thepillow
New member
- Joined
- Sep 12, 2012
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- 34
I was explaining to a friend how to calculate some probabilities having to do with playing cards and I came across a problem that I can understand numerically but not intuitively. I was hoping someone might have some insight.
Situation 1: The probability that, when dealt 2 random cards from a regular deck, you will get 2 aces given that at least 1 of the 2 cards is an Ace.
Situation 2: The probability that, when dealt 2 random cards from a regular deck, you will get 2 aces given that 1 of the 2 cards is the Ace of Spades.
The strange thing is that the probability of Situation 2 is nearly double the probability of Situation 1 despite the fact that Situation 2 is more specific.
Here are my calculations:
Situation 1
Let X = both cards are aces, and let Y = at least 1 of the cards is an ace
\[P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{\frac{\binom{4}{2}}{\binom{52}{2}}}{1-\frac{\binom{48}{2}}{\binom{52}{2}}}\frac{}{}\]
This comes out to 1/33
Situation 2
This is much easier. If one of the cards is the ace of spades then the probability that we have 2 aces is the same as the probability that the other card is an ace, which is 3/51 = 1/17
Can anyone provide an insight into why it would make intuitive sense for situation 2 to be essentially twice as likely as situation 1?
Much appreciated!
Situation 1: The probability that, when dealt 2 random cards from a regular deck, you will get 2 aces given that at least 1 of the 2 cards is an Ace.
Situation 2: The probability that, when dealt 2 random cards from a regular deck, you will get 2 aces given that 1 of the 2 cards is the Ace of Spades.
The strange thing is that the probability of Situation 2 is nearly double the probability of Situation 1 despite the fact that Situation 2 is more specific.
Here are my calculations:
Situation 1
Let X = both cards are aces, and let Y = at least 1 of the cards is an ace
\[P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{\frac{\binom{4}{2}}{\binom{52}{2}}}{1-\frac{\binom{48}{2}}{\binom{52}{2}}}\frac{}{}\]
This comes out to 1/33
Situation 2
This is much easier. If one of the cards is the ace of spades then the probability that we have 2 aces is the same as the probability that the other card is an ace, which is 3/51 = 1/17
Can anyone provide an insight into why it would make intuitive sense for situation 2 to be essentially twice as likely as situation 1?
Much appreciated!
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