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  #1  
Old 08-15-2005, 03:19 PM
LetYouDown LetYouDown is offline
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Default Another Boredom Filler

Well shuffled deck of cards. You deal the cards face up, one at a time. Every time you flip a card, you add one to the count, then repeat when you get to 13. So you count 1-13 four times. What are the odds that you will make it through the deck without ever saying a number that matches the rank of the card. Call Aces 1 and Kings 13 to make it easier to think about.
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  #2  
Old 08-15-2005, 07:44 PM
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Default Re: Another Boredom Filler

I can't believe you posted this problem. It has become a bit of an obsession of mine in that I don't know how to solve it analytically and neither does anyone else I know. I've asked some math professors, the wizard of odds (which by the way isn't too good at difficult problems despite having an otherwise nice gambling odds website) and my brother who has gotten an 800 on every SAT/GRE/LOGIC/SAT Subject Tests related to math/physics he ever took.

Anyway, you can write a Monte Carlo program to solve the problem and I am 100% positive the odds of making it through the deck without ever saying the same card that you flip over is 1:61

I have a spreadsheet of the odds of making it through X cards. So if you are interested in seeing the probability at different points in the deck I can send it to you.
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  #3  
Old 08-15-2005, 08:15 PM
elitegimp elitegimp is offline
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Default Re: Another Boredom Filler

I don't know if you can actually claim independence, but for any specific point in your count, there are 48 cards that don't match and 52 that do... so there's a 12/13 chance you don't match a specific time. To not match 52 times would therefore just be (12/13)^52, or 0.01557 (1.557% chance you make it through the deck). That works out to approximately 1 in 64.214, or 63.214:1.

The other thing I was gonna do was to just count the possible deck combinations, but that seems like a lot more work.

edit: what I want to know is what the expected number of cards you get through before the card = count... I've tried 5 times (sample size warning!!!) and didn't get past the 22 card any of those times.

edit 2: If my probability calculation is right, then
P(count = card i | no match in first i-1 cards) = (12/13)^(i-1) * (1/13) => P(get through the deck) = 1 - Sum (from i = 1 to 52) (12/13)^(i-1)*1/13 = 0.01557 as above

So EV = Sum (from i=1 to infinity... assuming an infinite deck so that eventually count = card) i*(12/13)^(i-1)*(1/13) = 12

So in the long run, you should average going through 12 cards per time (if I'm right, which is a big assumption)
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Old 08-15-2005, 08:48 PM
LetYouDown LetYouDown is offline
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Default Re: Another Boredom Filler

[ QUOTE ]
To not match 52 times would therefore just be (12/13)^52, or 0.01557 (1.557% chance you make it through the deck). That works out to approximately 1 in 64.214, or 63.214:1.

[/ QUOTE ]
This is only an approximation. I believe you'll need the inclusion/exclusion method to get an exact count...and I'm not sure that's all that's involved, or that it's even remotely as simple as that.
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Old 08-15-2005, 11:46 PM
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Default Re: Another Boredom Filler

[ QUOTE ]
[ QUOTE ]
To not match 52 times would therefore just be (12/13)^52, or 0.01557 (1.557% chance you make it through the deck). That works out to approximately 1 in 64.214, or 63.214:1.

[/ QUOTE ]
This is only an approximation. I believe you'll need the inclusion/exclusion method to get an exact count...and I'm not sure that's all that's involved, or that it's even remotely as simple as that.

[/ QUOTE ]

Yes, that person was incorrect on nearly all of their statements including parts you didn't quote.

LetYouDown, did you just come up with that problem or did you see it somewhere? Also, what do you do for a living (if that's not too personal)? I've noticed that you seem to answer many of the people's questions.

By the way, I'm still interested in the analytical solution. If anyone could show it to me I would be impressed.
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  #6  
Old 08-16-2005, 03:52 AM
hukilai hukilai is offline
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Default Re: Another Boredom Filler

inclusion/excursion is universal method and will work for this case as well, but it will be incredibly tedious.

To see it, you can try to do it for 13 cards (you also can find it books)...
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  #7  
Old 08-16-2005, 08:38 AM
donkeyradish donkeyradish is offline
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Default A Similar Problem

Shuffle 2 decks of cards

Then simultaneously deal from the top 1 card per deck at a time.

What is the probability that you will get through the whole deck without ever dealing the same 2 cards simultaneously?

Then for an encore what is the probability for n Decks
(1 < n < 52)
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  #8  
Old 08-16-2005, 08:47 AM
LetYouDown LetYouDown is offline
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Default Re: Another Boredom Filler

[ QUOTE ]
LetYouDown, did you just come up with that problem or did you see it somewhere? Also, what do you do for a living (if that's not too personal)? I've noticed that you seem to answer many of the people's questions.

[/ QUOTE ]
Computer Programmer/Software Engineer. Problems like this have always interested me, although I rarely need to use combinatorics for the "real world" applications I work on.

You can monte carlo this problem to get a pretty accurate answer, but I'd really really like to see a closed form solution. Inclusion/Exclusion would be ridiculously tedious, I concur...but I'm curious if that's really the only way to solve this. Granted, you could always write out all 8.0658175170943878571660636856404e+67 possible decks, and solve manually!
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  #9  
Old 08-16-2005, 09:18 AM
bobman0330 bobman0330 is offline
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Default Re: Another Boredom Filler

Could you sketch out how the I/E method would be applied to this problem?
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  #10  
Old 08-16-2005, 09:28 AM
LetYouDown LetYouDown is offline
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Default Re: Another Boredom Filler

Well, essentially you could turn it into an enormous tree diagram. The first card matching is obvious. On the second card you have a couple possibilities. If the first card was a 2, then you have 3/51. If it wasn't a 2, then you have 4/51. This problem is much more complex than I had originally thought. Although I'll admit, I didn't really think about it prior to posting it. It's been a while since I've messed with Mathematica, but I'd imagine it would be really useful here.
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