End
Let's see if we can pick it apart.
The probability of drawing two cards from a deck and their sum is > 16.
There's a trick here because of Jack, Queen, King also count as 10.
We can have Ace-Ace and Ace-not Ace. To sum over 16 we must have A6, A7, A8, A9, A10, AJ, AQ, AK, or AA. $\displaystyle \underbrace{(\frac{4}{52})}_{\text{Ace}}\overbrace {(\frac{32}{48})}^{\text{not Ace}}+\underbrace{(\frac{4}{52})}_{\text{Ace}}\ove rbrace{(\frac{3}{51})}^{\text{Ace}}=\frac{37}{663}$
For 10, J, Q, K we have $\displaystyle 4\left[(\frac{4}{52})(\frac{28}{48})+(\frac{4}{52})(\frac {3}{51})\right]=\frac{131}{663}$
For 9, we have $\displaystyle (\frac{4}{52})(\frac{24}{52})+(\frac{4}{52})(\frac {20}{48})=\frac{19}{442}$
For 8, 7, and 6 we have $\displaystyle (\frac{4}{52})(\frac{24}{48})+(\frac{4}{52})(\frac {20}{48})+(\frac{4}{52})(\frac{4}{48})=\frac{1}{13 }$
Add them up and get $\displaystyle \frac{165}{442}$
I could easily be over or under counting.
total more than 16 but not blackjack or double ace:
1st card A 2nd card 6, 7, 8, 9 (4/52)x(4/51)
1st card "10" 2nd card 7, 8, 9, "10" (4/52)x(27/51)
1st card 9 2nd card 8, 9, "10", A (4/52)x(27/51)
1st card 8 2nd card 9, "10", A (4/52)x(24/51)
1st card 7 2nd card "10" A (4/52)x(20/51)
1st card 6 2nd card A (4/52)x(5/51)
so probability is:
(4/52)x(4/51) + (4/52)x(27/51) + (4/52)x(27/51) + (4/52)x(24/51) + (4/52)x(20/51) + (4/52)x(5/51)
RonL
Hello, badboychow!
There are: .$\displaystyle {52\choose2} = 1326\text{ possible pairs.}$You draw 2 cards from a deck of 52 cards.
. . $\displaystyle \begin{array}{cc}\text{Blackjack or Double Aces:} & \text{win \$50} \\
\text{Total is more than 16:} & \text{win \$10} \\
\text{Otherwise:} & \text{lose \$5}\end{array}$
(Assume the face value of an Ace is 11)
Find the expected value.
Double Aces: .There are $\displaystyle {4\choose2} = 6$ ways.
Blackjack: .There are $\displaystyle {4\choose1} = 4$ ways to get an Ace.
. . . . . and $\displaystyle {16\choose1} = 16$ ways to get a card worth 10.
. . So there are: .$\displaystyle 4\cdot16 \:=\:64$ Blackjacks.
Hence, there are: .$\displaystyle 6 + 64 \:=$ 70 ways to win $50.
Sum greater than 16
We can have an Ace (4 choices) and any of 6, 7, 8, or 9 (16 choices).
. . There are: .$\displaystyle 4\cdot16 \,=\,64$ ways.
We can have a 7 (4 ways) and any "10" (16 ways).
. . There are: .$\displaystyle 4\cdot16\,=\,64$ ways.
We can have an 8 (4 ways) and a 9 or a "10" (20 ways).
. . There are" .$\displaystyle 4\cdot20 \,=\,80$ ways.
We can have two 9's. .There are: .$\displaystyle {4\choose2} = 6$ ways.
We can have a 9 (4 ways) and a "10" (16 ways).
. . There are: .$\displaystyle 4\cdot16 \,=\,64$ ways.
We can have two "10's". .There are: .$\displaystyle {16\choose2} = 120$ ways.
Hence, there are: .$\displaystyle 64 + 64 + 80 + 6 + 64+ 120 \,=$ 398 ways to win $10.
Then there are: .$\displaystyle 1326 - 70 - 398 \,=$ 858 ways to lose $5.
The expected value is: .$\displaystyle E \;=\;(50)\left(\frac{70}{1326}\right) + (10)\left(\frac{398}{1326}\right) + (-5)\left(\frac{858}{1326}\right)$
. . $\displaystyle E \;=\;\frac{3190}{1326}\;=\;2.405731523$
Therefore, we can expect to win an average of about $2.41 per game.