Mathematicians! Help!  | | | |  OK, so here's a really cool card trick that deals with odds. The trick is that if you name any two values in the deck (4, king, 6, jack, for example) and ignore the suit, then in a shuffled deck you will find your two values next to each other somewhere in the deck. This works almost all the time, but I was wondering what the exact odds are. I have no idea how to go about figuring this out, so any Mathematicians or computer programmers if you have some time could you try figuring the odds out? Thanks. Cheers!
P.S. for those of you who don't believe me, grab a deck of cards and try it, it blew my mind when I first did it.
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| | 1. takutaku (96)
|
3 years ago
| | umm dont know, will try and then think | | | | | | | | | | Custom Deck Specialists Find Five-Star Rated Deck Pros Backed by Our Service Guarantee! www.ServiceMagic.com | add comment | | | |
| 2.  owlwings (12342)
|
3 years ago
| |  The probability of two named values (1 through 13) being adjacent to each other at least once in a randomly shuffled pack of 52 cards is 0.735968, that is approximately 73.6% of the time or, very roughly, in 3 decks out of 4.
I did not actually calculate this myself but took 10 minutes or so to Google 'probability of two card values being adjacent'. The most suitable site I found was actually the second in the Google results (ignoring the sponsored links at the top). Here is the link: http://delphiforfun.org/P... .
You can read the logic and the math for yourself on that page. The calculation and result is actually provided by a program ("CardProbability.exe" - originally written in C but ported to Delphi) and available for download from the site.
As you are a 'speed lover', you may be interested to know that it took me about the same time (or a little longer) to write this response than it did to read your problem, begin to work it out, realise that the answer was very likely online, Google it, read the page, download and run the program and find the answer!
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| owlwings (12342)
|
3 years ago
| | I regret that I made an error! The probability I gave above was for 'adjacent or one intervening'. That is, any other card might come in between the two values, so if 6 and J were chosen, a sequence such as 6, 5, J would be allowable. This was not what you asked.
The correct probability that there would be at least one occurrence of the two chosen values occurring together (i.e. either as 6, J or J, 6) is rather lower at 0.486279 or slightly less than half the time. | | | |
 speedlover1994 (144)
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3 years ago
| | Wow thanks very much for the response, it was extremely thorough! That's so interesting that it's less than half, because every time I try it it seems to work very very often. Thanks again. Cheers! | | | | | | | Deck Drainage System Double the enjoyment of your deck! Creates dry space under raised deck www.DekDrain.com | add comment | | | |
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