# Thread: probability of a deck ofcards

1. ## probability of a deck ofcards

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You draw 2 cards from a deck of 52 cards and the return as follow:

Result
How much you win
Blackjack or Double Aces
$50 Total is more than 16$10

None of the above
- $5 (Assume the face value of an ace is 11) Blackjack or double aces [(16/52) x (4/51)] + [(4/52) x (3/51)] *is this correct? I'm not sure of the rest. Anybody help pls Blackjack or Double Aces. first card is a "10" second an ace (16/52)x(4/51) first card is an ace second card an ace or "10" (4/52)x(19/51) so probability is: (16/52)x(4/51) + (4/52)x(19/51) RonL 3. thanks but how about the probability where total is more than 16 and none of the above...i have no idea... 4. Let's see if we can pick it apart. The probability of drawing two cards from a deck and their sum is > 16. There's a trick here because of Jack, Queen, King also count as 10. We can have Ace-Ace and Ace-not Ace. To sum over 16 we must have A6, A7, A8, A9, A10, AJ, AQ, AK, or AA.$\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. 5. thanks thanks....now i playing cards...seeing how many probability....haha 6. Originally Posted by badboychow You draw 2 cards from a deck of 52 cards and the return as follow: Result How much you win Blackjack or Double Aces$50
Total is more than 16
$10 total more than 16 but not blackjack or double ace: 1st card A 2nd card 6, 7, 8, 9 (4/52)x(4/51) 1st card "10" 2nd card 7, 8, 9, "10" (4/52)x(27/51) 1st card 9 2nd card 8, 9, "10", A (4/52)x(27/51) 1st card 8 2nd card 9, "10", A (4/52)x(24/51) 1st card 7 2nd card "10" A (4/52)x(20/51) 1st card 6 2nd card A (4/52)x(5/51) so probability is: (4/52)x(4/51) + (4/52)x(27/51) + (4/52)x(27/51) + (4/52)x(24/51) + (4/52)x(20/51) + (4/52)x(5/51) RonL 7. Hello, badboychow! You draw 2 cards from a deck of 52 cards. . .$\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.
There are: .$\displaystyle {52\choose2} = 1326\text{ possible pairs.}$

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.

8. thank you so much....i think i know how to solve it....