

Slim King Eternal Order Orlando 17661 Posts 
Please help me figure the odds on getting either digit correct etc.
I'm ignorant. See the experiment here. http://www.themagiccafe.com/forums/viewt......forum=15 Thanks Dave
THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death.....

Nir Dahan Inner circle Munich, Germany 1390 Posts 
Can you post a full description of the problem. the link didn't help...

Slim King Eternal Order Orlando 17661 Posts 
There are three boxes...Each contain a two digit number 10 99. You have three guesses( RV's).
What are the odds of hitting the exact numbers in the exact boxes? What are the odds of hitting one of the two digits in each box? What are the odds of hitting any of the digits in any box? Those kinds of odds. Thanks Dave
THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death.....

Nir Dahan Inner circle Munich, Germany 1390 Posts 
Sorry dave it is not clear enough
if one box has 32 and you guess 23, does it consider you hit the digits? or position is also important. do you also need to have the raltive positions fit as well another example: boxes contain 23 45 57 you guess 25 is it 2 or 3 hits? 
Joshua Quinn Inner circle with an outer triangle 2049 Posts 
Someone correct me if I'm wrong, but I'm pretty sure it's like this:
Assuming all three numbers are different, and discounting any possible "psychological influence" factors (i.e. "it's under fifty, both digits are odd, but they're different")... >What are the odds of hitting the exact numbers in the exact boxes? 1/100 x 1/99 x 1/98 = 1 in 970,200 >What are the odds of hitting one of the two digits in each box? The odds of hitting one digit in one box would basically be like two tries at a 1 in 10 chance. So it would be 2 in 10 (or simplified, 1 in 5). So the odds of doing that for all three boxes would be: 1/5 x 1/5 x 1/5 = 1 in 125 >What are the odds of hitting any of the digits in any box? This depends on whether you're specifying the box or not, i.e. "There's a 6 in this box" vs. "There's a 6 in one of the boxes, somewhere." In the first case, there are a total of 2 digits to pick from, which means you've got two 1in10 shots, so the chances are 1 in 5. In the second case, there are a total of 6 digits to pick from, which means six 1in10 shots, so the chances are 6 in 10.
Every problem contains the seeds of its own solution. Unfortunately every problem also contains the seeds of an infinite number of nonsolutions, so that first part really isn't super helpful.

idris New user St. Louis, MO 38 Posts 
Dave,
Are we guarenteed that the three boxes have different values? I would assume so but it can make a difference. Quinn, The numbers are from 10 to 99, not 1 to 100. So the first part would be 1/90 x 1/89 x 1/88 or 1 in 704880. Haven't had time to play with the others yet. Nir's questions need answered as well.
Jerry

Slim King Eternal Order Orlando 17661 Posts 
I would think that the position would be important, but it would be nice to know the odds if simply the digits were correct....What if the numbers were reversed...etc...? Or the correct number in the wrong box?
Thanks Dave Posted: Jun 23, 2006 9:42am  Here's the results! The contents of the boxes were: Gargoyle....Number 16....sixteen The Angel....Number 22....twenty two The Sarcophagus....Number 43.....forty three In that order.....16,22,43 Here are some good attempts. 36, 12, 42 Kaytracy A correct digit in each number! 17, 26, 43 Harishjoe Off buy one, off by four and a direct hit! 13, 21, 34 Dr. Spector A correct digit in each number and the last one reversed. Only seven or eight people sumitted their thoughts. To me this seems prety good. Is it?
THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death.....

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