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December 21st, 2018, 07:26 PM   #11
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Quote:
Originally Posted by romsek View Post
seriously?
pick the 7 colors, you have 1 left with 7 colors to choose from.
Guess I wasn't too clear!
What I was doing was:
pick 7 only
are they all different color?

1 * 6/55 * 5/54 * 4/53 * 3/52 * 2/51 * 1/50
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December 21st, 2018, 08:44 PM   #12
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With a program, an average of 3 runs of a million trials each gives:

Probability of at least 1: 0.069954
Probability of exactly 1: 0.067655

My (perhaps rough) calculation for the probability of the first stack having all different colors - assuming 56 cards, 8 each of 7 colors, and arranged into a stack of 7 cards (per the OP in post #8 ):

$\displaystyle \large \frac{56}{56} \cdot \overbrace{\frac{48}{55}}^{\text{ cards of another color left}} \cdot \frac{40}{54} \cdot \frac{32}{53} \cdot \frac{24}{52} \cdot \frac{16}{51} \cdot \frac{8}{50}\approx 0.00904$

Now the following part definitely is rough, and I trust my program results more, but since there are 8 stacks:

$\displaystyle \approx 0.00904 \cdot 8 \approx 0.072$

which is in reasonable agreement with my program. I'm unsure how to get the exact number.

Edit: Oops had + instead of \cdot

Last edited by jks; December 21st, 2018 at 09:01 PM.
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December 31st, 2018, 12:56 PM   #13
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JKS:

I got the part you put down with (56/56)*(48/55).... And that works if I'm only making one stack. But once you throw in the 7 other stacks, those top numbers change. But to what, I'm unsure. My reasoning is this. By the time I get back to the first stack, I really don't have 55 cards left. I only have 48...and how many are of a different color from the original card in the first stack is really a variable that will change with each time you put that first card down...

I guess I'll have to be content with this "best guess" that you have provided me. It seems like such a high probability, though. Given how many times I've done this (granted, nowhere hear the million your program did three times), I should have had it happen at least once so far....

Thanks for your help!
Love and Light,
Namaste!
Shelly
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December 31st, 2018, 01:34 PM   #14
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Quote:
Originally Posted by jks View Post
With a program, an average of 3 runs of a million trials each gives:
.............
Now the following part definitely is rough, and I trust my program results more,
Curious. On your setup:
56 cards "put in" an array size 56 ?
1st pick at random from 56?
Then 56th array element swapped with element picked?
2nd pick at random from 55?
...and similarly afterwards?
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December 31st, 2018, 08:10 PM   #15
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With 1 representing one of the colors, 2 representing another, and so on,
then this interesting result (all stacks have the 7 colors) is possible:

1 2 3 4 5 6 7 8 : stacks
-----------------
1 2 3 4 5 6 7 1
2 3 4 5 6 7 1 2
3 4 5 6 7 1 2 3
4 5 6 7 1 2 3 4
5 6 7 1 2 3 4 5
6 7 1 2 3 4 5 6
7 1 2 3 4 5 6 7

However, probability of hitting that result is 1 in zillions!!

Happy new year!
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January 1st, 2019, 12:02 PM   #16
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Happy new year indeed!

Since I think that I can actually calculate the probability of all of the stacks containing all colors as in the post above, I will reply in reverse order to the posts. I think that the probability is:

$\displaystyle \large { \begin{align} & \left (\frac{56}{56} \cdot \frac{48}{55} \cdot \frac{40}{54} \cdot \frac{32}{53} \cdot \frac{24}{52} \cdot \frac{16}{51} \cdot \frac{8}{50} \right ) \\
\cdot & \left (\frac{49}{49} \cdot \frac{42}{48} \cdot \frac{35}{47} \cdot \frac{28}{46} \cdot \frac{21}{45} \cdot \frac{14}{44} \cdot \frac{7}{43} \right ) \\
\cdot & \left (\frac{42}{42} \cdot \frac{36}{41} \cdot \frac{30}{40} \cdot \frac{24}{39} \cdot \frac{18}{38} \cdot \frac{12}{37} \cdot \frac{6}{36} \right ) \\
\cdot & \left (\frac{35}{35} \cdot \frac{30}{34} \cdot \frac{25}{33} \cdot \frac{20}{32} \cdot \frac{15}{31} \cdot \frac{10}{30} \cdot \frac{5}{29} \right ) \\
\cdot & \left (\frac{28}{28} \cdot \frac{24}{27} \cdot \frac{20}{26} \cdot \frac{16}{25} \cdot \frac{12}{24} \cdot \frac{8}{23} \cdot \frac{4}{22} \right ) \\
\cdot & \left (\frac{21}{21} \cdot \frac{18}{20} \cdot \frac{15}{19} \cdot \frac{12}{18} \cdot \frac{9}{17} \cdot \frac{6}{16} \cdot \frac{3}{15} \right ) \\
\cdot & \left (\frac{14}{14} \cdot \frac{12}{13} \cdot \frac{10}{12} \cdot \frac{8}{11} \cdot \frac{6}{10} \cdot \frac{4}{9} \cdot \frac{2}{8} \right ) \\
\cdot & \left (\frac{7}{7} \cdot \frac{6}{6} \cdot \frac{5}{5} \cdot \frac{4}{4} \cdot \frac{3}{3} \cdot \frac{2}{2} \cdot \frac{1}{1} \right ) \\
& = \\
8^7 & \cdot (7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \\
\cdot 7^7 & \cdot (7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \\
& \vdots \\
\cdot 2^7 & \cdot (7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \\
\cdot 1^7 & \cdot (7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \ / \ 56! \\
& = \frac{(8!)^7(7!)^8}{56!} \approx 1.0 \times 10^{-13}
\end{align}}$


To check, I simplified to 3 stacks instead of 8 and I calculated the probability as:

$\displaystyle \large \frac{(3!)^7(7!)^3}{21!} \approx 0.000701$

I also modified my program to 3 stacks and to check only for all stacks containing all colors. An average of 3 runs of 10 million tries gave an average probability of $\approx 0.000699$ so hopefully the analysis is correct.
Thanks from romsek
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January 1st, 2019, 01:14 PM   #17
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Quote:
Originally Posted by Denis View Post
Curious. On your setup:
56 cards "put in" an array size 56 ?
1st pick at random from 56?
Then 56th array element swapped with element picked?
2nd pick at random from 55?
...and similarly afterwards?
You are essentially correct, I just don't do any swapping. First, partly due to the way Ruby works, I built a reference array. For each run I duplicate the reference array into a "cards" array and initialize a "shuffled_cards" array.

I pick a random number 0-55, then add that element to the shuffled array and delete it from the cards array. Then, as you state, I pick another random number from 0-54, and so on.

My code appears below w/o comments but I also attached an image with comments and color coding to make viewing it easier.

Some notes about the program (and Ruby):
  • If I say "cards = cards_ref" (both are Arrays) and then make changes to cards, the same changes get made to cards_ref. Hence, .dup is used (line 37 in image).
  • # denotes a comment.
  • The lines above 26 are just comments w/ the problem statement.
  • [] is an empty array and << adds an element to an array.
  • Comment out either line 50 or 51. Line 50 makes each stack one at a time, line 51 starts each stack and builds up. As romsek stated in post #5, it should not matter, and from running the program, it does not appear to.
  • "Colors" a - g are used, similar to post #6 by Denis, although I could have used anything, of course, including the numbers 1 - 7.
  • Ruby obviously makes some things rather easy (such as "letter.next", line 32), but it comes at the expense of speed. I can run about a million tries per minute. C would be much faster.
  • For simplicity, the basic program that just looks for at least 1 occurrence of all colors is shown.
  • % is the modulo operator.

Here is the code (also see the image):

Code:
num_tries = 1000000
num_all_colors = 0

letter = 'a'
cards_ref = Array.new(8, letter)
6.times do 
  letter = letter.next
  Array.new(8, letter).each{|el| cards_ref << el}
end

num_tries.times do
  cards = cards_ref.dup
  shuffled_cards = []

  for n in 0..55
    rand_ind = rand(56 - n)
    shuffled_cards << cards[rand_ind]
    cards.delete_at(rand_ind)
  end

  stacks = []
  8.times{stacks << []}

  for n in 0..55
    stacks[n/7][n%7] = shuffled_cards[n]
#  stacks[n%8][n/8] = shuffled_cards[n]
  end

  all_colors = false
  for n in 0..7
    if stacks[n].uniq.length == 7 
      all_colors = true
      break
    end
  end
  num_all_colors += 1 if all_colors
end

puts num_tries
puts num_all_colors.to_f / num_tries.to_f
Attached Images
File Type: jpg cards_2018_12_21_prog_img.jpg (95.3 KB, 2 views)
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January 1st, 2019, 02:31 PM   #18
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Quote:
Originally Posted by Rainbowlady View Post
JKS:

I got the part you put down with (56/56)*(48/55).... And that works if I'm only making one stack. But once you throw in the 7 other stacks, those top numbers change. But to what, I'm unsure. My reasoning is this. By the time I get back to the first stack, I really don't have 55 cards left. I only have 48...and how many are of a different color from the original card in the first stack is really a variable that will change with each time you put that first card down...

I guess I'll have to be content with this "best guess" that you have provided me. It seems like such a high probability, though. Given how many times I've done this (granted, nowhere hear the million your program did three times), I should have had it happen at least once so far....

Thanks for your help!
Love and Light,
Namaste!
Shelly
Hi Shelly,

I share your uncertainty of what the numbers should be and for the recognition of the difficulty of accounting for all of the possibilities as different cards are chosen. That is one of the reasons why I decided to try to program in the first place and come up with a rough calculation later (for at least some measure of checking).

Unfortunately, to more precisely calculate the probability, I would have to review some of the probability calculations that I have done in the past (I'm a bit out of practice, and frankly I was never really good at such problems).

I am not sure that I will be able to pursue it much further as work starts again tomorrow. However, if I do, I'm sure that I would start by simplifying to 2 or 3 stacks of 7 cards (similar to what I did with the 'all colors in all stacks' post above) and see if I could make some headway there. If I did, I would then try to extend it to 8, but even this may be difficult (I am unsure).

Of course, there is no reason why you cannot try this for yourself and if needed ask questions along the way. If I cannot help, there are some pretty sharp members who probably can.

Also, for the calculation for a single stack and for the 'all colors in all stacks' I think that you can see that I assumed that the 8 stacks are made one at a time. As romsek pointed out, it should not matter if the stacks are made one at a time or dealt out and built up individually. In my program, I can choose how the stacks are made and it indeed does not seem to matter. So it would be really neat to work the problem both ways and get the same answer, which of course, may be difficult.

Finally, how many times have you run the experiment? I modified my program to count how many tries it took to get at least one stack of all colors and the average was ~14.3. Of course, sometimes it took only 1 try, but sometimes it took over 100.
Thanks from Denis
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