise, if the offsuit
>flush comes four on the board, the KQs will lose, same regardless).
>Since suits only matter in flushes and straight flushes, I can't see
>how it really matters here whether the 5 or the 6 is suited with the
>KQ.... I would have to say that for the purpose of this drill (incited
>by Barbara Yoon, who made the same supposition that I do here in
>counting discrete starting hands), KxQx vs. 5x6y is utterly equivalent
>to KxQx vs. 5y6x.
>
>Now, KQs vs. A6o, that's another story, and its situations like these
>that make counting such hands VERY VERY difficult.
>
>I look forward to your explanation of why the two cited hands above must
>be considered unique.
>
>
>
>>
>> Two suited cards vs. two suited cards of the same suit (example: Kc Qc
>> vs. 6c 5c) (((13 x 12) / 2) x ((11 x 10) / 2)) = (4,290 / 2) = 2,145
>> (7,050 total).
>>
>> Two suited cards vs. two suited cards of a different suit where both
>> ranks are the same (example: ((12 x 11) / 2) (example: Kc Qc vs. Kd
>> Qd) (always 0.5 win rate) = 55.
>>
>> Two suited cards vs. two suited cards of a different suit where the
>> high rank…
>>
>> Again, there are errors in math and omissions in the list above. You
>> have to do your own calculations and list the categories (similar to
>> the way I have) to get the $100. It won't be worth your time, in a
>> money sense, but you'll probably get some obscure spiritual reward.
>> Maybe not.
>>
>> Straight Flushes,
>> Mike Caro
Steve Brecher
"Barbara Yoon" <by...@erols.com> wrote:
> Mike Caro wrote:
> > ...Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs. 6d 5c...
> Jeff -- Mike is right, but the differences there are very subtle, mainly in
> when FIVE DIAMONDS come up on the board, the 6d can more often improve the
> board and win, whereas the 5d more often has to settle for a
> playing-the-board tie... A complicated question, I know, but re-read my
> post carefully, and you will see that I did NOT make your "same
> supposition"...OK?
The numbers:
1,712,304 pots with board cards: (unspecified)
KcQc 6c5d
% chance of outright win 63.771386 35.513203
% chance of win or split 64.486797 36.228614
expected return, % of pot 64.129092 35.870908
fair pot odds:1 0.559355 1.787774
1,712,304 pots with board cards: (unspecified)
KcQc 6d5c
% chance of outright win 63.771328 35.519686
% chance of win or split 64.480314 36.228672
expected return, % of pot 64.125821 35.874179
fair pot odds:1 0.559434 1.787520
--
st...@brecher.reno.nv.us (Steve Brecher)
Bruce Schechter
Jeff Woods wrote in message <344781...@delta.com>...
>Barbara Yoon wrote:
>>
>> Mike Caro wrote:
>
>> > ...Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs. 6d 5c...
>
>> Jeff -- Mike is right, but the differences there are very subtle,
>> mainly in when FIVE DIAMONDS come up on the board, the 6d can more
>> often improve the board and win, whereas the 5d more often has to
>> settle for a playing-the-board tie... A complicated question, I
>> know, but re-read my post carefully, and you will see that I did NOT
>> make your "same supposition"...OK?
>
>I see that with this specific hand. You're the first to spell out why
>they're different (not even Mike did -- I'd missed that possibility).
>Thanks!
Yes, but is KcQc vs 2c3d different from KcQc vs 2h3d? After all, neither the
two or the three of diamonds will improve a diamond board to win. Except in
the case of making a straight flush and it seems to me that the 2d and 3d
will each make as many straight flushes as the other.
Mike Caro
Barry --
Oh, wait, sorry! Ignore my recent response. There's an entry contained
in your message. With my news reader, there's a lot of vertical space
between "POSSIBLE SPOILER" and the calculation, so I assumed that
marked the end of the message.
Thanks for your submission.
Straight Flushes,
Mike Caro
On Fri, 17 Oct 1997 13:01:19 GMT, bp...@hoflink.com (Barry Paul)
wrote:
>Sorry if you got this before but I am not sure my original post got
>through.
Mike Caro
9h 8h 5h 4h Ks, et al.
And anyway, categories with the same results for coincidental reasons
have been defined (for this challenge) to require separate
itemization.
Did you meant Kc Qc vs. either 3d 2c or 3c 2d? If so, the 2d and 3d
will NOT make the same number of straight flushes, and the 3d CAN play
with a five-diamond board, but the 2d can't...
Think about 9d 8d 5d 4d 2d (and you have 3d) vs. the equally likely 9d
8d 5d 4d 3d (and you have 2d). Thanks for responding.
Straight Flushes,
Mike Caro
Jeff Woods
Bruce Schechter wrote:
>
> >> > ...Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs. 6d 5c...
> >
> >> Jeff -- Mike is right, but the differences there are very subtle,
> >> mainly in when FIVE DIAMONDS come up on the board, the 6d can more
> >> often improve the board and win, whereas the 5d more often has to
> >> settle for a playing-the-board tie...
> Yes, but is KcQc vs 2c3d different from KcQc vs 2h3d? After all,
> neither the two or the three of diamonds will improve a diamond board
> to win. Except in the case of making a straight flush and it seems
> to me that the 2d and 3d will each make as many straight flushes as
> the other.
I was in the midst of writing a similar reply when I had a brain fart
and killed it, when I realized:
If you have 2d 3c, and five diamonds on the board, the board plays.
Chop the pot.
If you have 2c 3d, then five diamonds INCLUDING the 2d, and the board
does NOT play -- the 3d does, and the 23o scoops. Thus, the point I
raised earlier does seem to be irrelevant.
Steve Heston
Mike Caro wrote:
>How many discrete starting hand matchups are there in a heads-up hold 'em
game?
My answer is 41,548. I'll consider 5 categories of four cards according to
value (four of a kind, three of a kind, two pair, one pair, and four
different). Within each category there are several types according to the
number (and arrangement) of suits and how the four cards are dealt into two
hands of two.
Let (n,k) denote "n choose k" = n!/(k!*(n-k)!).
1) Four of a kind: 13 possibilities.
2) Two pair: (13,2) value combinations * 7 suit types = 546
possibilities.
a) 2 suits: 2 types.
i) both players have pairs.
ii) neither player has a pair.
b) 3 suits: 3 types.
i) neither player has a pair nor two suited cards.
ii) neither player has a pair but one player has two suited cards.
iii) both players have pairs.
c) 4 suits: 2 types.
i) both players have pairs.
ii) neither player has a pair.
3) Three of a kind: (13,2) value combinations * 3 suit types = 234
possibilities.
a) 3 suits: 2 types.
i) one player has suited cards.
ii) neither player has suited cards.
b) 4 suits: 1 type.
4) One pair: 13*(12,2) value combinations * 5 suit types * 2 deal types
= 8,580 possibilities.
A) 2 deal types
i) one player has a pair.
ii) neither player has a pair.
B) 5 suit types.
a) 2 suits: 1 type.
b) 3 suits: 3 types.
i) unpaired cards are suited.
ii) lowest unpaired card matches suit of a paired card.
iii) highest unpaired card matches suit of a paired card.
c) 4 suits: 1 type.
5) Four different values: (13,4) value combinations *15 suit types * 3
deal types = 32,175 possibilities.
A) 3 dealing types.
i) lowest and second lowest card in same hand.
ii) lowest and third lowest card in same hand.
iii) lowest and highest cards in same hand.
B) 15 suit types.
a) 1 suit: 1 type.
b) 2 suits: 7 types.
i) lowest card has distinct suit.
ii) second lowest card has distinct suit.
iii) third lowest card has distinct suit.
iv) highest card has distinct suit.
v) lowest and second lowest cards are suited.
vi) lowest and third lowest cards are suited.
vii) lowest and highest cards are suited.
c) 3 suits: 6 types.
i) lowest and second lowest cards are suited.
ii) lowest and third lowest cards are suited.
iii) lowest and highest cards are suited.
iv) second lowest and third lowest cards are suited.
v) second lowest and highest cards are suited.
vi) third lowest and highest cards are suited.
d) 4 suits: 1 type.
Bruce Schechter
Bruce Schechter:
Yes, but is KcQc vs 2c3d different from KcQc vs 2h3d?
After all, neither the two or the three of diamonds will
improve a diamond board to win. Except in the case of
making a straight flush and it seems to me that the 2d and 3d
will each make as many straight flushes as the other.
> [corrected post, clubs vice diamonds]
> The weaker hand's holding a club makes clubs
> on the board slightly less likely, thus diminishing
> slightly the relative strength of the KQs.
It's really all Schechter's fault anyway, as I think he
originally meant "2d 3c" (not "2h 3d")...
Yes, yes. Mea culpa, mea maximum culpa. I too had a major "brain fart" but
unlike Jeff Woods mine was public. Hope my stupidity helped illuminate some
of the subltle issues here...
Tom Weideman
Mike Caro wrote:
> As far as I can remember, I have made no definite assertion that the
> number is beyond 50,000 under the new definition -- but the messages
> are all here for anyone to review. However, it MIGHT be beyond 50,000,
> and I MIGHT still want to bet. I ain't sayin' nothin'. It's just that
> no bet has been made. It is a challenge, and more challenges may be
> forthcoming.
Hey wait a minute... This is starting to sound alot like something you
wrote in _Super System_ regarding either/or situations. WELL I'M ON TO
YOU.
Without doing a shred of math, I hereby assert that the correct number
is EXACTLY 50,000. Send the $100 check ASAP... I'll send you my snail
mail address by private email.
Tom Weideman
Barry Paul
Jeff Woods
Barbara Yoon wrote:
>
> Mike Caro wrote:
> > ...Kc Qc vs. 6c 5d is DIFFERENT than Kc Qc vs. 6d 5c...
> Jeff -- Mike is right, but the differences there are very subtle,
> mainly in when FIVE DIAMONDS come up on the board, the 6d can more
> often improve the board and win, whereas the 5d more often has to
Mike Caro
Tom --
Thanks for reminding me: I haven't written about either/or situations
in years, and they're just as valuable a psychological poker tool as
ever.
Alas, in this case no either/or deception was attempted. Plus, the
correct answer may have already been submitted. Our Honorable Chief
Justice (in fact, our ONLY justice for this challenge) Barbara Yoon
and I conferenced about this yesterday. If a suitable answer, with
breakdown and explanation of the math has already been received, we
will announce a winner tomorrow. If not, the quest continues.
Straight Flushes,
Mike Caro
Barry Paul
My previous post requires correction. The actual computation is
unchanged. The description of how suits are arranged in Standard Form
(RULE 1) got messed up. First of all "If the card does not match a
previous card in rank ..." should obviously say "suit" instead of
"rank".
Also I should have added: "If one hand is paired and the other not and
one of the unpaired cards is suited with one of the pair then their
suit is spades."
Gotta stop posting in the wee hours.
Barbara Yoon
Steve Heston:
> Let (n,k) denote "n choose k" = n!/(k!*(n-k)!).
> 3) Three of a kind: (13,2) value combinations * 3 suit types = 234
> a) 3 suits: 2 types.
> i) one player has suited cards.
> ii) neither player has suited cards.
> b) 4 suits: 1 type.
....and moving further down your list...three of a kind
combinations not "(13,2)" = 78, but 13 x 12 = 156...
Barbara Yoon
Barry Paul wrote:
> POSSIBLE SPOILER
> I make it 45,370
...but for one thing, aren't you overlooking
"pair against no pair, no pair SUITED"?
Will Hyde
Mike Caro wrote:
>
> I'm not going to take the time to carefully edit the list below, or
> even to check the numbers, so feel free to point out anything I may
> have omitted or miscalculated. This is not to be considered a proofed
> and published work. In fact, I've deliberately left out at least one
> category and deliberately miscalculated at least one figure, so you'll
> have to work for the $100…
>
> Pair vs. same Pair (example: Ac Ad vs. Ah As) (always 0.5 win rate) =
> 13 (13 total).
If there are 13 ways for Pair vs. same Pair to occur then you are
counting Ac Ad as a different hand from Ad Ac, right?
Please, somebody, tell me where I err here, but I only see 6 different
ways for Player A to hold AA, and whichever one he holds there remains
only one other AA combo ... so I only see 7 possibilities....
--
______________________________________________________w...@whitestone.com
(will hyde) http://www.whitestone.com
Erik Reuter
"Barbara Yoon" <by...@erols.com> wrote:
> Barry Paul wrote:
>
> > 4) Pair against no pair, no pair has two different suits. e.g.
> > KK vs J9. Pick two suits for the no pair then one of remaining
> > ranks for the pair. C(13,2) * 11 = 858. We can have the
> > following patterns for S1,S2 vs S3,S4 : SH vs SH, SH vs SD,
> > SH vs DS, SH vs DC. 4 * 858 = 3432
>
> > 5) No pairs, two matching ranks. e.g. JT vs JT. C(13,2) = 78 Rank
> > Forms. Patterns for suits may be SH vs HS, SH vs HD, SH vs DS,
> > SH vs DC, SS vs HH, SS vs HD. 6 * 78 = 468.
>
> "pair against no pair, no pair SUITED"?
Yes, 4) is missing SH vs DD, SH vs SS
Also, there is a duplicate in 5) since SH vs HD and SH vs DS are the same.
Mike Caro
Barbara Yoon has declared the winner to be Erik Reuter. Thanks to
everyone for participating. Please see separate post ("Yoon Declares
Reuter Winner") for details.
Barry Paul
On Wed, 15 Oct 1997 20:59:08 GMT, ca...@caro.com (Mike Caro) wrote:
Barbara Yoon and one other r.g.p'r (sorry I no longer have the article
or its author's name) have posted corrections to my computations.
I have redone it from scratch and now believe (well actually "believe"
is probably too strong a word) the answer to be 47,008. I believe I
have seen that number from a couple of other participants so -- who
knows -- it might even be correct.
One comment. A whole bunch of people who well understood how to answer
the question came up with a variety of answers. This suggests that for
a programmer to do this kind of computation and rely on the answer
could be dangerous. One way of avoiding the danger is to write a
program to solve the problem. Another is to have a lot of people do
the computation and compare results and methodology. As to how to get
a whole lot of people to work on such a problem ... {:-)
We write C(n,k) for "combinations of n things taken k at a time"
1) Pair against same pair e.g. AA vs AA: 13
2) Pair against different pair e.g. AA vs KK:
C(13,2) = 78 pair combos.
suit patterns: SH vs SH, SH vs SD, SH vs DC 78 * 3 = 234
3) Pair against no pair, one rank matches e.g. AA vs AK
13*12 = 156 rank combos
suit patterns: SH vs DS, SH vs DD, SH vs DC 156 * 3 = 468
4) Pair against no pair, no matching rank. e.g. AA vs KQ
Pick two ranks for the no pair then one of remaining
ranks for the pair. C(13,2) * 11 = 858 rank combos.
suit patterns: SH vs SS, SH vs SH, SH vs SD, SH vs DC,
SH vs DD, SH vs DC 6 * 858 = 5148
5) No pairs, two matching ranks. e.g. AK vs AK C(13,2) = 78 rank
combos. Suit patterns: SS vs HH, SS vs HD, SH vs HS, SH vs DS,
SH vs DC. 5 * 78 = 468.
6) No pairs, one matching rank. e.g. AK vs AQ. Pick the two
unmatched ranks then pick the matched rank from the remaining
ranks. C(13,2) * 11 = 858 rank combos. Suit patterns: SS vs HH,
SS vs HS, SS vs HD, SH vs HS, SH vs HH, SH vs HD, SH vs DS,
SS vs DH, SS vs DD, SS vs DC. 10 * 858 = 8580.
7) No pairs, four different ranks. e.g. AK vs QJ. Pick the four
ranks then pick one of the three lower ranks to join the highest
rank C(13,4) * 3 = 2145. Suit patterns: SS vs SS, SS vs SH,
SS vs HS, SS vs HH, SS vs HD, SH vs SS, SH vs SH, SH vs SD,
SH vs HS, SH vs HH, SH vs HD, SH vs DS, SH vs DH, SH vs DD,
SH vs DC. 15 * 2145 = 32,175
GRAND TOTAL: 13 + 234 + 468 + 5148 + 390 + 8580 + 32,175
= 47,008