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 December 14th, 2018, 05:58 PM #1 Newbie   Joined: Dec 2018 From: PA Posts: 5 Thanks: 0 Probability I have 8 cards of 7 different colors. I shuffle all these cards together. I make 8 stacks of 7 cards, placing one card in each stack before placing the second card in each stack. I take the cards from the top of the deck in the order they appeared after shuffling and I place them on the stack in the same order each time. (ie, I place the first card in stacks 1 through 8, then the second card in stacks 1-8, etc.) What is the probability that one of those stacks will have one card of each color?
December 14th, 2018, 06:30 PM   #2
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Quote:
 Originally Posted by Rainbowlady I have 8 cards of 7 different colors.
This means that you have a total of 8 cards, 2 of one color, and 1 each of six different colors.

Quote:
 I shuffle all these cards together. I make 8 stacks of 7 cards, placing one card in each stack before placing the second card in each stack.
Impossible. This requires 56 cards, and you have said that you "have 8 cards."

If this is a problem from a course that you are taking, please give the complete problem exactly as it is given.

If this is a problem that you have devised, try to make it make some sense.

 December 14th, 2018, 06:59 PM #3 Newbie   Joined: Dec 2018 From: PA Posts: 5 Thanks: 0 My apologies. I thought it was pretty clear since, as you pointed out, making 8 stacks of 7 cards each requires 56 cards and posting an impossible question would be a bit pointless. I have 56 cards: 8 each of 7 different colors. Does that make more sense? Last edited by Rainbowlady; December 14th, 2018 at 07:28 PM.
December 15th, 2018, 07:19 AM   #4
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 Originally Posted by Rainbowlady My apologies. I thought it was pretty clear since, as you pointed out, making 8 stacks of 7 cards each requires 56 cards and posting an impossible question would be a bit pointless. I have 56 cards: 8 each of 7 different colors. Does that make more sense?
It's considerable progress.

You have a deck of 56 cards. Each card is one of seven distinct colors. The number of cards of each color is eight.

You shuffle the deck thoroughly and then deal it out into 8 stacks of 7 cards each.

Is that it?

The question, however, remains unclear. Do you mean the probability that only one stack contains one card of each color or the probability that at least one stack contains one card of each color? Those are two different questions.

Are you studying probability? If so, what are your thoughts?

 December 15th, 2018, 08:35 AM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,370 Thanks: 1274 The problem setup is a bit of a red herring. This problem is equivalent to simply creating one stack of 8 cards and asking the probability of that stack having one of each color. That's the nature of shuffling. I'm also assuming the cards indistinct other than their color. So you figure out two numbers. How many different groups of 8 cards contain all 7 colors? How many different ways can you pick 8 cards from the deck of 56? The probability you're after is the ratio of these two numbers.
 December 15th, 2018, 09:04 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,112 Thanks: 1002 a,b,c,d,e,f,g representing the 7 colors: Code: 1 2 3 4 5 6 7 8 = stacks ? ? ? ? g ? ? ? ? ? ? ? a ? ? ? ? ? ? ? d ? ? ? ? ? ? ? e ? ? ? ? ? ? ? f ? ? ? ? ? ? ? b ? ? ? ? ? ? ? c ? ? ? What is probability that at least 1 stack (as example stack#5 above) will contain all the 7 colors? Is that what you mean ?
December 19th, 2018, 06:18 PM   #7
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Quote:
 Originally Posted by JeffM1 You shuffle the deck thoroughly and then deal it out into 8 stacks of 7 cards each. Is that it?
Essentially. I "deal" out the first card to each of the 8 stacks before dealing the second card to any of the stacks. And repeat that until I have seven cards in all 8 stacks.

Quote:
 Originally Posted by JeffM1 The question, however, remains unclear. Do you mean the probability that only one stack contains one card of each color or the probability that at least one stack contains one card of each color? Those are two different questions.
I was looking for the "at least one" option. But both would be interesting.

Quote:
 Originally Posted by JeffM1 Are you studying probability? If so, what are your thoughts?
I am not. This is purely an exercise in curiosity. While I have my degree in civil engineering (although I never got my PE) and a minor in math, probabilities were always a weakness. I get the basics but when you get into situations like this, where the outcome of the second choice relies on the outcome of the first and continues that way (for example, if I deal out the first 4 cards and happen to deal all blue cards, then a random color for the next four cards and then start dealing the second card for each of the 8 stacks and happen to deal the remaining 4 blue cards to the 4 stacks that already have blue cards, there is no chance that any of the 8 stacks will have one of each color card. I get lost trying to translate those options into numbers.

December 19th, 2018, 06:20 PM   #8
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 Originally Posted by Denis a,b,c,d,e,f,g representing the 7 colors: Code: 1 2 3 4 5 6 7 8 = stacks ? ? ? ? g ? ? ? ? ? ? ? a ? ? ? ? ? ? ? d ? ? ? ? ? ? ? e ? ? ? ? ? ? ? f ? ? ? ? ? ? ? b ? ? ? ? ? ? ? c ? ? ? What is probability that at least 1 stack (as example stack#5 above) will contain all the 7 colors? Is that what you mean ?
Essentially, yes.

December 21st, 2018, 03:32 PM   #9
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Quote:
 Originally Posted by romsek The problem setup is a bit of a red herring. This problem is equivalent to simply creating one stack of 8 cards and asking the probability of that stack having one of each color. That's the nature of shuffling.
I'm no expert at this stuff; is it:
(6! * 49!) / 55! = 1 / 28,989,675 ?

December 21st, 2018, 04:27 PM   #10
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Quote:
 Originally Posted by Denis I'm no expert at this stuff; is it: (6! * 49!) / 55! = 1 / 28,989,675 ?
seriously?

pick the 7 colors, you have 1 left with 7 colors to choose from.

$p = \dfrac{7}{\dbinom{56}{8}}$

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