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 08-08-2012, 05:19 PM #1 Carpal \'Tunnel     Join Date: Sep 2002 Posts: 8,907 Old Problem Revisited Try to do this without performing any calculations (or simulations). Starting from the top of a full deck, I will turn over the cards one at a time until I turn over the first ace. Then I will turn over one more card. Is it more likely that card will be the ace of spades or the deuce of clubs? What are the probabilities for each? I answered this years ago, and I recently happened across it again. I was struck at how it affords an extremely simple and elegant solution that is related to the solutions to some other problems that we've had here lately. Last edited by BruceZ; 08-08-2012 at 05:24 PM.
 08-08-2012, 05:47 PM #2 Pooh-Bah     Join Date: Sep 2004 Posts: 4,327 Re: Old Problem Revisited Spoiler: Should be 1 in 52 for both. Relative to the order of the other cards in the deck, any one card is equally likely to be in one of 52 possible slots (after 0..51 cards), and one is a winner and the rest are losers.
 08-08-2012, 08:13 PM #3 adept   Join Date: Aug 2011 Location: Nova Posts: 737 Re: Old Problem Revisited Spoiler: The first ace could have been the ace of spades, so the chance that it's the card after the first ace is only (3/4)*(1/51). Whereas the deuce of clubs has a probability of 1/51.But I have my doubts, since I feel like you wouldn't be asking this if that were the answer.
 08-08-2012, 09:29 PM #4 adept   Join Date: Aug 2011 Location: Nova Posts: 737 Re: Old Problem Revisited Oh I didn't take into account the location of the first ace. Extreme example of how location affects the probabilities: if the first ace isn't until the 49th card, the final 3 cards must be all be aces. So I guess the challenge now is to handle that in the simplest way possible.
 08-09-2012, 06:52 AM #5 journeyman     Join Date: May 2003 Location: Welcome to the top of my range Posts: 214 The As always has exactly a 0.75 chance of being in the stub. The 2c always has exactly a 1/51 chance of being the "next card after first Ace". I'm struggling to figure out an elegant solution but am guessing the As is a big favourite.
 08-09-2012, 07:04 AM #6 journeyman     Join Date: May 2003 Location: Welcome to the top of my range Posts: 214 Hold on. I just imagined a 3 card deck. As 2c Ac. There's six ways to arrange these three cards. And two arrangements have 2c in middle (after first ace) and two arrangements have As after Ac (with the 2c either first or last). So it's the same, or have I horribly oversimplified?
 08-09-2012, 09:18 AM #7 centurion   Join Date: Sep 2010 Posts: 100 Re: Old Problem Revisited It seems kinda hard to do without any calculations, but since its possible there has to some sick simple logic solution. I'd think that given that the 2 of clubs is still in the deck once we've reached the first ace, 3/4 P( 2clubs )= P( Ace of spades ). So the problem kinda boils down to whats the probability for the 2 of clubs to be thrown out of the deck before the first ace is reached which. If we give the cards a number between 1 ( bottom of deck ) and 52 ( top of deck), whats the probability the 2 of club have a higher number than all of the four aces? I think its 1/5, allthough Im not sure which is kinda embarassing since it shouldnt be too hard to figure out. If we assume its 1/5 I think we'll get P( 2clubs ) = 4/5 * 4/3 P( Ace of spades ) = 16/15 P( Ace of spades ) which means getting the 2 of clubs is slightly more likely. However I dont think this is the way to go, feels like Ive missed something and made it too easy to myself =) Last edited by Frallan10; 08-09-2012 at 09:20 AM. Reason: I sucks at speling
 08-09-2012, 10:34 AM #8 centurion   Join Date: Sep 2010 Posts: 100 Re: Old Problem Revisited Once you know x in P( 2clubs ) = x * P( Ace of spade ) I think u can determine the probabilitys using 48*P( 2clubs ) + 4*P( Ace of spade )=1
 08-09-2012, 11:53 AM #9 Pooh-Bah     Join Date: Jul 2006 Posts: 4,340 Re: Old Problem Revisited we know that after the first ace there are 3 aces remaining in the deck. what we really try to figure out is whether there will be more or less than 3 deuces left in the deck (on avg) after the first ace is turned. if its more than 3 then the 2c is more likely otherwise the As is more likely. since were not supposed to calculate anything this cant be right though. edit: the chance that we first draw an ace or first draw a deuce are the same. however we dont stop drawing when we draw a deuce. therefore the 2c is slightly more likely since there there must be slightly more deuces remaining in the deck.
08-09-2012, 02:11 PM   #10
journeyman

Join Date: Sep 2004
Posts: 360
Re: Old Problem Revisited

Quote:
 Originally Posted by David Lyons I just imagined a 3 card deck. As 2c Ac.
This leads you to the correct solution (however with calculations):

Spoiler:

 08-09-2012, 02:22 PM #11 adept   Join Date: Aug 2011 Location: Nova Posts: 737 Re: Old Problem Revisited Tom and cyberfish, I don't see how either card can be 1/52. We're not just talking about a spot in the deck. It has to be after an ace. Focusing on the 2c, we already know that a non-2c has been removed from the deck (since it was an Ace), so if anything it would be 1/51. It's 2 adjacent spots in the deck. There are 51 pairs of adjacent spots. edit -- oh, there's also a chance the 2c came before the first ace. I guess you're saying that when you multiply 1/51 by P(not before ace), it comes out to 1/52. I'll have to see if that's true. edit #2 -- That can't be it either. P(not before 1st ace) = 4/5. That times 1/51 doesn't equal 1/52. Last edited by heehaww; 08-09-2012 at 02:30 PM.
 08-09-2012, 03:08 PM #12 journeyman   Join Date: Sep 2004 Posts: 360 Re: Old Problem Revisited If you draw an ace as the first card it is 1/51 (for 2c being the second card) but if you draw the first ace later on the probability gets lower for 2c. Imagine this: Starting from the top of a full deck, I will turn over the cards one at a time until I turn over a red card. Then I will turn over one more card. Is it more likely that card will be a red card or a black card?
 08-09-2012, 04:18 PM #13 adept   Join Date: Aug 2011 Location: Nova Posts: 737 Re: Old Problem Revisited Nevermind, the solution came to me while I was running. Earlier I stopped halfway through my reasoning, all I had to do was finish my train of thought. There are 51 ways for the 2c to come right after the first ace, Out of a possible 52P2 permutations of Ax2c. = 51 / (52*51) = 1/52 (To be more explicit, one can also multiply the numerator and denominator by 4 since there are 4 possible aces it can be. The denominator can also be 4P2 * 52C2.) Turning our attention to the ace of spades: (3 * 51) / (3 * 52*51) = 1/52 Cool
08-10-2012, 11:22 AM   #15
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,907
Re: Old Problem Revisited

Quote:
 Originally Posted by BruceZ The answer to cyberfish's problem isn't as obvious by Tom's method. He asked about dealing to the first red card, and asked for the probability that the next card is red.
Of course you can do that by Tom's method too. Just find the probability that the next card is the ace of diamonds, which is 1/52, and multiply that by 26 since the red cards are mutually exclusive.

If you have absorbed these methods, then you can certainly answer these:

Starting from the top of a full shuffled deck, I will turn over the cards one at a time until I have turned over 3 aces. Then I will turn over one more card. Is it more likely that card will be an ace or a deuce? What are the probabilities for each?

Spoiler:

Starting from the top of a full shuffled deck, I will turn over the cards one at a time until I have turned over 12 spades. Then I will turn over one more card. Is it more likely that card will be a spade or a heart? What are the probabilities for each?

Spoiler:

Starting from the top of a full shuffled deck, I will turn over the cards one at a time until I have turned over 2 aces. Then I will turn over two more cards. What is the probability that both of those cards will be aces?

Spoiler:

Starting from the top of a full shuffled deck, I will turn over the cards one at a time until I have turned over 7 spades, 5 clubs, and 2 aces. Then I will turn over 2 more cards. What is the probability those cards will be 2 black aces?

Spoiler:

Last edited by BruceZ; 08-10-2012 at 11:40 AM.

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