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09-14-2010 , 01:26 PM
Quote:
Originally Posted by youtalkfunny
OK, if we're doing puzzles, here's two:

1. Playing Omaha high, you are dealt 9333, including two spades. There are three hands you can make that would be the nuts. Name all three.
Spoiler:

You'd have the nuts if the board read:

999XX

AKQJTs

22223
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09-14-2010 , 01:41 PM
Quote:
Originally Posted by Lyrrad
If a + b + c + d + e = 20, then 35a + 17b + 11c + 8d + 5e would be 40 more than a multiple of 3.
Er, then the previous guy was wrong. And for that matter, if the bet is 3, then the payout is surely a multiple of it. Hmm. Well, it can't be 500, it seems, I'm pretty sure of that.
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09-14-2010 , 02:01 PM
Quote:
Originally Posted by pfapfap
Answers for ytf:

Spoiler:
99345
Spoiler:
You lose to 95?
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09-14-2010 , 02:33 PM
Quote:
Originally Posted by Meraxes
Spoiler:

You'd have the nuts if the board read:

999XX
Spoiler:
Wouldn't that lose if the xx was a pair bigger than 9s (TT, JJ, QQ, KK, AA) to a villain with a pair of those as their hole cards?
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09-14-2010 , 02:45 PM
Quote:
Originally Posted by NYCNative
Spoiler:
Wouldn't that lose if the xx was a pair bigger than 9s (TT, JJ, QQ, KK, AA) to a villain with a pair of those as their hole cards?
Spoiler:

You're absolutely right, XX would have to exclude pairs bigger than 9.


This line of replies is starting to feel very clandestine.
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09-14-2010 , 02:51 PM
Quote:
Originally Posted by youtalkfunny
OK, if we're doing puzzles, here's two:

1. Playing Omaha high, you are dealt 9333, including two spades. There are three hands you can make that would be the nuts. Name all three.
Spoiler:
Community of AKQJ10 suited would not be the nuts. You can't play the board in Ohama. If it was the board, you would with have a flush AKQ93 or a pair of three with AKQ kickers.

But I agree a board of 999xx, provided xx are not paired, would the best hand (99993) you can hope for. Can't think of what the other two could possibility be.
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09-14-2010 , 02:54 PM
Quote:
Originally Posted by pfapfap
Okay, I suck. I can do the elementary stuff of getting it down to four variables and each as an expression of the other, but after that I keep ending up with 0=0, so I don't know what else to do. But here's some stuff if someone else wants a crack:

35a + 17b + 11c + 8d + 5e = 500
a + b + c + d + e = 20

a = 40/3 - 2b/5 - c/5 - d/10
b = 100/3 - 5a/2 - c/2 - d/4
c = 200/3 - 5a - 2b - d/2
d = 400/3 - 10a - 4b - 2c

As with most riddles, I'm sure there's an easy obvious answer here. Knowing that they need to be integers may be a key. Or maybe we just misunderstand the problem, since now he's saying outside bets. (Or maybe I don't know what outside bets means.)

EDIT: Tho' looking at these, it's impossible for them to be integers. I think the simple "always one fewer than a multiple of 3" answer may be correct, meaning that it can't be done.
A system of equations can only be uniquely solved if you have at least as many equations as you have variables. There are infinite answers here. However, most will not fit the criteria of natural number solutions.
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09-14-2010 , 03:10 PM
Quote:
Originally Posted by The Big K
Spoiler:
Community of AKQJ10 suited would not be the nuts. You can't play the board in Ohama. If it was the board, you would with have a flush AKQ93 or a pair of three with AKQ kickers.

But I agree a board of 999xx, provided xx are not paired, would the best hand (99993) you can hope for. Can't think of what the other two could possibility be.
Definitely still the nuts.. Play the 9 3 of spades.. what flush can beat you?

999xx and 22223 are the other two.. Meraxes ftw!
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09-14-2010 , 03:10 PM
Quote:
Originally Posted by Nutcracker19
ive been assured that the payout of exactly 500 from 20 chips IS possible. outside bets only. all chips must be winning chips. 501 is defo possible (will post solution to that soon). 500 is the payout and does not include the 20 wagered.

ps: thx for efforts
It seems that your friend is jerking your chain. I've tried every way I can think of and the closest I can get is 499. I can only get 500 with 19 chips or on an american (double zero) wheel.
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09-14-2010 , 03:17 PM
This was an awesome thread until the math nerds started nitting it up.
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09-14-2010 , 03:24 PM
Quote:
Originally Posted by Dealer-Guy
Think about how the kitten must feel every single day?
Dealer-Guy FTW
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09-14-2010 , 03:25 PM
Quote:
Originally Posted by YouCheckRaise

Question 2

Spoiler:
3 of a kind 10's?
If that was top set, then a straight is possible, and you don't have the nuts.

Meraxes got the Omaha question correct; pfap the hold'em question.
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09-14-2010 , 03:30 PM
Quote:
Originally Posted by Meraxes
Spoiler:

You're absolutely right, XX would have to exclude pairs bigger than 9.


Spoiler:
This line of replies is starting to feel very clandestine.
Spoiler:
FYP
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09-14-2010 , 03:33 PM
Quote:
Originally Posted by ButTheyreSuited!!
Definitely still the nuts.. Play the 9 3 of spades.. what flush can beat you?
Sorry, you are correct. Failed to see beyond the board while ironically stating you can't play the board.
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09-14-2010 , 04:20 PM
Quote:
Originally Posted by pfapfap
Or maybe we just misunderstand the problem, since now he's saying outside bets.
Typo. He meant "inside" bets. The problem is impossible without inside bets since 20 wagers ain't gonna get you back 500 at 2:1.

It ain't rocket science. It's computer science. It's only 3.2 million possibilities.

% cat roulette.c
#include <stdio.h>
main() {
int a,b,c,d,e,sum,tot;
for (a = 0; a < 21; a++ )
for (b = 0; b < 21; b++ )
for (c = 0; c < 21; c++ )
for (d = 0; d < 21; d++ )
for (e = 0; e < 21; e++ ) {
sum = a + b + c + d + e;
tot = 35*a+17*b+8*c+11*d+5*e;
if ( sum == 20 && tot > 495 && tot < 504 )
printf("%d %d %d %d %d %d\n",tot,a,b,c,d,e);
}
}

% gcc -O3 roulette.c
% ./a.out |& sort
496 10 6 0 4 0
496 10 7 0 2 1
496 10 7 2 1 0
496 10 8 0 0 2
496 11 2 0 7 0
496 11 3 0 5 1
496 11 3 2 4 0
496 11 4 0 3 2
496 11 4 2 2 1
496 11 4 4 1 0
496 11 5 0 1 3
496 11 5 2 0 2
496 12 0 0 6 2
496 12 0 2 5 1
496 12 0 4 4 0
496 12 1 0 4 3
496 12 1 2 3 2
496 12 1 4 2 1
496 12 1 6 1 0
496 12 2 0 2 4
496 12 2 2 1 3
496 12 2 4 0 2
496 12 3 0 0 5
496 13 0 0 1 6
496 13 0 2 0 5
496 9 10 0 1 0
499 10 7 1 2 0
499 10 8 1 0 1
499 11 3 1 5 0
499 11 4 1 3 1
499 11 4 3 2 0
499 11 5 1 1 2
499 11 5 3 0 1
499 12 0 1 6 1
499 12 0 3 5 0
499 12 1 1 4 2
499 12 1 3 3 1
499 12 1 5 2 0
499 12 2 1 2 3
499 12 2 3 1 2
499 12 2 5 0 1
499 12 3 1 0 4
499 13 0 1 1 5
499 13 0 3 0 4
502 10 7 0 3 0
502 10 8 0 1 1
502 10 8 2 0 0
502 11 3 0 6 0
502 11 4 0 4 1
502 11 4 2 3 0
502 11 5 0 2 2
502 11 5 2 1 1
502 11 5 4 0 0
502 11 6 0 0 3
502 12 0 0 7 1
502 12 0 2 6 0
502 12 1 0 5 2
502 12 1 2 4 1
502 12 1 4 3 0
502 12 2 0 3 3
502 12 2 2 2 2
502 12 2 4 1 1
502 12 2 6 0 0
502 12 3 0 1 4
502 12 3 2 0 3
502 13 0 0 2 5
502 13 0 2 1 4
502 13 0 4 0 3
502 13 1 0 0 6
502 9 11 0 0 0

Took 0.008 seconds to determine it's not possible.
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09-14-2010 , 04:25 PM
Too much math, but I got the Omaha and Hold'em questions right, even though I initially forgot about the 999xx where x<9.
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09-14-2010 , 04:28 PM
Quote:
Originally Posted by bav

Took 0.008 seconds to determine it's not possible.
Are you claiming to have entered that code in .008 seconds?
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09-14-2010 , 04:32 PM
I was going to try something like that in C++ but I wasn't sure how to approach the nested FOR loops. FWIW, what you did is kind of what I was thinking, but I shyed away from it thinking there had to be a better, more efficient way. I guess I didn't take in to account that even a few million iterations isn't much on a newer processor.
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09-14-2010 , 04:48 PM
Spoiler:
For the second question, at first I thought it was a set of Jacks on a board of JT762, but a straight is possible with 89. The answer is on a set of Queens on a board of QJ762 (several other queen-high non-connected boards will satisfy this too).
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09-14-2010 , 05:04 PM
Quote:
Originally Posted by Meraxes
Spoiler:

You'd have the nuts if the board read:

999XX

If the xx are 78 and they're suited and someone holds the JTxx of the same suit wouldn't a straight flush beat 4 of a kind?
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09-14-2010 , 05:06 PM
Quote:
Originally Posted by psandman
Are you claiming to have entered that code in .008 seconds?
Overhead costs are amortized.

I actually did the code a couple days ago before my first post saying it was impossible--didn't really figure I needed to post it. But somehow the debate continued, so fine... there wasn't much effort to rerun it and cut/paste.

If you add "f" and "g" bets at 2:1 and 1:1, it'll spit out lots of ways to hit 500 while taking a whopping 1.8 seconds to run--takes a bit longer to enumerate 1.28 billion possibilities (kinda mind blowing for someone who's been doing this computer stuff for over 30 years that this 3-year-old Intel CPU can compute 1.28E9 roulette payouts of 7 bets in less than 2 seconds using one of four CPU's--there was a time that 10 billion calculations might take a fair while).

Also note you can optimize the loops trivially and make it run a gazillion times faster since I'm not using the sum having to be 20 to stop extra loops--if a = 20, then b,c,d,e must be 0 and there's no point iterating over all of them. But when it takes <2 seconds to run unoptimized, worrying about making it run 1000x faster ain't real important.

This stuff is SOOO off topic.

Last edited by bav; 09-14-2010 at 05:18 PM.
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09-14-2010 , 05:09 PM
You mean most casino dealers don't daylight as computer programmers?
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09-15-2010 , 12:51 AM
Quote:
Originally Posted by C Bids
If the xx are 78 and they're suited and someone holds the JTxx of the same suit wouldn't a straight flush beat 4 of a kind?
Oh, FFS!

EVERYONE FORGET THAT YOU EVER READ THE TEXT "999xx"!

The answer is:

--quad 9's (the easiest answer)

--flush, if board is AKQJTs (the answer that takes a little thought)

--top boat with board of 32222 (the answer that takes more thought)
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09-15-2010 , 03:59 AM
That is pretty typical of me. I screwed up the easy one, but got the hard one right!
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09-15-2010 , 04:25 AM
Quote:
Originally Posted by YouCheckRaise
Spoiler:
You lose to 95?
Yeah, well... shut up.

Quote:
Originally Posted by bav
It ain't rocket science. It's computer science. It's only 3.2 million possibilities.
I enjoyed my paper and pencil approach, so nyeah.

Tho' it did take me a while to notice that it was impossible for the result to be integers, but I was too focused on trying to isolate variables in a situation where they couldn't be isolated.
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