Coin Community Family of Web Sites Join Thousands of Coin, Bullion, & Money Collectors
300,000 items to help build your collection! Vancouvers #1 Coin and Paper Money Dealer Royal Canadian Mint products, Canadian, Polish, American, and world coins and banknotes. Join Thousands of Coin, Bullion, & Money Collectors Specializing in Modern Numismatics Royal Estate Auctions - $1 Coin AuctionsCoin, Banknote and Medal Collectors's Online Mall








Username:
Password:
Save Password
Forgot your Password?


This page may contain links that result in small commissions to keep this free site up and running.

Welcome Guest! Registering and/or logging in will remove the anchor (bottom) ads. It's Free!

Introducing Scooby Snacks!

To participate in the forum you must log in or register.
Author Previous TopicReplies: 678 / Views: 16,148Next Topic
Page: of 46
Pillar of the Community
Scooby Due's Avatar
United States
4000 Posts
 Posted 10/24/2010  1:39 pm  Show Profile   Bookmark this reply Add Scooby Due to your friends list Get a Link to this Reply
We are given 2 cards to find.

There are 13*12 = 156 combinations that they can choose to give us.
Let's say J & 6.

So, we have 4 opportunities to win the game: J6x, 6Jx, Jx6 and 6xJ.
Each one of those give us 4(J) * 4(6) = 16 possibilities IN THAT POSITION.

So, winning combinations include:

J6xxxxxxx......
6Jxxxxxxx......
Jx6xxxxxx......
6xJxxxxxx......
.
.
xxxxJ6xxx......
xxxx6Jxxx......
xxxxJx6xx......
xxxx6xJxx......
.
.
.
Jxxxxx............xxxxxxxx6
Jxxxxx............xxxxxxx6x
.
.
.
etc.


So, if J6x and 6Jx = 4*4*2 = 32 chances of being next to each other,
and Jx6 and 6xJ = 4*4*2 = 32 chances of being one card apart,
and there are 52 "starting positions", then we have (32+32)*52 = 3328 chances to find their cards.

If they can only choose 156 to start with, then do we have 3328 / 156 = 21.333 to 1 ratio of finding their card? Should we expect to find 21 out of every 22 tries? Or 95.5%?
Pillar of the Community
xshift's Avatar
United States
2669 Posts
 Posted 10/24/2010  2:25 pm  Show Profile   Bookmark this reply Add xshift to your friends list Get a Link to this Reply
There are 78 combinations, as order doesn't matter.

64 chances at 78 combinations is a ration of 1.21875 to 1. Pretty good odds.
Pillar of the Community
Scooby Due's Avatar
United States
4000 Posts
 Posted 10/24/2010  2:29 pm  Show Profile   Bookmark this reply Add Scooby Due to your friends list Get a Link to this Reply
That sounds a lot closer to what I've experienced.

How are you getting 78?
Pillar of the Community
Scooby Due's Avatar
United States
4000 Posts
 Posted 10/24/2010  2:31 pm  Show Profile   Bookmark this reply Add Scooby Due to your friends list Get a Link to this Reply
And wouldn't 64/78 = 82% = 4:1 odds?
Pillar of the Community
xshift's Avatar
United States
2669 Posts
 Posted 10/24/2010  2:40 pm  Show Profile   Bookmark this reply Add xshift to your friends list Get a Link to this Reply
yep.. it's 82%.

Out of 15 times today, I'm at 60%.
Overall (due to more than one in a hand), it's at 86.7%.

Next to each other has shown up 33.33% of the time. (5/9 or 55% of winning hands)
One between has shown up 53.33% of the time. (8/9 or 88% of winning hands)
Pillar of the Community
xshift's Avatar
United States
2669 Posts
 Posted 10/24/2010  2:41 pm  Show Profile   Bookmark this reply Add xshift to your friends list Get a Link to this Reply
and now.. I'm afraid my brain has melted.
Pillar of the Community
xshift's Avatar
United States
2669 Posts
 Posted 10/24/2010  2:44 pm  Show Profile   Bookmark this reply Add xshift to your friends list Get a Link to this Reply
oh - 78 is half your 156. The order doesn't matter, so it's half of 13*12.

You can see the difference using:
=COMBIN(13,2) and
=PERMUT(13,2) (order matters in permutations)
Valued Member
rikcando's Avatar
Canada
287 Posts
 Posted 10/24/2010  5:50 pm  Show Profile   Bookmark this reply Add rikcando to your friends list Get a Link to this Reply

Quote:
13*12 = 156 combinations

Maybe I spent too much time with the paint fumes today but I cannot see this. You do not have 13*12. You have 4 cards and 4 cards in a deck of 52. Where does everyone get this 13?
Valued Member
rikcando's Avatar
Canada
287 Posts
 Posted 10/24/2010  5:59 pm  Show Profile   Bookmark this reply Add rikcando to your friends list Get a Link to this Reply
If you pick any one card in the deck, the one next to it can be any one of the other 51. Not just 12. If anything it would be one of 13 cards since there are duplicates.
Pillar of the Community
Scooby Due's Avatar
United States
4000 Posts
 Posted 10/24/2010  6:01 pm  Show Profile   Bookmark this reply Add Scooby Due to your friends list Get a Link to this Reply
Since they are only naming 2 cards from Ace through King as the parameters. For example, they are not naming 2 Jacks to find. Then it would matter. (But, not allowed ).

It doesn't matter that there are 4 of each in the deck.

At least, that's how I keep leaning.
Edited by Scooby Due
10/24/2010 6:02 pm
Valued Member
rikcando's Avatar
Canada
287 Posts
 Posted 10/24/2010  6:11 pm  Show Profile   Bookmark this reply Add rikcando to your friends list Get a Link to this Reply

Quote:
It doesn't matter that there are 4 of each in the deck.

I don't see how you can say this as JJ or 66 is a possible combination that must be counted as an incorrect combination. You cannot simply exclude combinations from the deck because it is easier for you, or because you didn't ask for them. I didn't ask for my gas bill this month but I got it anyways, and you know I cannot just ignore it because I didn't want it.
Edited by rikcando
10/24/2010 6:21 pm
Valued Member
rikcando's Avatar
Canada
287 Posts
 Posted 10/24/2010  6:13 pm  Show Profile   Bookmark this reply Add rikcando to your friends list Get a Link to this Reply
You cannot calculate the odds without knowing the number of total possibilities. JJ and 66 are in that list.
Pillar of the Community
Scooby Due's Avatar
United States
4000 Posts
 Posted 10/24/2010  6:22 pm  Show Profile   Bookmark this reply Add Scooby Due to your friends list Get a Link to this Reply
But they can still only name 1 of 13*12 possiblities for you to look for.

It is YOUR advantage that there are 4 of each in the deck.
Valued Member
rikcando's Avatar
Canada
287 Posts
 Posted 10/24/2010  6:50 pm  Show Profile   Bookmark this reply Add rikcando to your friends list Get a Link to this Reply
Yeah, it must be the paint fumes because there is no way I can work anything out with 13*12. I ust cannot understand where it comes from. There is no such thing in my deck of 52 cards. I think I'll sit on the sidelines and see what develops.
Valued Member
Manuel's Avatar
United States
163 Posts
 Posted 10/24/2010  8:24 pm  Show Profile   Bookmark this reply Add Manuel to your friends list Get a Link to this Reply
No. They have only sixteen chances of being one card apart, then sixteen chances of being next to each other. You only have 5.3333333334 to 1 odds. Also, you shouldn't expect to find something. Look at it like roll hunting. I don't know about you guys, but I hope to find errors and old coins ('50s and older), and I hardly ever do.
I really hope I worked it out right since I'm half asleep because it's the weekend.
I don't mean to sound short tempered if I do. I'm actually not, and I'm trying to figure out how to type this without all kinds of random stuff. It's so tempting......
  Previous TopicReplies: 678 / Views: 16,148Next Topic
Page: of 46

To participate in the forum you must log in or register.



    




Disclaimer: While a tremendous amount of effort goes into ensuring the accuracy of the information contained in this site, Coin Community assumes no liability for errors. Copyright 2005 - 2026 Coin Community Family- all rights reserved worldwide. Use of any images or content on this website without prior written permission of Coin Community or the original lender is strictly prohibited.
Contact Us  |  Advertise Here  |  Privacy Policy / Terms of Use

Coin Community Forum © 2005 - 2026 Coin Community Forums
It took 0.41 seconds to rattle this change. Forums