easy poker question, debate ensues

[LongLiveYorke 38]

Just pretend that you have a deck made out of four cards, two aces and two kings. If your opponent has a pocket pair, you are 100% to have a pocket pair. If your opponent is unpaired, you are 0% to have a pocket pair. The same idea holds for a bigger deck. Interesting question, Tim.

[NickG 0]

If you have AA, there is 1 combination of card where your opponent could have AA, and 6 combinations for each of 22-KK, for a total of 1 + (6*12) = 73 combinations.If you have AK, there are 3 combinations of cards where you opponent could have each of AA and KK, and 6 combinations for each of 22-QQ, for a total of 3 + 3 + (6*11) = 72 combinations.So it is very slightly more likely that your opponent will hold a pocket if you do also.

[doublemeup 0]

If you have AA, there is 1 combination of card where your opponent could have AA, and 6 combinations for each of 22-KK, for a total of 1 + (6*12) = 73 combinations.If you have AK, there are 3 combinations of cards where you opponent could have each of AA and KK, and 6 combinations for each of 22-QQ, for a total of 3 + 3 + (6*11) = 72 combinations.So it is very slightly more likely that your opponent will hold a pocket if you do also.

ok his example is pathetic because theres no deck that has 4 cards. Think of this: How many pocket pairs are? Now, how many upaired combinations are there?

[Abbaddabba 0]

This is the right answer:

if you have unpaired cards, the probability of him having a PP is(44/50)x(3/49)+(6/50)x(2/49) If you have paired cards....(48/50)x(3/49)+(2/50)x(1/49)I'm too lazy to find a calculator.

Unpaired cards 0.0589Paired cards .119

This is not:

48/50 * 3/49 + 2/50 * 1/49144/2450 + 2/2450 = 146/2450versus44/50 * 3/49 + 6/50 * 2/49132/2450 + 12/2450 = 146/2450OOPS! It's the exact same.

Err... The sum of two fractions with the same denominator is just the sum of the two enumerators over that denominator, is it not? Yeah, i realize the simple addition error, but the figure is only off by a tiny bit. You bolded the one where we had (essentiallky) the same figure.It's the other figure where there's a big discrepency.

(44/50)x(3/49)+(6/50)x(2/49) If you have paired cards....(48/50)x(3/49)+(2/50)x(1/49)I'm too lazy to find a calculatorUnpaired cards 0.0589Paired cards .119

(48/50)x(3/49)+(2/50)x(1/49) does not equal .119, unless i just had a huge brain fart.As it turns out, it's 5.8% versus 5.9%. Definitely not 5.8 versus 11.9% though. On intuition alone, you can see that it would never make it TWICE as likely to be dealt a pocket pair for something so trivial.

[doublemeup 0]

so your conclusion from all that pointless math is......?

[Suited_Up 2]

so your conclusion from all that pointless math is......?

You might be the most thick headed, ignorant, flat out stupid person I've ever come across in my life.

[speedz99 145]

so your conclusion from all that pointless math is......?

I'd like to point out the fact that you suck at life.

[Wingmaster05 0]

so your conclusion from all that pointless math is......?

...That it doesn't matter.

[doublemeup 0]

yea thats what i thought.

[princeof56k 0]

I love how tim posted this in general instead of strat.In strat this thread would be over quickly as someone would figure out the correct answer, people would check it over and that would be it.In general this is destined to become a 3 page thread where people continue to misunderstand the question and will defend their incorrect answere on a clear cut issue til the end of time.You have a strange sense of humor tim.

[SapphireTiger 0]

so your conclusion from all that pointless math is......?

hey fucker! You dont' know what you're talking about! Have a good day!

[SapphireTiger 0]

I love how tim posted this in general instead of strat.In strat this thread would be over quickly as someone would figure out the correct answer, people would check it over and that would be it.In general this is destined to become a 3 page thread where people continue to misunderstand the question and will defend their incorrect answere on a clear cut issue til the end of time.You have a strange sense of humor tim.

i think that general needs more strategy and that it's good for people who don't go to strat to actually think for once. No problem with posting in general.

[Petoria 0]

I got the answer on the 5th response, that's pretty quick for general. Especially, considering the 4 flame minimum before a serious post.

[DDiabolical 0]

wow you people are retards.

[Hand_Cracker 0]

If you are playing hold'em heads up, is you opponent more likely to have a pocket pair when you have unpaired cards or when you have a pocket pair?I know the answer, but thought it was interesting.

more likely

[mrdannyg 274]

If you have unpaired cards, each card blocks three possible PP combinations, so you're blocking 6 total.If you have paired cards, there's only one possible remaining PP combination of the card you hold, so you're blocking 5 total.  It's more likely your opponent has a PP.

it's so short and simple, most of us skipped right past it.thanks garamond.times new roman!

[NickG 0]

If you have AA, there is 1 combination of card where your opponent could have AA, and 6 combinations for each of 22-KK, for a total of 1 + (6*12) = 73 combinations.If you have AK, there are 3 combinations of cards where you opponent could have each of AA and KK, and 6 combinations for each of 22-QQ, for a total of  3 + 3 + (6*11) = 72 combinations.So it is very slightly more likely that your opponent will hold a pocket if you do also.

ok his example is pathetic because theres no deck that has 4 cards. Think of this: How many pocket pairs are? Now, how many upaired combinations are there?

What do you mean no deck has 4 cards? Where are you getting 4 cards from?In a 52 card deck, there are 78 possible combinations that create a pocket pair...6 for each pocket pair.With any pocket pair taken out of the deck, 5 of those combinations become impossible...thus 73 remaining.With any non-pair, 6 of those combinations become impossible...thus 72 remaining.

[princeof56k 0]

What do you mean no deck has 4 cards?  Where are you getting 4 cards from?

LoL. That guy doublemeup quoted the wrong post. If you look at the post just before the one he quoted (he quoted your post), you'll see that there was a guy using 4 cards as an example.

[doublemeup 0]

so your conclusion from all that pointless math is......?

hey fucker! You dont' know what you're talking about! Have a good day!

wow that was such an insightful post.

[Jeepster80125 0]

If you are playing hold'em heads up, is you opponent more likely to have a pocket pair when you have unpaired cards or when you have a pocket pair?I know the answer, but thought it was interesting.

You, sir, are not Tucker Max. Stop trying to write like him.

[dead money 1]

the question started with, you have a pocket pair. so you can leave out the 1/17 you get dealt a pocket pair. The example states that we have it.

Nowhere does the example state that we have a pocket pair.

It's hard for me to comprehend how wrong you actually are

Show me where Im wrong. Those are the correct odds.

[SapphireTiger 0]

so your conclusion from all that pointless math is......?

hey fucker! You dont' know what you're talking about! Have a good day!

wow that was such an insightful post.

so was this one. please don't try and be witty. it hurts when you do.

[Guest Zach6668]

This is the right answer:

if you have unpaired cards, the probability of him having a PP is(44/50)x(3/49)+(6/50)x(2/49)  If you have paired cards....(48/50)x(3/49)+(2/50)x(1/49)I'm too lazy to find a calculator.

Unpaired cards 0.0589Paired cards .119

This is not:

48/50 * 3/49 + 2/50 * 1/49144/2450 + 2/2450 = 146/2450versus44/50 * 3/49 + 6/50 * 2/49132/2450 + 12/2450 = 146/2450OOPS!  It's the exact same.

Um, I think 132 +12 = 144. Since 144/2450 < 146/2450, therefore the chances are better that he has a pair when you do.Zach