emeraldcw
- Thread starter emeraldcw
- Start date
Unknown008
Junior Member
- Joined
- Sep 6, 2011
- Messages
- 148
78 cards? What kind of deck is that?
Hello, emeraldcw!
I agree with Unknown008 . . . We need more information.
I will take a calculated guess . . .
Since \(\displaystyle 78 \:=\:52 + 26\), I assume that half the cards are duplicated.
For example: all the red cards \(\displaystyle (\heartsuit,\:\diamondsuit)\) have duplicates;
. . . . . . . . . .the black cards \(\displaystyle (\spadesuit,\:\clubsuit)\) are normal.
The first person must draw a red card: .\(\displaystyle \dfrac{52}{78}\,=\,\dfrac{2}{3}\)
The second must draw its duplicate: .\(\displaystyle \dfrac{1}{77}\)
Therefore: .\(\displaystyle \text{Prob.} \:=\:\dfrac{2}{3}\cdot\dfrac{1}{77} \:=\:\dfrac{2}{231}\)
I agree with Unknown008 . . . We need more information.
I will take a calculated guess . . .
Since \(\displaystyle 78 \:=\:52 + 26\), I assume that half the cards are duplicated.
For example: all the red cards \(\displaystyle (\heartsuit,\:\diamondsuit)\) have duplicates;
. . . . . . . . . .the black cards \(\displaystyle (\spadesuit,\:\clubsuit)\) are normal.
The first person must draw a red card: .\(\displaystyle \dfrac{52}{78}\,=\,\dfrac{2}{3}\)
The second must draw its duplicate: .\(\displaystyle \dfrac{1}{77}\)
Therefore: .\(\displaystyle \text{Prob.} \:=\:\dfrac{2}{3}\cdot\dfrac{1}{77} \:=\:\dfrac{2}{231}\)
