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 08-10-2012, 12:18 PM #16 Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,423 Re: Old Problem Revisited Does it in any way help to think of the deck of cards as a wheel rather than a deck with a defined top and bottom? You pick a random slot to start from and just keeping going one direction until you reach a result. I don't know why, but thinking of it that way somehow helped me with this problem.
 08-10-2012, 12:54 PM #17 journeyman   Join Date: Sep 2004 Posts: 360 Re: Old Problem Revisited I think it doesn't matter because you reach the condition (trigger) always before you reach the end of the deck. In the next scenario it matters: Starting from the top of a full deck, I will turn over the cards one at a time until I turn over Kh. Then I will turn over one more card. Is it more likely that card will be 2c? Spoiler: 1/52 But: If Kh is the last card there's no next card (this equals to no 2c to get 1/52) If you use a wheel (or put turned cards at the bottom) it's something different. Last edited by cyberfish; 08-10-2012 at 01:04 PM.
08-10-2012, 01:01 PM   #18
Carpal \'Tunnel

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Re: Old Problem Revisited

Quote:
 Originally Posted by cyberfish I think it doesn't matter because you reach the condition (trigger) always before you reach the end of the deck.
Right, that's the key. As long as condition X always happens before the last N cards, the probability that Y will happen on the next N cards after X happens is the same as the probability of Y happening on the last N cards.

08-10-2012, 02:47 PM   #19
Carpal \'Tunnel

Join Date: Sep 2002
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Re: Old Problem Revisited

Quote:
 Originally Posted by Sherman Does it in any way help to think of the deck of cards as a wheel rather than a deck with a defined top and bottom? You pick a random slot to start from and just keeping going one direction until you reach a result. I don't know why, but thinking of it that way somehow helped me with this problem.
Note we aren't saying that no matter where the first ace occurs the probability for the next card is 1/52. The probability changes depending on where the first ace occurs. If we get the first ace on the first card, the probability of 2c on the next card is 1/51. If we get the first ace on card number 49, the probability that the next card is 2c is zero. But wherever we get the first ace, the probability that the next card is 2c is the same as the probability that the last card in the deck is 2c. All remaining unseen cards are equivalent. So we can just look at the overall probability that the last card in the deck is the 2c, and that's the probability that the card after the first ace will be the 2c. That is also true for the As.

08-10-2012, 03:36 PM   #20
Carpal \'Tunnel

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Location: Psychology Department
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Re: Old Problem Revisited

Quote:
 Originally Posted by BruceZ So we can just look at the overall probability that the last card in the deck is the 2c, and that's the probability that the card after the first ace will be the 2c. That is also true for the As.
Unless the first A you draw is the As. Then the probability the next card is the As is zero. This happens with a 1/4 probability. This is the part that is confusing to me. The rest of it all makes sense.

That is, if you changed the question so that it was "when I draw a K what is the probability the next card is either the 2c or the As?" it would obviously be the same. The part that tricks me is the fact that the As will be the card that gives you a probability of 0 for the As and (on average) a probability > 0 for the 2c when the As is draw first. This must be counteracted by the fact that the 2c can come up with a probability 0 when it is removed before an A occurs. The same cannot be said for the when the As occurs (i.e., it is never removed prior to an A coming) because it is already an A and requires that we stop.

08-10-2012, 04:54 PM   #21
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Re: Old Problem Revisited

Quote:
 Originally Posted by Sherman Unless If the first A you draw is the As, then the probability the next card is the As is zero. This happens with a 1/4 probability. This is the part that is confusing to me. The rest of it all makes sense.
The only thing different about the As is that the probability that it is the card after the first ace depends on whether the first ace was the As. But that doesn't matter for our argument because whatever the probability that the next card is the As is, it will always be the same as the probability that the last card in the deck is the As, even when that probability is zero. The probability that the last card in the deck is the As is 1/52, so the probability that the next card after the first ace will be the As must also be 1/52.

Last edited by BruceZ; 08-10-2012 at 07:24 PM.

 08-11-2012, 01:29 AM #22 veteran   Join Date: Jun 2011 Posts: 2,263 Re: Old Problem Revisited Sadly my brain wanted to convince me the answer wasn't 1/52. Thank you Bruce, Tom, and cyber for trying to explain why it is.
 08-15-2012, 08:33 PM #23 journeyman   Join Date: Sep 2010 Posts: 205 Re: Old Problem Revisited If we change a problem a little: If we know that there is an ace on the top of an otherwise shuffled deck, what is the chance that the second card is the ace of spades, and what is the chance that the second card is the two of clubs. In this case, it looks like we have 1/51 chance for the 2 of clubs, and 3 / 4 * 1 / 51 chance for the ace of spades. For the ace, the chance is lower, since we do have some new information about the cards in the deck. If the As is on the top, which it is one quarter of the time, we have zero chance that it is second, and if it is not, there is 1/51 chance that it is second. Combining those two, we get that 3/4 * 1/51 chance. Now, if we deal from a random deck until we reach any ace, when we do hit an ace, it seems like we are exactly at the start of the first problem. However, we are not. In the first problem, we have a random deck of 51 cards below our first ace. In the second case, the dealing procedure changes the distribution of aces in the rest of the deck, and an ace is much more likely to follow the first ace. This effect increases the chance from 3/4*1/51 up to exactly 1/52 for the ace of spades, and also ups the two of clubs from 1/51 to 1/52. Why is an ace, any ace more likely to follow the first ace? Because, if two aces are close together, we are more likely to hit the first ace more often. More the aces are grouped together, more this effects the chance that you will hit the first ace in the group. Also really strange how this effects the chance of seeing the two of clubs as the next card. Really weird result. Just to check, I did a 200 million deals simulation and I got those results: As pct = 0.019125465 2c pct = 0.01920807 1 / 52 = 0.0192307692307692 1 / 51 = 0.0196078431372549
 08-18-2012, 02:35 AM #24 adept   Join Date: Mar 2009 Posts: 981 Re: Old Problem Revisited I was a skeptic until I imagined a deck of 2 aces and 2 deuces. Then by enumeration it is straightforward to tally that the ace of spades and deuce of clubs each follows the first ace in the deck exactly 6 of the 24 possible deck orderings. A very good problem.

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