Question

. In a 7 card hand, what is the probability of

a. three cards of one rank two of another rank and one each of two more different ranks?

b. three cards of one suit, two of a second suit and one each of the remaining suits?

c. three of one rank, three of a second rank

Answer 1

number of ways to pick 7 cards from 52 =(₅₂C₇) =133784560

a)

number of ways three cards of one rank two of another rank and one each of two more different ranks

=N(select one denomination from 13 and then pick 3 cards from 4 of this , then select 1 denomination from 12 and pick 2 card from this and then pick two denomination from 11 and pick 1 card from each)

=(₁₃C₁)*(₄C₃)*(₁₂C₁)*(₄C₂)*(₁₁C₂)*(₄C₁)*(₄C₁)=13*4*12*6*55*4*4 =3294720

therefore probability =3294720/133784560 =0.0246

b)

N(3 cards of one suit , two of 2nd and one of remaining) =(₄C₁)*(₁₃C₃)*(₃C₁)*(₁₃C₂)*(₂C₂)*(₁₃C₁)*(₁₃C₁) =4*286*3*78*1*13*13 =45240624

therefore probability=45240624/133784560 =0.3382

c)

number of ways =(₁₃C₂)*(₄C₃)*(₄C₃)*(₁₁C₁)*(₄C₁) =78*4*4*11*4=54912

therefore probability=54912/133784560 =0.0004