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Old 06-21-2013, 05:32 PM
 
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Here's a simpler example that corresponds to the "pick up 5 cards and say 'I have at least 3 aces'" situation: I flip a coin 3 times and tell you that heads came up at least twice. What's the probability that heads came up all 3 times? Using H to symbolize heads and T to symbolize tails, the possible flip sequences are HHH, HHT, HTH, and THH. In this situation, we can agree that there's a 1/4 chance that heads came up all 3 times, right?

Now let's discuss a situation that corresponds to the "pick up 3 cards and they're all aces" situation: I flip a coin twice and tell you that heads came up twice. If I flip the coin a third time, what's the probability that heads will have come up on all 3 flips? It's 1/2.

So we have two coin-flip situations that seem to share the same variables (3 independent flips, at least 2 heads come up), and yet the chances of getting 3 heads is much greater in the second example because we changed the parameters of the experiment.
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Old 06-21-2013, 08:54 PM
 
Location: Victoria TX
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Originally Posted by pcity View Post

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)
If it is 2/49, explain why, of the hands that occur naturally with exactly the same configuration (at least 3 aces), only 1/95 contain that fourth ace? Why is the fourth ace so much more probable when the three aces occur at random instead of by design?
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Old 06-21-2013, 10:20 PM
 
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It's not a case of random vs. design. It's just that if you pick up 3 cards and they're all aces, this is different from picking up 5 cards and getting at least 3 aces. Look at the coin flip example above.

Let's go through sequences that could occur with each instance. We'll symbolize aces with A and non-aces with X.

In the case where you pick up 3 aces immediately and show them to me, the sequence of your first three cards is AAA.

In the case where you pick up 5 cards and tell me you have at least 3 aces, the sequence of your first three cards could have been AAA, AAX, AXA, XAA, AXX, XAX, or XXA.

If your first 3 cards are AAA, you have 2/49 chance of getting 4 of a kind. If your first 3 cards are AAX, AXA, or XAA, and your fourth card is an A, you have 1/48 chance of getting 4 of a kind. If your first 3 cards are AAX, AXA, or XAA, and your fourth card is an X, you have 0 chance of getting 4 of a kind. If your first 3 cards are AXX, XAX, or XXA, you have 0 chance of getting 4 of a kind.

All of the situations I describe in the previous paragraph are possible if you draw 5 cards, look at them, and then tell me you have 3 A's. That's why the probability of 4 of a kind in this example is different (and less) than if you just draw 3 A's in sequence and show them to me.
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Old 06-22-2013, 05:21 PM
 
Location: Mississippi
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Originally Posted by pcity View Post
It's not a case of random vs. design. It's just that if you pick up 3 cards and they're all aces, this is different from picking up 5 cards and getting at least 3 aces. Look at the coin flip example above.

Let's go through sequences that could occur with each instance. We'll symbolize aces with A and non-aces with X.

In the case where you pick up 3 aces immediately and show them to me, the sequence of your first three cards is AAA.

In the case where you pick up 5 cards and tell me you have at least 3 aces, the sequence of your first three cards could have been AAA, AAX, AXA, XAA, AXX, XAX, or XXA.

If your first 3 cards are AAA, you have 2/49 chance of getting 4 of a kind. If your first 3 cards are AAX, AXA, or XAA, and your fourth card is an A, you have 1/48 chance of getting 4 of a kind. If your first 3 cards are AAX, AXA, or XAA, and your fourth card is an X, you have 0 chance of getting 4 of a kind. If your first 3 cards are AXX, XAX, or XXA, you have 0 chance of getting 4 of a kind.

All of the situations I describe in the previous paragraph are possible if you draw 5 cards, look at them, and then tell me you have 3 A's. That's why the probability of 4 of a kind in this example is different (and less) than if you just draw 3 A's in sequence and show them to me.
This is pretty simple. It's a hypergeometric formula. It goes as such:

Take the population size and call it 'N'.

Then, take the number of success in the population (there are 4 aces in the population of 52 cards). Call that 'm'.

Take a sample size of 'N' and call it 'n'. In this case, our sample size will be 5 for a five-card deck and 3 for a three-card deck.

Now, call 'k' the number of successes in your sample size. For example, if you are calculating for three aces in a five card hand then k will be three. For four aces, k will be four.

The formula will be set up as follows:



So... We'll set it up with a fifty-two card deck, the cards are drawn at random for a five-card hand, and we're looking for three aces in that five card hand.

N = 52
m = 4
n = 5
k = 3

Because this forum doesn't support LaTeX, I can't show how to plug and chug. But, you can plug your numbers into this calculator to see the probabilities.

Now, if you've been dealt a five-card deck and three of them were Aces, that means there are 47 cards left, one of which is an ace. The odds, if you discard one card (and don't put it back in the deck) and draw another are 1/47 that the next card will be an Ace.

HOWEVER,

If you were to ask beforehand what the odds are that you'll be dealt a five-card hand with three aces, that you'd discard one card, and then pick up an Ace the answer would be the previously discussed hypergeometric formula MULTIPLIED by 1/47.

So, it's different depending on when you look at it.

For example, I'm dealt a five card hand with three aces in it. The probability of that happening was .00174 (or .174%). If I discard one card (it is not placed back in the deck) and I draw another card at random, the odds of that card being an ace are 1/47 or .0212... Otherwise read as 2.12%

However, before we ever deal the cards, if I ask you what the odds are of dealing me three aces in a five card hand, that I would discard one card, and pull another to make it my fourth ace then it would be the results of the hypergeometric formula multiplied by 1/47.

So...


.00174 * .0212 = .000036888 or .0036888%.
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Old 06-22-2013, 05:27 PM
 
Location: Mississippi
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Quote:
Originally Posted by pcity View Post
Here's a simpler example that corresponds to the "pick up 5 cards and say 'I have at least 3 aces'" situation: I flip a coin 3 times and tell you that heads came up at least twice. What's the probability that heads came up all 3 times? Using H to symbolize heads and T to symbolize tails, the possible flip sequences are HHH, HHT, HTH, and THH. In this situation, we can agree that there's a 1/4 chance that heads came up all 3 times, right?

No... It's actually:

1/2 * 1/2 * 1/2 = 1/8 or 12.5%

You forgot about TTH, THT, HTT, and TTT

And again, it depends on when you decide to calculate the probability. If you ask "What are the odds I will flip this coin three times in a row and get heads each time?" Then, it's simply 1/2 * 1/2 * 1/2. If you ask "What are the odds I will flip this coin three times in a row, get heads each time, and then flip a tail on the next flip" then it's (1/2)^4. But, if after having flipped three times, you had three heads and then you asked "What are the odds of me flipping a (insert side here) on the next throw?" Then, it would be 50% or 1/2 because the next throw you make is independent of all the other throws. Prior to flipping the coins, however, you've asked about a very deterministic effect that has only a 6.25% chance of happening. Whereas stopping midway through and making a determination on only one independent event is different.

Last edited by GCSTroop; 06-22-2013 at 05:35 PM..
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Old 06-22-2013, 06:06 PM
 
Location: Victoria TX
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Explaining the mathematical formulas of the odds calculations does not explain the apparent paradox. There is a simple way, and here it is.

Take out 3 aces (any three, lets say clubs, diamonds, hearts), and put them back into the deck face up. Then keep reshuffling and examining the pack. Each time all three face-up hearts are among the five top cards, then look at the other two that are in the top five. Would you expect the ace of spade to be one of them 2/49 of the time, or 1/95 of the time? Of course, the answer is 2/49.

However, here is the explanation of the paradox. There are also shuffles in which only two face-up aces appear, and you are ignoring those and reshuffling -- but many of those will contain the face-down ace of spades, satisfying the sample of 3-ace deals, and satisfying the 2/49 chance that it is in the top five, but are being abandoned and not participating in the test sample. So, 3/4 of the time, the deck contains a 3-ace hand, which is being ignored in the test because only two of the aces are signalling their presence. All of those will be 3-ace hands, and none of them will be 4-ace hands, so that multiplies by four the number 3-ace hands, but does not affect the number of 4-ace hands.

So, for every time you see a 3-ace hand, you have a 2/49 chance that it is a 4-ace hand. But for every one of those, there are three 2-ace hands which could contain the ace of spades, too, but they are being ignored. So there is 2/49 chance that there is an ace of spades, but 3/4 of those are being disregarded, leaving a chance of 1/95 that both conditions are met: 1) The spade is present, and 2) the other three are face-up in the hand to signal our operator that the spade is to be checked for. Three times out of four, when the ace of spades meets the 2/49 chance, it is being disregarded.
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Old 06-23-2013, 12:00 AM
 
Location: Mississippi
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Originally Posted by jtur88 View Post
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?
A full house is actually defined as a three of a kind and a two of a kind, not two pairs with a fifth card that matches one of the two pairs. The reason being is that a hand of cards is a set of five - not four. However, if you're dealt four cards, you pick them up and look at them, and you have two pairs, then the odds of the fifth card matching one of the two pairs is 4/48.

However, if you are dealt five cards at random and you do not look at the cards until all five are dealt, the number of permutations possible for a full house is: (13choose1)(4choose3)(12choose1)(4choose2) =3744...

You can read that as (((13! / 12!) / 1!) * ((4! / 1!) / 3!) * ((12! / 11!) / 1!) * ((4! / 2!) / 2!)) = 3744

However, if you have two pairs and no fifth card yet, the number of permutations for those two pairs is:

(13choose 2)(4 choose 2)(4 choose 2)(11 choose 1)(4 choose 1).

Or...

(((13! / 11!) / 2!) * ((4! / 2!) / 2!) * ((4! / 2!) / 2!) * ((11! / 10!) / 1!) * ((4! / 3!) / 1!)) = 123552

But, that doesn't really matter because the next card that comes out has an 8.333% (4/48) chance of matching one of your pairs. The key factor here is defining what a full house is in the proper fashion and when you choose to look at the cards. I'm pretty sure it would be a fallacy to state that it is two pairs followed by a fifth card matching one of those two pairs because you are then looking upon it with the two pairs contingent on the entire thing - not as a separate five card system.

In other words, once you've distinguished the fact that you have one of the 123552 possible permutations of pairs, the entire system could be thought of as "zeroed" and waiting upon the fifth card whose chances are either 44/48 that it won't match one of your pairs or 4/48 that it will.

However, to look at the hand as a whole... That is, all five cards, there are only 3744 possible permutations that can give you a full house.

Last edited by GCSTroop; 06-23-2013 at 12:11 AM..
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Old 06-23-2013, 06:49 AM
 
Location: Victoria TX
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Quote:
Originally Posted by GCSTroop View Post
A full house is actually defined as a three of a kind and a two of a kind, not two pairs with a fifth card that matches one of the two pairs. .
Then, what do you call a hand containing two pairs and a fifth card matching one of the pairs?

The only reason I used that perfectly valid definition was to show that, once a full five-card hand is completed, the Full House is in fact "a hand that contains two pairs", and subject to the probability laws of a hand that contains two pairs, but also falling into a subset of those in which a fifth card matches one of the pairs. Which can then be rearranged by the user in such a way as to exclaim, if he wishes, "Look I have three of a kind and two of a kind", which does not change the configuration of his hand at all.
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Old 06-23-2013, 02:26 PM
 
Location: Mississippi
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Quote:
Originally Posted by jtur88 View Post
Then, what do you call a hand containing two pairs and a fifth card matching one of the pairs?
I call it four card draw plus another one after you've looked at the first four cards.

Quote:
Originally Posted by jtur88 View Post
The only reason I used that perfectly valid definition was to show that, once a full five-card hand is completed, the Full House is in fact "a hand that contains two pairs", and subject to the probability laws of a hand that contains two pairs, but also falling into a subset of those in which a fifth card matches one of the pairs. Which can then be rearranged by the user in such a way as to exclaim, if he wishes, "Look I have three of a kind and two of a kind", which does not change the configuration of his hand at all.
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. The odds of having two pairs with four cards having been dealt are less than having two pairs with five cards having been dealt. As I said before, once you have examined a four card hand with two pairs in it, you have effectively zeroed everything out. Because the next card being dealt, the fifth card, is independent of the first four cards, the odds of getting a card that matches one of your pairs is 4/48.

For a four card hand dealt with two pairs, there are (13 choose 2 = 78) ways to represent the ranks and (4 choose 2 = 6) ways to arrange two cards. Since you have two pairs, the 6 will be squared, so (78 * 6^2 = 2808). This means that in a four card hand, there are only 2808 permutations possible to arrive at a two pair hand.


This makes sense in light of the fact that we previously determined that there are 3744 ways to get a full house in a five card hand.


But, BECAUSE YOU ARE PLAYING FIVE CARD POKER, when you do the calculations for two pairs, you have (13 choose 2 = 78) ways to represent the ranks, (4 choose 2 = 6) ways to arrange two cards, and then there are 48 cards left (44 if you're actually playing poker correctly and the other guy has four cards dealt to him). So, 78 * (6^2) * 44 = 123552 and, if you're just messing around with cards by yourself, it'll actually be 78 * (6^2) * 48 = 134784.

It is the fifth card in five card poker that increases your odds for two pairs by an exponential amount and that is because the order of the fifth and final card does not matter. The fifth and final card could be the one that gives you your second pair or it could be nothing at all. But, because the fifth card is inclusive in the hand (the system, if you will) it MUST be taken into consideration not as a separate entity but as part of the system.

If we are to examine the first four cards only as a set of pairs (or whatever) we must use the odds of what we are looking at for four card draw - not for five card draw.
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Old 06-24-2013, 11:40 AM
 
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Originally Posted by GCSTroop View Post
No... It's actually:

1/2 * 1/2 * 1/2 = 1/8 or 12.5%

You forgot about TTH, THT, HTT, and TTT
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.

Quote:
And again, it depends on when you decide to calculate the probability. If you ask "What are the odds I will flip this coin three times in a row and get heads each time?" Then, it's simply 1/2 * 1/2 * 1/2. If you ask "What are the odds I will flip this coin three times in a row, get heads each time, and then flip a tail on the next flip" then it's (1/2)^4. But, if after having flipped three times, you had three heads and then you asked "What are the odds of me flipping a (insert side here) on the next throw?" Then, it would be 50% or 1/2 because the next throw you make is independent of all the other throws. Prior to flipping the coins, however, you've asked about a very deterministic effect that has only a 6.25% chance of happening. Whereas stopping midway through and making a determination on only one independent event is different.
I fully agree with the rest of this.
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