In the casino card game blackjack, the dealer gets two cards, one of which you c

Question

In the casino card game blackjack, the dealer gets two cards, one of which you can see and one that is upside down. Each player also gets two cards. When the dealer’s visible card is an ace, she offers you to take out “insurance”: you can bet $1 that the invisible card is a face card or a 10. If is is, you get $2, otherwise you lose your $1. Let X denote your net winnings. It takes values +$1 or -$-1. What is the expected value of X if

(a) if we assume the dealer’s other card was chosen from a second deck of 52?

(b) if we know that both the dealer’s cards and our own 6 and 8 come from the same deck of 52 cards.

Explanation / Answer

X takes 2 values : +$1 and -$1

X takes the value +$1 when the dealer's visible card is an ace and the invisible card is a face card or a 10

X takes the value -$1 when the dealer's visible card is an ace and the invisible card is neither a face nor a 10

a) when dealer's other card was chosen from a second deck of 52 cards

then the invisible card is chosen from a deck which has all the 52 cards out of which there are 4 cards of 10 and for each denomination there are 3 face cards: king,queen,jack. so all total 3*4=12 cards.

so if the invisible card is from these 12+4=16 cards then X=$1

so P[X=+$1]=16C1/52C1=16/52 so P[X=-$1]=1-16/52

so E[X]=1*16/52-1(1-16/52)=4/13-1+4/13=-5/13 [answer]

b) if we know that both the dealer’s cards and our own 6 and 8 come from the same deck of 52 cards.

which means one ace and one 6 and 8 card is missing from the deck of 52 cards. hence it has 52-1-1-1=49 cards

with 3*4=12 face cards all total and 4 10 cards

so X=$1 if the invisible card is any one from these 12+4=16 cards

so P[X=$1]=16C1/49C1=16/49

so P[X=-$1]=1-16/49=33/49

hence E[X]=1*16/49-1*33/49=(16-33)/49=-17/49 [answer]

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