The following cards are dealt to three people at random, so that everyone gets the same number of cards. What is the probability that everyone gets a red card?

Guest Feb 12, 2022

#1**+4 **

Hint: out of 6 total cards, 3 of them are red. For everyone to get at least one red card, what has to happen?

dolphinia Feb 12, 2022

#3**+1 **

this is what I tried.

I am not sure if it is riight

Chose which color to give out first 2 choices

Dole out the red cards first

person one could get any of 3 cards

person 2 can get ether of the other 2

person 3 gets the last on

So that is 3! ways to dole out the red cards. There will be 3! ways to dole out the yellw cards 2* 3!*3!=6*6 = 72

Now you need to work out how many ways they can be given out if there are no restrctions.

(for this method getting a R1Y1 is different from Y1R1)

6! i think

72/ 6! = 0.1 or 10%

Melody Feb 12, 2022

#4**+1 **

**Hi Melody,**

**You correctly solved this problem here:** https://web2.0calc.com/questions/probability-question_88#r3

**Confirmation:** \(\large (2^3) / (nCr(6,3) = \dfrac {2}{5} = 40\%\)

GA

--. .-

GingerAle
Feb 13, 2022

#6**+1 **

**Hi Melody,**

**The graphic below depicts the** \(2^3 = 8\) **arrangements of (statistical) success, where each person receives a red card. I used this formula here: **https://web2.0calc.com/questions/pleaze-help#r9 **where there are three sets of three colors.**

\(\hspace {.15em}\left[ {\begin{array}{ccc} \scriptsize \hspace {.02em} P_1 & \scriptsize P_2 & \scriptsize P_3 \hspace {.2em} \\ \end{array} } \right] \small \hspace {.2em} \text {Persons Horizontal, sets Vertical. }\small \text{ Though identified by number, persons are indistinguishable. }\\ \left[ {\begin{array}{ccc} R & R & R \\ Y & Y & Y \\ \end{array} } \right] \text {First arrangement of (R} \scriptsize{s} \normalsize{) } {\; \mathrm and \; \mathrm (Y} \scriptsize{s} \normalsize{)} \\ \left[ {\begin{array}{ccc} R & R & Y \\ Y & Y & R \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{ccc} R & Y & R \\ Y & R &Y \\ \end{array} } \right] \text {Third arrangement} \\ \left[ {\begin{array}{ccc} R & Y & Y \\ Y & R & R \\ \end{array} } \right] \text {Fourth arrangement} \\ \left[ {\begin{array}{ccc} Y & R & R \\ R & Y & Y \\ \end{array} } \right] \text {Fifth arrangement} \\ \left[ {\begin{array}{ccc} Y & R & Y \\ R & Y & R \\ \end{array} } \right] \text {Sixth arrangement} \\ \left[ {\begin{array}{ccc} Y & Y & R \\ R & R & Y \\ \end{array} } \right] \text {Seventh arrangement} \\ \left[ {\begin{array}{ccc} Y & Y & Y \\ R & R & R \\ \end{array} } \right] \text {Eighth arrangement} \\ \, \\ \textbf {. . .} \hspace {1em} \textbf {. . .} \hspace {1em} \textbf {. . .}\\ \, \\ \small \text {For all other arrangements: }\\ \small \text {one person has zero red (0) cards, one person has one (1) red card, and one person has two (2) red cards.} \\ \)

GA

--. .-

GingerAle
Feb 14, 2022

#7**0 **

Thanks Ginger,

I still have trouble getting my head around your method. I don't like working with formulas unless I develop them myself.

This is just becasue I have a bad memory and don't trust myself to apply other people's logic properly.

I usually just prefer to think of all cards or what ever being unique and work back from there.

Let see if i can do it again now, without looking back, and come up with the same answer.

There are 6! ways to line up the cards, but it doesn't matter what order the pairs are in

so that makes 6!/(2*2*2)= 90 ways

There are 3!*3! ways that they can get each get a red and a yellow = 6*6=36

36/90 = 40%

Great, now we are all in total agreement. :) (I hope guest actually learns something.)

Melody Feb 15, 2022