+586 A certain deck of cards contains 16 cards. Each card is red, yellow, green, or blue. Also, each card has a number from 1 to 4. There is a card for each color-number combination, e.g. there is a yellow card that has a 3 on it.
These 16 cards are dealt at random to four different people, with each person getting four cards. What is the probability that each person gets a card with a number 1 on it?
+2511 Solution:
\(\small \text{There are } \dbinom{16}{4} = 1820 \text{ sets of 4-cards. Four of these sets contain exactly one (1) card with a one.}\\ \text {Probability: } \dfrac{4}{1820} = \dfrac{1}{455}\\\)
GA
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I’m leaving this train wreck of a solution here as a sign post to slow down the trains of thought, before changing tracks.
I initially solved this using (4^4) as the number of arrangements where the “1 cards” could occupy any row in the four columns of four rows (this is easy to see). Then dividing this by the total number of combinations,(44) / (nCr(16,4)), gives (64/445) = 0.1406593406593407, the probability.
For some reason, I thought this was too high of a probability....
So, what did I do to “fix” it? I’m not sure ... took the fourth root before dividing it (maybe –IDFK). After changing tracks, I didn’t realize the train actually jumped the tracks until the caboose ended up where the engine should be. This is a simple, high school level probability problem that my cat could solve when buzzed on catnip, yet I turned it into a bloody train wreck.
I am a much better troll than I am a mathematician.
Melody’s solution is defined and nuanced at every step. It’s even better than my cat could do.
GA
+118687 Next attempt.
There are 16 cards. All cards are dealt.
[comment deleted]
16C4*12C4*8C4*1 = 1820*495*70*1 = 63 063 000 ways to choose 4 groups of 4
12C3*9C3*6C3*1 ways to chose 4 groups of 3
12C3*9C3*6C3*1 *4*3*2*1 ways to chose 4 groups of three with no 1s and then add a 1 to each group.
\(\text{Prob(1 in each group)}=\frac{12C3*9C3*6C3*1 *4!}{16C4*12C4*8C4*1}\\ \text{Prob(1 in each group)}=\frac{12C3*9C3*6C3 *4!}{16C4*12C4*8C4}\\ \text{Prob(1 in each group)}=\frac{369600*4!}{63063000}\\ \text{Prob(1 in each group)}=\frac{3696*4!}{630630}\\ \text{Prob(1 in each group)}=\frac{616*4!}{105105}\\ \text{Prob(1 in each group)}=\frac{14784}{105105}\\ \text{Prob(1 in each group)}=\frac{4928}{35035}\\ \text{Prob(1 in each group)}\approx 0.14065\)
check
((nCr(12,3)*nCr(9,3)*nCr(6,3)*1*4!)/(nCr(16,4)*nCr(12,4)*nCr(8,4)*1) = 0.1406593406593407((nCr(12,3)*nCr(9,3)*nCr(6,3)*1*4!)/(nCr(16,4)*nCr(12,4)*nCr(8,4)*1 = 0.1406593406593407
So I have an answer. I have little idea about whether it is correct.
+6250 well your answer agrees with sim!
what's the extra factor of 4! for?
oh nm, it's because you are having order matter.
(sorry I edited accidentally, I think it is back to how it was now)
+118687 No the 4! is not for order mattering. Actually I do not think I have made order matter not within each hand anyway.
The 4! was just for the placement of the cards with 1 on them
The first hand had 4 possibilities, the second hand had 3 possibilities the 3rd hand had 2 possibilities and the last hand took the remaining card.
+118687 XxXTenTacion
When you get the answer can you let us know please. (Answer with logic)
+6250 Here is basically Melody's solution but rearranged a bit to make more intutive sense.
\(\text{First off there are $N=\dbinom{16}{4}\dbinom{12}{4}\dbinom{8}{4} =63063000 $ possible hands}\\ \text{First person gets their cards. There are 4 1's to choose from and $\dbinom{12}{3}$ ways to choose the other 3 cards}\\ \text{Second person has 3 1's and $\dbinom{9}{3}$ ways to pick the other cards}\\ \text{Third person has 2 1's and $\dbinom{6}{3}$ other cards}\\ \text{This gets us the total number of hands $n$ where each player has a 1 in their hand}\\ n = 4\dbinom{12}{3}\cdot 3\dbinom{9}{3}\cdot 2\dbinom{6}{3}\\ p = \dfrac n N = \dfrac{\dbinom{12}{3}\dbinom{9}{3}\dbinom{6}{3} 4!}{\dbinom{16}{4}\dbinom{12}{4}\dbinom{8}{4}} = \dfrac{8870400}{63063000 } = \dfrac{64}{455}\)
