a poker hand contains five cards find the mean of each of th

a poker hand contains five cards find the mean of each of the following

1.the number of different suits in a poker hand

2.the number of different face values in a poker hand

Solution

a poker hand contains five cards. these 5 cards from a deck of 52 cards can be selected in ⁵²C₅ ways.

1. let X denotes the number of different suits in a poker hand.

now there are all total 4 different suits in a deck of cards(spade,diamond,heart,club) each containing 13 cards

so X can be either 1 [all 5 cards are from the same suit] or 2[some cards are from one suit and others from another suit] or 3[some cards are from one suit,some from another suit and the rest are from another suit] or 4[ four cards are from 4 different suits and the 5th one is from one of the 4 suits]

so P[X=1]=P[all 5 cards are from the same suit]=P[out of 4 suits, one suit is selected and from that suit all 5 cards are selected]=⁴C₁*¹³C₅/⁵²C₅

P[X=2]=P[out of 4 suits 2 suits are selected and then from first suit 3 cards are selected and from the second suit 2 cards are selected or from first suit 2 cards are selected and from the second suit 3 cards are selected or 4 cards are selected from the first suit and 1 card is from the second suit or vice versa]=2*⁴C₂*(¹³C₂*¹³C₃+¹³C₄*¹³C₁)/⁵²C₅

P[X=3]=P[out of 4 suits 3 suits are selected and then from the first suit 2 cards are selected,from the 2nd suit 2 are selected and from the third suit 1 is selected or from the first suit 2 cards are selected,from the 2nd suit 1 are selected and from the third suit 2 is selected or from the first suit 1 cards are selected,from the 2nd suit 2 are selected and from the third suit 2 is selected]=3*⁴C₃*¹³C₂*¹³C₂*¹³C₁/⁵²C₅

P[X=4]=P[from the 4 different suits 1 card is selected from three different suits and 2 cards are selected from the 4th suit]=4*¹³C₁*¹³C₁*¹³C₁*¹³C₂/⁵²C₅

so mean of X is

E[X]=1*⁴C₁*¹³C₅/⁵²C₅+2*2*⁴C₂*(¹³C₂*¹³C₃+¹³C₄*¹³C₁)/⁵²C₅+3*3*⁴C₃*¹³C₂*¹³C₂*¹³C₁/⁵²C₅+4*4*¹³C₁*¹³C₁*¹³C₁*¹³C₂/⁵²C₅

       =2.44 [answer]

2. there are 13 different face values each containing 4 cards

let Y denote the number of different face values in a poker hand

so Y can take the values 2,3,4,5

P[Y=2]=P[2 faces are selected from 13 faces and there 4 cards from first face and the other one is from second face or, 4 cards from second face and the other one is from first face or 3 cards from the first face and 2 from the second or vice versa]=2*¹³C₂*(⁴C₄*⁴C₁+⁴C₃*⁴C₂)/⁵²C₅

P[Y=3]=P[3 faces are selected from 13 faces and first face 2 cards are selected,from the 2nd face 2 are selected and from the third face 1 is selected or from the first face 2 cards are selected,from the 2nd gace 1 are selected and from the third face 2 is selected or from the first face 1 cards are selected,from the 2nd face 2 are selected and from the third face 2 is selected]=3*¹³C₃*⁴C₁*⁴C₂*⁴C₂/⁵²C₅

P[Y=4]=P[4 faces are selected from 13 faces and from one face 2 cards are selected and from the others 1 card os selected]=4*¹³C₄*⁴C₂*⁴C₁*⁴C₁*⁴C₁/⁵²C₅

P[Y=5]=[ 5 faces are selected from 13 faces and from each face 1 card is selected]=¹³C₅*(⁴C₁)⁵/⁵²C₅

so mean of Y is

E[Y]=2*2*¹³C₂*(⁴C₄*⁴C₁+⁴C₃*⁴C₂)/⁵²C₅+3*3*¹³C₃*⁴C₁*⁴C₂*⁴C₂/⁵²C₅+4*4*¹³C₄*⁴C₂*⁴C₁*⁴C₁*⁴C₁/⁵²C₅+5*¹³C₅*(⁴C₁)⁵/⁵²C₅=2.47298 [answer]