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 Replies: 678 / Views: 16,161 |
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Pillar of the Community
United States
602 Posts |
13 spiders. WOLF
Edited by wolf-n-wa
10/21/2010 6:13 pm
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Pillar of the Community
 United States
4000 Posts |
Quote:
spiders and flies  |
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Pillar of the Community
United States
602 Posts |
Ya I caught that and was editing as you were answering. HA! to funny. I'll take the hit and pass then. WOLF
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Pillar of the Community
United States
539 Posts |
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Pillar of the Community
United States
4000 Posts |
weavus, is that your first snack?
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Pillar of the Community
United States
2605 Posts |
That's one crowded web! There are fewer spiders than flies but not enough to have a decent meal. It's an elementary algebra so I disqualified myself.
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Pillar of the Community
United States
4000 Posts |
Woohoo! Getting off of work early tonight!  Give me a bit to get home and I'll see if I can scrape up a puzzle. Stay tuned! |
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Pillar of the Community
United States
539 Posts |
yes it is my first snack. I couldn't believe I actually got in on one of these. Usually folks are all over these way quicker than I can get to them. I had to build the spreadsheet first and then decrypto the puzzle but I still got it done quick enough to win!!
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Pillar of the Community
United States
539 Posts |
going home early, scrapping up a puzzle? ....you do know the Giants are playing and they are going to win the NLCS tonight - right?
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Pillar of the Community
United States
4000 Posts |
I have two eyes and the Marty Feldman thing down pat!
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Pillar of the Community
United States
4000 Posts |
This one isn't too bad, for 3 Scooby Snacks:  |
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Pillar of the Community
United States
602 Posts |
Adrian, in the 2nd truck carrying wallpaper
Harry in the 5th truck carrying paint
Mitch in the 4th truck carrying shoes
Steve in the 1st truck carrying clothing
Will in the 3rd truck carrying carpet.
HOWS that gonna get me some. Oh yeah. WOLF
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Pillar of the Community
United States
4000 Posts |
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Pillar of the Community
United States
602 Posts |
Yeah now my brain hurts again. WOLF
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Pillar of the Community
United States
4000 Posts |
OK guys. I need your help.
There is a card trick I seen a few years back that I have used with great success.
It involves taking a shuffled deck of cards and have a spectator name two "face values" of some cards. Suits do not matter.
You will then proceed to find their named cards either a) right next to each other, or b) only one card apart.
It is important to keep in mind that if the first or second card in the deck is one of their cards,
and the 52nd and 51st card in the deck are one of the named cards, this is considered a win.
It's as if you cut the deck (or even can), you would find the cards within 2 of each other.
Also note, if they are cards 2 and 51, this is not a win, as they are not within 2 cards of each other. Again, cut the deck and you will see that they are 2 cards apart and do not count.
I want to know what the formula is to calculate the expected winning % of this card trick.
It must be ridiculously high. I have done this trick probably a couple of hundred times and only had if fail about a dozen times.
Anyone care to put this into a formula and tell me the expected winning %?
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Replies: 678 / Views: 16,161 |
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Posted 10/21/2010 6:13 pm
