06242013, 01:10 PM



Location: Mississippi
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Quote:
Originally Posted by pcity
In my example, I specifically said there were at least two heads in the sequence, which means that TTH, THT, HTT, and TTT are not possible.

You're right. I missed that part about the heads coming up at least two times.

06262013, 05:21 PM



Location: Victoria TX
42,668 posts, read 71,538,289 times
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Quote:
Originally Posted by GCSTroop
I call it four card draw plus another one after you've looked at the first four cards.
It is NOT a perfectly valid definition. The reason being is that your definition requires you to examine the first four cards (before the fifth is dealt) and you are therefore examining a four card hand. .

Really. Look. Deal me five cards. Then ask me "do you have two pairs of cards of the same rank?" I either do or I don't. Maybe I have a full house and maybe I don't, but if I have two 3s and 2 Js, after looking at all five cards, I will say "Yes. My hand contains two pairs, exactly as you asked". If I've only looked at four cards, I will say "Wait a minute, I haven't looked at all my cards yet, don't rush me." If you want to ask me "Do you also have a third 3 or a third J?" then that is a different question. But I cannot lay "Yes, I also have a full house" if I don't first have two 3s and 2 Js. No matter what you want to call it, a full house is a hand of five cards in which, after examining all five cards, it is found that there are two pairs, and it is called a full house only because one of the two pairs is matched by another card. Why do you keep trying to say "But you only looked at four cards", or some other nonsense that does not comport with the facts at the table? I just explained to you exactly what I did. I looked at all five cards, and I reported to you the fact that two of them were 3s and two of them were Js, exactly as you asked. At no time did it ever matter what order the cards were dealt in, nor what order I looked at them, nor what order I rearranged them for examination.
So it is a perfectly valid definition to say that a full house is a hand of five cards in which, at a minimum, there are two matching pairs (there are always two matching pairs in a full house, always), in addition to being in the subset of such hands which also has a third card matching one of the pairs. How is that definition not correct?
OK, then. After you ask if I have two pairs, and I say yes, you will quite correctly conclude that there is a 4/48 chance that the card that we have not yet talked about will be a 3 or a J, because I have a poker face and have not tipped you off. It could be any card in the deck. Therefore, 1/12 of all hands to which I announce that two pairs are present, will also turn out to be a full house. I've looked at all five, this is not draw poker, we are playing the five I have. Two of them are 3s and two of them are Js, and 4/48 of the time, the other one will be a 3 or a J. 4/48 of the time it will be an ace, 4/48 of the time it will be a 2, etc., which added together accounts for 48/48 of the equal possibilities. What makes the possibilities unequal, reducing the chance of a 3 or J to less than 4/48, and increasing the chance that will be some other card to better than 4/48? Please do not introduce any other factors that did not occur in the representation I described, such as only looking at four cards, or drawing a card, or whatever.
Last edited by jtur88; 06262013 at 06:09 PM..

06282013, 11:09 AM



86 posts, read 223,998 times
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Quote:
Three guests check into a hotel room. The clerk says the bill is $30, so each guest pays $10. Later the clerk realizes the bill should only be $25. To rectify this, he gives the bellhop $5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn't know the total of the revised bill, the bellhop decides to just give each guest $1 and keep $2 for himself. Each guest got $1 back: so now each guest only paid $9; bringing the total paid to $27. The bellhop has $2. And $27 + $2 = $29 so, if the guests originally handed over $30, what happened to the remaining $1?

Both are examples of just twisting words around to trick people. There is no paradox.

06282013, 11:42 AM



3,137 posts, read 7,885,316 times
Reputation: 1946


Quote:
Originally Posted by dogbert_2001
Both are examples of just twisting words around to trick people. There is no paradox.

Yup. There is no "remaining $1". Of the original $30, the hotel got $25, the guests got $3, and the bellhop got $2.
There's no reason the total paid by guests plus the bellhop's share should equal $30. Rather, the total paid by guests minus the bellhop's share should equal $25, which it does ($27$2).

06292013, 03:20 PM



Location: Mississippi
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Reputation: 4273


Quote:
Originally Posted by jtur88
Really. Look. Deal me five cards. Then ask me "do you have two pairs of cards of the same rank?" I either do or I don't. Maybe I have a full house and maybe I don't, but if I have two 3s and 2 Js, after looking at all five cards, I will say "Yes. My hand contains two pairs, exactly as you asked". If I've only looked at four cards, I will say "Wait a minute, I haven't looked at all my cards yet, don't rush me." If you want to ask me "Do you also have a third 3 or a third J?" then that is a different question. But I cannot lay "Yes, I also have a full house" if I don't first have two 3s and 2 Js. No matter what you want to call it, a full house is a hand of five cards in which, after examining all five cards, it is found that there are two pairs, and it is called a full house only because one of the two pairs is matched by another card. Why do you keep trying to say "But you only looked at four cards", or some other nonsense that does not comport with the facts at the table? I just explained to you exactly what I did. I looked at all five cards, and I reported to you the fact that two of them were 3s and two of them were Js, exactly as you asked. At no time did it ever matter what order the cards were dealt in, nor what order I looked at them, nor what order I rearranged them for examination.
So it is a perfectly valid definition to say that a full house is a hand of five cards in which, at a minimum, there are two matching pairs (there are always two matching pairs in a full house, always), in addition to being in the subset of such hands which also has a third card matching one of the pairs. How is that definition not correct?
OK, then. After you ask if I have two pairs, and I say yes, you will quite correctly conclude that there is a 4/48 chance that the card that we have not yet talked about will be a 3 or a J, because I have a poker face and have not tipped you off. It could be any card in the deck. Therefore, 1/12 of all hands to which I announce that two pairs are present, will also turn out to be a full house. I've looked at all five, this is not draw poker, we are playing the five I have. Two of them are 3s and two of them are Js, and 4/48 of the time, the other one will be a 3 or a J. 4/48 of the time it will be an ace, 4/48 of the time it will be a 2, etc., which added together accounts for 48/48 of the equal possibilities. What makes the possibilities unequal, reducing the chance of a 3 or J to less than 4/48, and increasing the chance that will be some other card to better than 4/48? Please do not introduce any other factors that did not occur in the representation I described, such as only looking at four cards, or drawing a card, or whatever.

Dude... I don't know why you're trying so hard to get this to work. For starters, you've posted absolutely no link, no source material, or anything of the sort to suggest that this is anything other than your fanciful imagination at work.
Here's the bottom line:
If you have five cards in your hand, the odds of having two pairs is greater than if you have just four cards in your hand. I've even laid the math out for you on this... I mean, I literally, handed it to you on a platter. What you're talking about is like saying "Let's call a car a vehicle with an engine and transmission, has three tires, but needs a fourth to drive." Twisting the words around to exclude a fundamental operating platform isn't a paradox  it's just stupid mental masturbation.
Again, you wrote:
Quote:
Originally Posted by jtur88
I just explained to you exactly what I did. I looked at all five cards, and I reported to you the fact that two of them were 3s and two of them were Js, exactly as you asked.

This is exactly my point. The odds of looking at your hand with five cards in it and picking two pairs out of it are much greater if you have five cards than if you have four cards. Let me do the math for you real slowly...
In a five card hand, the number of ways to get two pairs is as follows. Let me lay this out a little different.
(13 choose 2) = Thirteen ranks, choose two of them
(4 choose 2)^2 = Each pair can have any two of four suits
(11 choose 1) = The remaining card CANNOT BE one of the previous two chosen ranks
(4 choose 1) = The final card can be of any suit
So:
(13 choose 2) * (4 choose 2)^2 * (11 choose 1) * (4 choose 1) = 123552 (Feel free to copy and paste the equation into WolframAlpha and it will give you the answer)
In a hand in which four cards have been dealt, the number of way to get two pairs is as follows:
(13 choose 2) = Thirteen ranks, choose two of them
(4 choose 2)^2 = Each pair can have any two of four suits
So:
(13 choose 2) * (4 choose 2)^2 = 2808
Now, if we say that we are looking for a full house, the calculation is done in this way:
(13 choose 1) = One set is going to be a triple and it can be any of the thirteen ranks
(4 choose 3) = In order to make the triple, three of the four suits must be utilized
(12 choose 1) = There is a remaining pair that must be made. That remaining pair must be of a different rank than the triple
(4 choose 2) = The remaining pair comprises two of four suits.
(13 choose 1) * (4 choose 3) * (12 choose 1) * (4 choose 2) = 3744
If we first deal four cards, there are ( 52 choose 4 = 270725) ways to receive your cards. Of those 270725 ways to receive those cards, 2808 of them will be two pairs. Another way to look at this is that you are being dealt a five card hand. The dealer throws you the first four cards and then time stops. At that point where time stops, there are 270725 possible ways your hand is configured at that very moment in time. 2808 of those configuration include two pairs. OK. What makes a difference is whether you look at your four cards or not! If you sneak a peak at your first four cards and see two pairs there, then one of only 2808 configurations has been met. If you are hoping for a full house at this point and you know there are only 3744 ways to get a full house in a five card hand, and 37442808 = 936. In other words, there are only 936 ways the fifth card can be dealt to you and match either of the two pairs in your hand.
The fifth and final card makes a difference. The dealer throws you the fifth card and NOW you decide to sneak a peak. As we discussed before, there are 3744 ways to get a full house but since there are five cards, there are ( 52 choose 5 = 2598960) ways to be dealt a hand. Now, if we look at our cards and have two pairs, we realize that that could have been one of 123552 combinations  and the way to calculate those odds is contingent upon the fifth card NOT being a match to the other two pairs. Since there are a finite number of ways to be dealt a hand in five card poker, all the possible combinations must add up to (52 choose 5).
Let's do that:
Royal Straight Flush = 4
Straight Flush = 36
Four of a Kind = 624
Full House = 3744
Flush (no straight) = 5108
Straight (no flush) = 10200
Three of a kind = 54912
Two pair = 123552
One pair = 1098240
No Pair = 1302540
Add them all up: 4 + 36 + 624 + 3744 + 5108 + 10200 + 54912 + 123552 + 1098240 + 1302540 = 2598960
I don't know what else to say. If you're still not getting it then I suggest taking a probability course.

06302013, 10:27 PM



Location: Mississippi
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I got to thinking about this some more and wanted to clarify something. When you are being dealt five cards and time stops after the first four, there are two configurations in which the fifth card can make a full house. The first scenario is that you have two pairs, as we've been discussing, and the fifth card matches one of the two pairs. But, what keeps getting forgotten about, and perhaps I could have done a better job mentioning it, is the scenario whereby you are dealt a three of a kind and a card that does not match the three of a kind. If the fifth card matches the standalone card, you still have a full house.
From a probabilistic standpoint, you cannot use the definition of a full house to simply be two pairs where the fifth card matches one of the two pairs because there is more than one way to get a full house. Ignoring the other way to get a full house conveniently skews all the numbers.
Once the cards are in your hand, if I am to ask you "Do you have two pairs?" and you say "Yes," then there are 123,552 ways you could have two pairs. But, the calculation for two pairs specifically includes the fifth card NOT matching one of the two pairs. So, in order for me (the unwitting observer) to get an accurate gauge of this, I must temporarily eliminate the fifth card from my calculations and then calculate the odds that you have two pairs, and then the probability of you having a fifth card matching or not. But, then I also have to calculate the odds that you have three of a kind with a standalone and the fifth card matching the standalone.
We've established that when dealt four cards there are 2,808 ways to have two pairs. There are:
13 choose 1 = Thirteen ranks. Choose one rank for our triple
4 choose 3 = Four suits. Choose three of them for the triple.
12 choose 1 = Twelve ranks left for our standalone card. Choose one of them.
4 choose 1 = Four suits. Choose one of them for our standalone card.
(13 choose 1) * (4 choose 3) * (12 choose 1) * (4 choose 1) = 2496 ways to get three cards and a standalone in a four card hand.
So, there are 2,496 ways that I can have three cards with a standalone and 2,808 ways that I can have two pairs. Now, in order for the fifth and final card to complete my full house, I have to do the following calculation: I have to determine what the odds are that you have two pairs and need a fifth and final card to match one of the two pairs, and ALSO what the odds are that you have three of a kind with a standalone and the fifth and final card will match the standalone. Keep in mind, that there are only 3,744 ways to have a full house in five card draw and I currently have 2496 + 2808 = 5304 four card configurations. Not all of these four card configurations will lead to a full house, though, since we've determined there are only 3,744 ways to get a full house. So, 5304  3744 = 1560 ways that our fifth and final card will not help us  or about 29.412% of the time.
An easy way to set this is up is to look at the situation in which I have two pairs:
(13 choose 1) = Thirteen ranks choose one of them
(4 choose 2) = Four suits, choose two of them
(12 choose 1) = Twelve ranks, choose one of them for the second pair
(4 choose 2) = Four suits, choose two of them for the second pair
(4 choose 1) = Four cards left that can match one of our two pairs
(4/48) or (1/12) = Chances that next card will be a match to the second pair
(13 choose 1) * ((4 choose 2)^2) * (12 choose 1) * (4 choose 1) * 1/12 = 1872
That means there are 1872 ways that you can have two pairs with the initial first four cards and then the fifth one matches. If there are 3744 ways to get a full house then the remaining number must be with a three of a kind and a fourth card that doesn't match. This is pretty cool because 3744  1872 = 1872.
Thus, there are 1872 ways you can have a two pair with a fifth card that matches and 1872 ways you can have a triple with a standalone fourth card and then have a fifth card that matches the standalone. Notice also that the number of possible two pair hands in is 123552.
Notice that (1872 * 66) = 123552 and (1872 * 33) = 61776 and that (3744 * 33) = 123552. In your OP, you asked:
Quote:
Originally Posted by jtur88
Let us define a Full House as a subset of all TwoPair hands in which the fifth card matches one of the pairs. The number of such hands ought to be 4/48 (= 1/12) of all 2P hands. There are 123,552 possible 2P hands, so the number of hands in the FH subset ought to be 10,296. But it's not, it is only 3,744, and only 1/33 of all 2P hands have one of those four cards (out of 48) that will match a pair. Where is the fallacy in this approach?

You are correct in that a five card hand gives you 123,552 possible twopair hands. Here's where things get tricky, though. The fallacy is that when you describe a full house as ONLY a two pair with the fifth card matching one of the pairs, you are forgetting about the situation where you have three of a kind and a standalone. Remember, there are 1872 ways to have a pair after four cards are dealt and then have a fifth card match. You stated only 1/33 of all 2P hands have one of those four cards that will match a pair. Multiply 1872 * 33 and you get 61,776  EXACTLY half of 123,552 which is the number of possible configurations for a two pair hand. But, now, you have 1872 ways to get three of a kind with a standalone and then have a fifth card match. Again, 1872 * 33 = 61,776 which is the other half of your twopair set. Now that you have five cards, looking backwards in time and calling it two pairs with a fifth card matching would ALSO include all the situations in which your first four cards were comprised of a triple and a standalone which would incorrectly call it (depending on your time perspective) two pairs with a fifth matching.
Bayesian trees actually do a better job at analyzing situations like this backwards in time but the forum isn't really configured to express them very well. The thing I can't express enough is how and when you are doing the analysis is important. If you are being dealt five cards and time completely stops after the fourth card is dealt (you are somehow unaffected by this time stoppage) you look at your cards and exclaim "I have two pairs!" then that means I KNOW that when your fifth card comes, you have 1872/61776 (1/33) chance of getting a fifth and final card that matches.
Similarly, if you are dealt you first four cards and time stops (you're still unaffected by this strange timestoppage), you look at your cards and exclaim "I have three of a kind and a fourth card!") then I also know that you have a 1872/61776 (1/33) chance of getting a fifth and final card that will match your standalone.
However, when you have five cards that comprise a full house, and you declare that they absolutely must be two pairs with a fifth and final card that matches, this ignores the way in which the cards fell into place because they could have just as easily been a triple, a standalone, and a fifth card that matches the standalone. In effect, there is no paradox, only the fact that by restricting the definition of a full house to two pairs with a fifth matching card, is only taking into consideration HALF of the ways that a full house can arrive.
Edit: I think I may have screwed up the odds for two pair hands with four cards but it still doesn't change much (I kept getting interrupted when I was writing this and the distractions didn't help). I believe there are actually 5616 configurations that would net us four cards with two pairs. This would mean that there would be a total of 5616 + 2496 = 8112 possible full house potential four card hands. Since there are only 3744 ways to get a full house, that would mean that our fifth and final card would not help 8112  3774 = 4338 and 4338/8112 = 54.48% of the time which seems a lot more accurate.
Last edited by GCSTroop; 06302013 at 11:21 PM..

07022013, 07:59 AM



Location: Victoria TX
42,668 posts, read 71,538,289 times
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Show me a Full House that does not meet this configuration:
Two cards of the same rank.
Two more cards of another same rank.
Another card.
Every Full House, once it has been completely dealt face down and then examined, can be disassembled to meet exactly that definition. All hands that meet that configuration can be sorted into two subsets. Those called "full house" in which "another card" matches one of the ranks above, and those called "Two pairs" in which "another card" does not match one of the ranks above. There are no other permutations  "another card" either matches, or it doesn't. For visualization, deal all five cards face down, then look at them and lay the cards out on the table as follows: Any two cards of the lower rank on the left, side by side. Any two cards of the higher rank on the right, side by side. The remaining card in between. (If you cannot find two pairs, reshuffle the cards and try again, for your hand does not meet the minimum examination criteria.)
In four cases out of 48, it will be found that "another card", in the center of our tableau, matches a paired rank. In 44 out of 48 cases, it will be found that "another card" does not match a paired rank.
Last edited by jtur88; 07022013 at 08:09 AM..

07022013, 01:08 PM



5,092 posts, read 4,359,069 times
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Quote:
Originally Posted by jtur88
Show me a Full House that does not meet this configuration:
Two cards of the same rank.
Two more cards of another same rank.
Another card.

GCSTroup has shown you a Full House that does not meet the above configuration:
Three cards of the same rank.
One card of another rank.
Another card.
Perhaps the difference between the probabilities posted and the probabilities that you expect is that you haven't accounted for the other means by which a full house can be obtained.

07022013, 01:54 PM



Location: Mississippi
6,715 posts, read 12,038,211 times
Reputation: 4273


Quote:
Originally Posted by jtur88
Show me a Full House that does not meet this configuration:
Two cards of the same rank.
Two more cards of another same rank.
Another card.
Every Full House, once it has been completely dealt face down and then examined, can be disassembled to meet exactly that definition. All hands that meet that configuration can be sorted into two subsets. Those called "full house" in which "another card" matches one of the ranks above, and those called "Two pairs" in which "another card" does not match one of the ranks above. There are no other permutations  "another card" either matches, or it doesn't. For visualization, deal all five cards face down, then look at them and lay the cards out on the table as follows: Any two cards of the lower rank on the left, side by side. Any two cards of the higher rank on the right, side by side. The remaining card in between. (If you cannot find two pairs, reshuffle the cards and try again, for your hand does not meet the minimum examination criteria.)
In four cases out of 48, it will be found that "another card", in the center of our tableau, matches a paired rank. In 44 out of 48 cases, it will be found that "another card" does not match a paired rank.

You're not getting it and you still haven't shown me any sort of evidence that this is anything other than your fanciful imagination at work. I explained to you in mathematical detail why what you're talking about does NOT work. Yes, every Full House once it has been completely dealt face down CAN BE disassembled to meet that definition but that does not mean that it ARRIVED that way. As I explained to you, HALF of the time, the first four cards dealt to you (when you have a Full House) will be two pairs and then the fifth card will match one of those pairs. The other HALF of the time, you will be dealt three cards that match, a standalone card, and then the fifth card which matches your standalone.
When you demand that the definition of a Full House strictly implies two pairs with a fifth card matching, this means that you are taking ALL the scenarios that unfolded wherein you had a triple, a standalone, and a fifth card matching the standalone into the fold and including them too. There IS A DIFFERENCE!!!
Now, if you want to show me something that states otherwise, like some actual math or some links to some actual math explaining why I'm wrong then I'll be happy to read it and even admit that I'm wrong. But, I'm not going to entertain your fanciful notions just so you can protect your ego. As I suggested before, you may want to look at taking a probability course and it will clear some things up for you.

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