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06-20-2013, 08:19 AM
 Location: Victoria TX 42,668 posts, read 71,574,557 times Reputation: 35869
Let us define a Full House as a subset of all Two-Pair 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?

Similarly, why does only one 3-of-a-Kind hand out of 88 (624/54,912) have the other card that makes 4-of-a-Kind, when there appear to be two chances in 49 for it to turn up? Isn't 4-Kind just a subset of all 3-Kind hands in which the fourth matching card occurs?

Let's say I have my hand and you have the deck. I show 3 Aces, and you want to know if I'm bluffing. What are the odds that you have the other ace among the 47 cards still in the deck? 87/88? Yes, according to the mathematical odds, it is. The chances are 46/49 that you have the fourth ace, and 1/88 that I have it, which adds up to less than one. Where the hell did that ace go?

Last edited by jtur88; 06-20-2013 at 08:54 AM..

06-20-2013, 09:47 AM
 3,137 posts, read 7,887,844 times Reputation: 1946
On the last one, where are the 87/88 and 46/49 coming from?

If there are 49 cards in a deck, and you're looking for one specific one that exists in the deck, the odds of you having that specific card are 1/49 for each card you have. So if you have two cards in your hand that you haven't shown me and I have the remainder of the deck, there's a 2/49 chance that you have that ace and a 47/49 chance that I have the ace.

06-20-2013, 10:01 AM
 Location: Victoria TX 42,668 posts, read 71,574,557 times Reputation: 35869
Quote:
 Originally Posted by pcity On the last one, where are the 87/88 and 46/49 coming from? If there are 49 cards in a deck, and you're looking for one specific one that exists in the deck, the odds of you having that specific card are 1/49 for each card you have. So if you have two cards in your hand that you haven't shown me and I have the remainder of the deck, there's a 2/49 chance that you have that ace and a 47/49 chance that I have the ace.
The 87/88 is the ratio of the poker odds table, on the likelihood that a 3K hand is also a 4K hand. The 46/49 is what common sense would suggest is the chance that the fourth matching card is NOT in your hand -- it' still in the deck or in somebody else's hand.

if, as you say, there is a 2/49 chance that you have the fourth ace to go with the other three, and a 46/49 chance that the ace is not in your hand, (49/52 less the three known cards) why is the chance of 4K only 1/88 of the chance for a 3K hand?

Using the ratio of 3K and 4K odds, the odds table is telling us that if you have 3 aces, then out of 88,000 trials the fourth ace is in your hand 1,000 times, and in the undealt cards 87,000 times. How is that possible?

http://en.wikipedia.org/wiki/Poker_p...rd_poker_hands

Last edited by jtur88; 06-20-2013 at 10:09 AM..

06-20-2013, 02:46 PM
 Location: Victoria TX 42,668 posts, read 71,574,557 times Reputation: 35869
Let me try to put it more simply. Say you take the three aces (or whatever -- just three of the same rank) out of a deck of cards and place them in front of me. Then you reshuffle the cards, and give me two more, keeping 47 for yourself. How is it possible that 87 times out of 88, the fourth ace will remain in your deck. 54,912 chances that the ace is still in the deck, and only 624 chances it is one of the two cards you give me to go along with my three aces.

(Never mind that we have pre-arranged the three aces. We could have programmed a computer to keep dealing hands and ignoring all deals in which I do not get three aces, with the same result. It doesn't matter, the ratio of 3-ace hands to 4-ace hands is still going to be 1:88. It also doesn't matter if the five of diamonds is the first, second, third, fourth or fifth card dealt, as long as we count only the hands with at least 3 aces.)

06-20-2013, 04:54 PM
 3,137 posts, read 7,887,844 times Reputation: 1946
Quote:
 Originally Posted by jtur88 Let me try to put it more simply. Say you take the three aces (or whatever -- just three of the same rank) out of a deck of cards and place them in front of me. Then you reshuffle the cards, and give me two more, keeping 47 for yourself. How is it possible that 87 times out of 88, the fourth ace will remain in your deck. 54,912 chances that the ace is still in the deck, and only 624 chances it is one of the two cards you give me to go along with my three aces.
OK... now I get where you're getting the numbers from.

First, the 54,912 refers to hands that are 3-of-a-kind and only 3-of-a-kind. If you wanted the ratio of 4-of-a-kind hands to all hands where 3 cards match, the calculation would be 4-of-a-kind hands/(4-of-a-kind hands + 3-of-a-kind hands + full-house hands). This is 624/(624 + 54,912 +3744) = 1/95.

Second, the conditions of your first example (I give you the three aces and two random cards, keeping the other 47) are not governed by the 1/95 ratio. Here's why:

The 1/95 ratio means that of all the 3-cards-match hands, 1 out of 95 will be a 4-of-a-kind hand. There are a number of situations where you would get a 3-cards-match hand:

1) Out of your first 3 cards, all 3 match (3 aces, for example).
2) Out of your first 3 cards, 2 match, and then at least 1 of the next 2 you draw also matches those 2.
3) Out of your first 3 cards, 0 match, but the next 2 you draw match 1 of your first 3.

Your first example only deals with situation 1, where you know you have 3 aces already but no one knows the last 2 cards. If the first 3 cards you draw are 3 aces, the chances of you getting 4 aces total are much higher than if the first 3 cards contain only 2 aces and there are still 2 unknown cards. In situation 3, where none of your first 3 cards match each other, 4-of-a-kind is impossible.

If you look at situation 1 exclusively, the numbers I gave before hold: There's a 2/49 chance that you have that 4th ace in your other 2 cards and a 47/49 chance that the 4th ace stays in the deck.

If you look at situations 1, 2, and 3 collectively, then you have a 1/95 chance that your 3-cards-match hand ends up being a 4-of-a-kind hand.

Last edited by pcity; 06-20-2013 at 05:03 PM..

06-20-2013, 05:17 PM
 3,137 posts, read 7,887,844 times Reputation: 1946
A quick sum up of what I wrote above would be that the probability changes based on what you know.

Think of an example where I deal you cards and I'm allowed to look at them but you aren't.

If I look at all 5 of your cards and tell you you have a hand where at least 3 cards match, you know there's a 1/95 chance you have 4-of-a-kind.

If you then pick up 3 of your cards and they all match, you know there's a 2/49 chance that you have 4-of-a-kind.

If you pick up 3 of your cards and none of them match, you know there's 0 chance that you have 4-of-a-kind.

06-20-2013, 10:39 PM
 Location: Victoria TX 42,668 posts, read 71,574,557 times Reputation: 35869
OK, let's say I deal you five cards, and you look at them and announce "I have at least 3 aces". This is not contingent on which 3 of your cards are aces. nor what order they were dealt, that is irrelevant. To a third person, who knows only that you have at least 3 aces, what is the probability that I still have one ace in among the 47 undealt cards? Of the 49 cards that are not your three aces, there is a 2/49 chance that you have the fourth ace, and a 47/49 chance that I have it. Which means, in any hand that has at least three aces, there is a 2/49 chance that it has four aces, not a 1/88 chance, as predicted by the table of poker odds comparing 3-kind and 4-kind probability. Our third person would be very smart to put up a dollar, at 88:1 (or even 50:1), on you having four aces, with a 49:2 chance of winning.

06-21-2013, 08:18 AM
 3,137 posts, read 7,887,844 times Reputation: 1946
If I look at all five of my cards, and then choose to reveal that I have 3 aces, then you have to account for my motives and this becomes more complicated than a random probability problem.

If there's a rule of the game that says one must reveal when they look at their 5 card hand and it contains at least 3 aces, then we're back in the realm of random probability. In this instance, you should calculate that I have 1/95 chance of having 4-of-a-kind. The reason is because my 3-ace hand could fall under any of the situations I described in post #5 of this thread.

If I choose to reveal 3 cards of my 5 card hand at random, and I show 3 aces, then you should calculate that there's a 2/49 chance one of the other 2 cards in my hand is an ace.

06-21-2013, 03:15 PM
 Location: Victoria TX 42,668 posts, read 71,574,557 times Reputation: 35869
Quote:
 Originally Posted by pcity If I look at all five of my cards, and then choose to reveal that I have 3 aces, then you have to account for my motives and this becomes more complicated than a random probability problem. If there's a rule of the game that says one must reveal when they look at their 5 card hand and it contains at least 3 aces, then we're back in the realm of random probability. In this instance, you should calculate that I have 1/95 chance of having 4-of-a-kind. The reason is because my 3-ace hand could fall under any of the situations I described in post #5 of this thread. If I choose to reveal 3 cards of my 5 card hand at random, and I show 3 aces, then you should calculate that there's a 2/49 chance one of the other 2 cards in my hand is an ace.
Not at all. You can program a computer to do it, so there is no motive taken into account except to determine the frequency of a given event over a series of randomized trials. The only purpose for declaring whether there are at least three aces, is to quickly eliminate all hands that do not meet the qualification of "at least 3 aces" and are therefore irrelevant to the calculation.

We want to know the frequency of 3-ace hands meeting other criteria, so all we are doing is throwing out the hands that are not 3-ace hands, and are therefore meaningless to the test. That fact that I designated a human operant, without any ulterior motive, instead of a computer, to throw out the meaningless hands has no bearing on the results.

Thee only "other" result that can arise is the Full House. Here is the number of chances"

4-Kind 624
Full house 3,744
3-Kind 54,912.

Total 59 280, of which only 1 out of 95 becomes 4-Kind by the presence of another card, and a Full House does not meet our criteria, so is the same as 3-Kind for purposes of the 4-Kind discussion.

That weighs even more in my favor, for now there are even more "at least 3 of a kind" possibilities, from which there are only 624 containing that elusive fourth ace. One out of 95. Which means 94 out of 95 trials, the fourth ace is one of the 47 undealt cards, and not one of the hidden two. Why is that fourth ace so rarely among the 2/49 remaining cards held in the dealt hand? (And registered by a computer, since you don't trust the motives of a human to look at the cards and signal compliance with the criteria.)

At the risk of further complicating it, there is even another paradoxical phenomenon. Of the two cards besides the 3-Kind, it is about twice as likely to have any pair (1/13), than the fourth ace paired with any other card (2/49). So why does the Full House turn up six times as often as the 4-Kind, instead of twice as often?

Let's do four exercises:

1. Take a deck of cards, shuffle it, turn up two cards. What are the chances one of them is Ace of Spades? Do we agree it is 2/52? OK.

2. Remove three cards and toss them aside without looking at them. Turn up two cards. Still 2/52? OK.

3. Remove three cards, look at them, and make sure none of them is the Ace of Spades. It doesn't matter if any of them are aces. Turn up two cards. Now, 2/49, right? OK.

4. Remove the other three aces from the full deck and set them aside. You still have a deck with three cards missing, none of them the Ace of Spades. Turn up two cards. What are the chances one of them is the Ace of Spades? 2/49? Or 1/95? Why?

Let's muddy it further. Say you're playing draw poker, and you get three aces, and draw two. What is your chance of drawing the fourth ace? Of course, 2/49. Right. Then, why is it four times more likely to get the ace on the draw, than having had it dealt to you in the first place?

Last edited by jtur88; 06-21-2013 at 04:18 PM..

06-21-2013, 05:18 PM
 3,137 posts, read 7,887,844 times Reputation: 1946
Quote:
 Originally Posted by jtur88 That weighs even more in my favor, for now there are even more "at least 3 of a kind" possibilities, from which there are only 624 containing that elusive fourth ace. One out of 95. Which means 94 out of 95 trials, the fourth ace is one of the 47 undealt cards, and not one of the hidden two. Why is that fourth ace so rarely among the 2/49 remaining cards held in the dealt hand?
I don't think this is a paradox. It's just looking at probability in two different situations.

Picking up 5 cards at random and declaring that at least 3 of them are aces is a different situation than picking up 3 cards at random and declaring that all 3 of them are are aces.

Quote:
 Let's do four exercises: 1. Take a deck of cards, shuffle it, turn up two cards. What are the chances one of them is Ace of Spades? Do we agree it is 2/52?
Yes.

Quote:
 2. Remove three cards and toss them aside without looking at them. Turn up two cards. Still 2/52?
Yes.

Quote:
 3. Remove three cards, look at them, and make sure none of them is the Ace of Spades. It doesn't matter if any of them are aces. Turn up two cards. Now, 2/49, right?
Yes.

Quote:
 4. Remove the other three aces from the full deck and set them aside. You still have a deck with three cards missing, none of them the Ace of Spades. Turn up two cards. What are the chances one of them is the Ace of Spades? 2/49? Or 1/95? Why?
2/49, because the other aces being removed do not affect the probability of the Ace of Spades being drawn.

But this isn't the same as drawing a 5 card hand, looking at all 5 cards, and then telling me you have 3 aces in it. Perhaps I can illustrate this better with coin flips... (to the next post)
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