Probability improves?

Question

You have a total of 3n cards, n red cards, n blue cards and n yellow cards. What is the probability of drawing 3 cards one of each color? Now if you have 3n+3 cards instead, n+1 of each color, will the probability compared to the last case improve? Thanks!

Answer 1

For 3n cards: P(x)=[C(n,1)C(n,1)C(n,1)]/C(3n,3)

For 3n+3 cards: P(x)=[C(n+1,1)C(n+1,1)C(n+1,1)]/C(3n+3,3)

You can simply plot the above function on wolfram alpha by typing

C(n,1)^3/C(3n,3) plot (n from 2 to 10) in the search box and verify it is decreasing.

Answer 2

Case 1

Experiment : Drawing three cards out of 3n Cards (n red, n blue & n yellow) Sample Space : (3n)C3 = (3n)(3n-1)(3n-2) / 3!

Event A : Getting three cards of different color. Favorable Case : nC1 * nC1 * nC1 = n^3

P(A) = Favourable Case / Sample Space = {(n^3)} * 3! / (3n)(3n-1)(3n-2)

Case 2

Experiment : Drawing three cards out of 3n+3 Cards (n+1 red, n+1 blue & n+1 yellow) Sample Space : (3n+3)C(3) = (3n+3)(3n+2)(3n+1) / 3!

Event A : Getting three cards of different color. Favorable Case : (n+1)C(1) * (n+1)C(1) * (n+1)C(1) = (n+1)^3

P(A) = Favorable Case / Sample Space = {((n+1)^3)} * 3! / (3n+3)(3n+2)(3n+1)

As compared to Case 1, We see that in Case 2, the numerator factor in increased by 1 whereas the denominator factor is raised by 3. Hence the probability in Case 2 is less than Case 1 i.e., the probability is not improved.

Answer 3

The probability of picking one of each in three draws is the number of combinations that result in one red, one blue, and one yellow over the total cards each time: 3n/3n*2n/(3n-1)*n/(3n-2). Which reduces to 6n^2/(3n(9n^2-9n-2)) and then to 2n/(9n^2-9n-2). Next you want to know if adding one to n changes the probability so:

2n/(9n^2-9n-2)=2(n+1)/(9(n+1)^2-9(n+1)-2)

2n/(9n^2-9n-2)=2n+2/(9(n^2 +2n +1)-9n-9-2)

2n/(9n^2-9n-2)=2n+2/(9n^2 +18n +9-9n-9-2)

2n/(9n^2-9n-2)=2n+2/(9n^2+9n-2)

n/(9n^2-9n-2)=(n+1)/(9n^2+9n-2)

(n+1)(9n^2+9n-2)=(9n^2-9n-2)n

9n^3+9n^2-2n+9n^2+9n-2=9n^3-9n^2-2n

9n^3+18n^2+7n-2=9n^3-9n^2-2n

27n^2+9n-2=0

n=-0.48581,0.15248

so at -0.48581 and 0.15248 1+n has the same probability as n and since we are dealing with n>=1 ill plug in 1 to the equation to see if it the difference in probability is greater than or less than 0 for n>=0.15248.

27(1)^2+9(1)-2?=0
27+9-2?=0
34>0

meaning that the probability of drawing one of each card does not increase as you add one more of each card but, in fact, it decreases.

note: remember that the left side of the equation represents n and the right side represents n+1

also this answer makes sense because if n=1 the probability has to be 100% because you are drawing every card, but not if n>1 because you are not drawing all of the cards.