Poker math problem

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.

[Ragnar4]

1 in 4?

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

[ponyboy]

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.

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.

[Ragnar4]

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.

[swiggidy]

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.

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

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.

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.

[Ragnar4]

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.

[daven]

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.

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.

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.

[settecba]

58.82% ???

I wont post how i did it until you post the answer...or should I?

[settecba]

no more answers?

will you post the answer iopq?

[swiggidy]

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.

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%

no more answers?

will you post the answer iopq?

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

[pilipolio]

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

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

[Cardsharp]

25%

25%

wrong

[lolzzz_321]

50/50 either you get two diamonds, or you don't

50/50 either you get two diamonds, or you don't

oh, you

[settecba]

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...

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

[settecba]

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?

that's correct

[swiggidy]

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.

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

[sarbox68]

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...