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10-18-2008, 06:48 PM
Post subject: Poker math problem
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#1 (permalink)
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Doing mathematical calculations in poker is important. We did some math problems in Ventrilo, but I still liked the first problem we discussed the best.
What is the chance to get at least two diamonds in your five card poker hand? Assume you just got dealt five cards in draw poker. Answer will be posted after everyone posts theirs.
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Ragnar4
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Full House
Join Date: Sep 2005
Location: Billings, Montana
Posts: 1,284
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1 in 4?
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The older I get, the more I start wondering; Just what in the hell is going on here? 
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How is that even logical?
you have 5 cards
you have 1 in 4 chance of getting one diamond for each card
you need two diamonds
should be closer to 1/2 (roughly) but I'm looking for exact answers
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ponyboy
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10-18-2008, 10:19 PM
Post subject: Re: Poker math problem
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#4 (permalink)
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Flush
Join Date: Jan 2005
Posts: 379
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Quote:
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Originally Posted by iopq
Doing mathematical calculations in poker is important. We did some math problems in Ventrilo, but I still liked the first problem we discussed the best.
What is the chance to get at least two diamonds in your five card poker hand? Assume you just got dealt five cards in draw poker. Answer will be posted after everyone posts theirs.
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Wouldn't it be 13/52 (odds of first card) times 12/51 (odds if you get first card you get second card)? That would be 156/2652 or 5.8%. Or loosely one in twenty.
Edit: Sorry, that would be for two diamonds dealt first two cards. Not sure how that translates into a five card hand.
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Ragnar4
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Full House
Join Date: Sep 2005
Location: Billings, Montana
Posts: 1,284
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2 or more diamonds.. IE you could land 2 diamonds, 3 diamonds, 4 diamonds or 5 diamonds and satisfy your need for a girls best friend.
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The older I get, the more I start wondering; Just what in the hell is going on here?
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swiggidy
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10-18-2008, 11:59 PM
Post subject: Re: Poker math problem
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#6 (permalink)
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4-of-a-Kind
Join Date: Sep 2005
Location: Waiting in the shadows ...
Posts: 3,777
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Quote:
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Originally Posted by iopq
What is the chance to get at least two diamonds in your five card poker hand? Assume you just got dealt five cards in draw poker. Answer will be posted after everyone posts theirs.
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This really isn't relevant to any calculation you would ever do at the table.
There's lots of cool math here:
http://www.math.sfu.ca/~alspach/
that won't help your game in anyway
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(\__/)
(='.'=)
(")_(")
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10-19-2008, 12:04 AM
Post subject: Re: Poker math problem
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#7 (permalink)
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Quote:
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Originally Posted by swiggidy
Quote:
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Originally Posted by iopq
What is the chance to get at least two diamonds in your five card poker hand? Assume you just got dealt five cards in draw poker. Answer will be posted after everyone posts theirs.
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This really isn't relevant to any calculation you would ever do at the table.
There's lots of cool math here:
http://www.math.sfu.ca/~alspach/
that won't help your game in anyway |
So what? I'm sure you could do this problem, but getting to the point where you can solve poker probabilities will let you analyse AWAY from the table.
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Ragnar4
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Full House
Join Date: Sep 2005
Location: Billings, Montana
Posts: 1,284
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The thing that's throwing me the most, is that the order doesn't matter. As long as you end up with 20% of your cards being diamonds after having been dealt those cards, you don't care how you got them. That means we have to devide our permutations somewhere down the road... but I for the life of me can't remember when or what.
Blah. Stupid probability. If I ever go back to school, I so totally taking a probability class.
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The older I get, the more I start wondering; Just what in the hell is going on here?
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daven
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10-19-2008, 12:22 AM
Post subject: Re: Poker math problem
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#9 (permalink)
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Straight Flush
Join Date: Aug 2007
Location: soaking up ethanol, moving on up
Posts: 5,805
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Quote:
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Originally Posted by iopq
What is the chance to get at least two diamonds in your five card poker hand? Assume you just got dealt five cards in draw poker. Answer will be posted after everyone posts theirs.
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good question. Problem is that those who can answer it easily won't bother, and those who can't do so easily won't bother either cos they're too lazy to get better - instead they'll read the answer and fool themselves that they understand.
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I don't think anyone can answer this "easily" it takes a few minutes anyway to get the correct answer. I encourage everyone to take a shot at it. If you're sure your answer is 100% correct don't post it until everyone else tries it.
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settecba
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Flush
Join Date: Aug 2008
Location: stealing blinds from UTG
Posts: 326
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58.82% ???
I wont post how i did it until you post the answer...or should I?
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Quote:
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Originally Posted by ISF
Getting good at poker is like that scene in the matrix where Neo suddenly sees that everyone is just a bunch of structured numbers and then he starts bending those numbers in really weird ways.
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settecba
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Flush
Join Date: Aug 2008
Location: stealing blinds from UTG
Posts: 326
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no more answers?
will you post the answer iopq?
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Quote:
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Originally Posted by ISF
Getting good at poker is like that scene in the matrix where Neo suddenly sees that everyone is just a bunch of structured numbers and then he starts bending those numbers in really weird ways.
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swiggidy
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10-20-2008, 03:55 AM
Post subject: Re: Poker math problem
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#13 (permalink)
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4-of-a-Kind
Join Date: Sep 2005
Location: Waiting in the shadows ...
Posts: 3,777
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Quote:
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Originally Posted by iopq
So what? I'm sure you could do this problem, but getting to the point where you can solve poker probabilities will let you analyse AWAY from the table.
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Knowing how to find 2 or more of a specific suit in a 5 card hand does not help solve any poker problem.
I get ~36%
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(\__/)
(='.'=)
(")_(")
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Quote:
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Originally Posted by settecba
no more answers?
will you post the answer iopq?
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yes I'll post it
but solving this hones your MATH skills, not poker skills
but those MATH skills let you do some poker analysis
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pilipolio
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Join Date: Dec 2007
Posts: 50
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That's usual combinatorics, but I always sucked at it 
Chance to be dealt a first diamond : 13/52
Chance to be dealt the second one : 12/51
The three other cards don't matter as they can be diamond or whatever
The chance that at least the two first card of your 5 cards hand are 13/52 * 12/51,
But those 2 diamonds can actually appear in any of the 5 spots and the number of possible combinations is (without repetitions) 5*4 / 2 = 10
My result is therefore 10 * 13/52 * 12/51 = 58,82% as settecba said. That seems a bit high at the fist sight, so I might be off the point.
Thanks iopq for posting math problem, but it is true that I don't see a big interest of such calculation while playing
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Here's the answer:
We can calculate it two different ways, either by directly calculating the chance of getting 2,3,4,5 diamonds, or by calculating the chance of NOT getting 1 or 0 diamonds. As you can see, the second way is twice faster.
1-(C(39,5)/C(52,5) + C(13,1)*C(39,4)/C(52,5)) which is 36.70%
or you could do it the long way which is
C(39,3)*C(13,2)/C(52,5) + C(39,2)*C(13,3)/C(52,5) + C(39,1)*C(13,4)/C(52,5) + C(13,5)/C(52,5)
also giving us the same result
swiggidy and spoon are the only people who got it right
pilipolio and settecba did it wrong, because they overcounted
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Cardsharp
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Join Date: Aug 2008
Posts: 17
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25%
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Quote:
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Originally Posted by Cardsharp
25%
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wrong
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lolzzz_321
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NO YOU
Join Date: Oct 2004
Location: My ice is polarized
Posts: 2,797
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50/50 either you get two diamonds, or you don't
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Quote:
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Originally Posted by Triptanes
50/50 either you get two diamonds, or you don't
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oh, you
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settecba
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Flush
Join Date: Aug 2008
Location: stealing blinds from UTG
Posts: 326
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thanks iopq, i also think solving math problems IS going to help our game.
i had trouble at first figuring out where pilipolio and i messed up, i can see it now...
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Quote:
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Originally Posted by ISF
Getting good at poker is like that scene in the matrix where Neo suddenly sees that everyone is just a bunch of structured numbers and then he starts bending those numbers in really weird ways.
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here's how you guys messed up:
imagine a simpler problem:
what's the chance of getting 1 black card in a hold'em hand?
you say, "well, the chance of getting a black card is 1/2, there's two cards so we will get a black card 100% of the time"
DUCY this is wrong
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settecba
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Flush
Join Date: Aug 2008
Location: stealing blinds from UTG
Posts: 326
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yes..of course i see...thanks again, great example
BTW...to see if i get this right...the chance of getting at least 1 black card in a hold´em hand would be:
1-C(26,2)/C(52,2)=75.49%
right? or did i mess up again?
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that's correct
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swiggidy
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4-of-a-Kind
Join Date: Sep 2005
Location: Waiting in the shadows ...
Posts: 3,777
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I like counting like IOPQ did. You can also use the probabilities
0.25 diamond
0.75 not a diamond
Exactly 2:
0.25^2 * 0.75^3 * 5! / (2! * 3!)
Seriously check out the link I posted. Counting the number of unique Omaha hands is fun.
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(\__/)
(='.'=)
(")_(")
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the problem of course is after you have one diamond the probability of getting a second one decreases so your answer is not exact, but of course not far off
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sarbox68
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Full House
Join Date: Sep 2006
Location: wondering where the 3 extra chairs at my 6max table came from
Posts: 871
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42!
Oh f-ck .... wrong question....
Dude, last time I did this kinda math was 10 years ago studying for the f-in GMAT. I get your point but I'ma have to find another option than mental combinatronics where close guesstimation is good enough... 
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