OK

[Jenisi]

I need somone's help with this.. You have a deck of 52 cards... 4 suites.. club, diamond, heart and spade... if you shuffle the cards, and draw the top two cards (ignoring that you already drew one of a current suite) what is your probability or having the the suite you want (chose beforehand) drawn on the top two cards.

[Bobmuhthol]

I think you have a 50:50 chance.

14\52 on the first one and 14\51 on the second one. I'm pretty sure for both together its
14\52 + 14\51

[Bobmuhthol]

13/52 + 13/51. 52 / 4 = 13.

[Jenisi]

so who's right??? lol

[Jenisi]

basically, would it be closer to a 50% chance or a 25% chance?

[Bobmuhthol]

It would be within a percent of 50.

yea its 13\52 + 13\51. My bad

[Jenisi]

So it would only be a 2/4 chance to get it even with there being 52 total cards? not a 2/8 chance ??

[Edited on 1-4-2004 by Jenisi]

[Xcalibur]

I still think it's a bit less than 25%

[Edited on 5-1-04 by Xcalibur]

[Bobmuhthol]

There are 4 suits. Your chances are 1/4 to get the suit you want on the first try. On the second try, your chance is 1/4 again. 1/4 + 1/4 = 2/4 or 1/2.

[Xcalibur]

so by your logic, if you took 4 cards, you have 100% of chance?

Hmm...

[Bobmuhthol]

Correct. Doesn't mean it will happen. That's the logical chance that it will.

[Jenisi]

that what I'm not getting

[Bobmuhthol]

Explain your answer then, because me and Yarans have the same answer and we've both explained it. All you've done is disagree and try to disprove us.

I think i'm wrong, but i need to hop in the shower then i'll think about it.

[peam]

/lights up the Kranar-signal.

[Xcalibur]

You have LESS chance to be right on each pick

By your logic, at 8 cards you are 200% sure to be right to take 8 cards of the same suit from a deck of 52 cards.

Illogical

My high school years are far from me, but i'm sure you don't add the variable in that case.

Actually, you have mor eof a chance with each proceeding card, assuming you don't pick your suit. I forget what you do to the 2 different numbers but i'll think of it when i finish my business.

My first reaction was you would add them and get an average, but for some reason that sounds wrong. I'll have to go through and test it down to where i know its 100% and see if it follows through.

[Edited on 1-5-2004 by RangerD1]

[Bobmuhthol]

Shut the hell up Xcalibur, I just told you explain your answer and you try to disprove mine again.

1. Fuck you.
2. By YOUR 1/4 logic, 8 cards is 200%. You're disproving the wrong answer.

[Xcalibur]

Bob, shut up you 31 years old liar.

n!/n!*(n-x)!

That is the correct formula

[Edited on 5-1-04 by Xcalibur]

[Bobmuhthol]

Let's see if Xcalibur can try to turn this into an 'illogical' answer...

Your chances are indeed close to 25% for two cards. For the first card, your chance was exactly 25%. But when you pull the second card, there's one less card in the deck, so your chances are 13/51, ~25.4%. For each subsequent card, your chances increase in an irregular but linear pattern. The chance of the first and second pick are completely irrelevant to each other, and at your 40th card, there's a 100% chance of the suit you wanted.

Edited to add: Xcalibur is dumb.

Correct formula:
x = 13/(53-x)
0 < x < 41

A bit simpler than n!/n!*(n-x)! I think.

[Edited on 1-5-2004 by Bobmuhthol]

Originally posted by Xcalibur
Bob, shut up you 31 years old liar.

n!/n!*(n-x)!

That is the correct formula

[Edited on 5-1-04 by Xcalibur]

What do the variables stand for.

Okay i read it wrong. Its if you wan thearts and you want the top 2 cards to be hearts. Thats completely different then what i was thinking.

It would be 13\52 *12\51. However, thats the proability that *both* cards would be hearts. But your question was whats the probability of picking the second card if you had already picked a heart. So your answer is 12\51.

[Xcalibur]

My first thought was 13/52*12/51

AS you want to find the % of chance you have the 2 cards of the same suits chosen before.

[Bobmuhthol]

The question is what's the probably of picking the top card's suit again within the next two cards? WTF!

[Edited on 1-5-2004 by Bobmuhthol]

[Sean]

but if i read the rules correctly X arent you supposed to ignore that you drew a card of the current suit for the second one.

so wouldnt it be (13/52)*(13/51)

Okay, you need to clarify the question. fuck this

[Xcalibur]

so to have, for exemple, 4 cards from heart chosen from top

(13/52)*(12/51)*(11/50)*(10*49)

chance DECREASES

and parantheses are important :lol:

[Edited on 5-1-04 by Xcalibur]

[Soulpieced]

I'm confused on the actual question. Do you want to know the probability that BOTH cards are the suit guessed, or that either of them is appropriate.

You have a deck of cards and guess spades.
Do you want to know if both of the 2 cards are spades
OR
Do you want to know if either of the 2 cards are spades.

[Edited on 1-5-2004 by Soulpieced]

[Bobmuhthol]

I've already answered the one of two cards being the suit question, so if it's both.. then Jenisi doesn't know how to write a question.

[Bobmuhthol]

<<so to have, for exemple, 4 cards from heart chosen from top

13/52*12/51*11/50/10*49

chance DECREASES>>

That completely destroys the law of probability.

You can't put in a scenario and just say "You picked all hearts, the chance of picking another decreases."

Also, that has nothing to do with the question.

[Xcalibur]

I did not understood her question well then, as many people here.

My understanding was from a deck of 52 cards, what is the probability of having (choising it before) 2 cards of the same suit

[imported_Kranar]

It's a tad bit harder than the answers provided. I will give you the general answer, so regardless of what you asked you can use this information to answer your question, because I too am a bit confused.

First card you pick you have a 13 / 52 chance that it will be the suit you want. That's Term1

Second card you pick will have a (13 / 51) chance that it will be the suit you want, this is Term2. This happens only if the first card you picked was not the suit you wanted (which happens 39 / 52 times which will be known as Term3).

Otherwise the second card you pick will have a 12 / 51 to be the suit you want. This is Term4. The 12 / 51 chance happens if the first card you picked was actually the suit you wanted. This happens 13/52 times which is Term1.

You now do the following:

Odds that atleast one card will be the suit you want is:

Term1 + Term2*Term3 + Term1*Term4

13 / 52 + (13 / 51)*(39/52) + (13 / 52)*(12/51) = 50 percent.

Now the odds that BOTH cards will be the same suit, and the suit that you want are as follows:

Term1*Term2

(13/52)*(12/51) = 5.88235 percent.

Edited to fix a calculator error. I hate using Windows Calculator.

[Edited on 1-5-2004 by Kranar]

[Bobmuhthol]

So I was right to begin with.

GOOD ONE XCALIBUR.

[imported_Kranar]

<< Correct. Doesn't mean it will happen. That's the logical chance that it will. >>

That's wrong. If something has a 100 percent chance of happening, then it MUST happen 100 percent of the time.

You can prove it to yourself by remembering that the sum of all possible outcomes is always 100 percent. If only one outcome has a 100 percent chance of occuring, then that means there are no other possible outcomes.

[Xcalibur]

You were wrong in the first place by adding each probability, BOBMUHTHOL.

I (with ranger) were right after, and you came after, and we're still not knowing what was the real question

[imported_Kranar]

It's not a pissing contest.

[Bobmuhthol]

<<You were wrong in the first place by adding each probability, BOBMUHTHOL.>>

Doesn't matter how I got it, I said the answer was within 1 percent of 50%. You lose, you lose, you lose.

It is a pissing contest now. Also..

I won.

[Xcalibur]

Eh, all male confrontations is a pissing contest :lol:

[Drew2]

Both of you grow up please. The question is answered, drop it.

[Solkern]

Theres 13 of each suit, so in turn you'd have a 25% chance of drawing the suit you want.

[Bobmuhthol]

Wow did you come up with that yourself or did the last 25 posts give you a hint?

Be careful bob. Hes cooler than you.

[Solkern]

I haven't posted here in the past 2 days, I have a life bob, thank you

[Bobmuhthol]

Okay, you haven't posted for 2 days. What does that have to do with the fact there are 25 posts about the question, and it's already been solved, and the answer is not 25%?

Solkern is a n00b from Needs to STFU World.

[Solkern]

I'm a newbie? heh riight.

[Bobmuhthol]

<<I haven't posted here in the past 2 days, I have a life bob, thank you>>

If you cared to notice, there's a gap between my posts. September 28 through December 27 I didn't post at all. I have 45x the life you do I guess.

[Solkern]

The gap? was it cause maybe you were banned?

[Solkern]

BOb, I've been around gs alot longer then you have, so refrain from making comments that aren't true.

[Soulpieced]

Both of you grow up please. The question is answered, drop it.

.

Sweet mother of mercy, Tayre telling someone to shut up. The world is about to end.

[Bobmuhthol]

http://forum.gsplayers.com/viewthread.php?tid=3828

It already has.

[Soulpieced]

Point noted.

[Drew2]

Whens the last time I sat here and argued with anyone over anything dumb.

I think I reserve that right.

That some sort of trick question?

[Drew2]

I say dumb things. I don't argue over dumb things. Big difference.

Or maybe not so big, but the point still remains.

[Edited on 1-5-2004 by Tayre]

[Wezas]

A bit late to the conversation, but the formula someone posted is correct

Drawing one card of a suit is 25%

Drawing 2 cards:
(13/52)*(12/51) = 5.8824% chance (or roughly 1 in 17 chance)

[Edited on 1-5-2004 by Wezas]