"What is the probability of randomly drawing two cards from a standard deck of playing cards and getting two aces of spades?"
 |
| Validity of mathematics not guaranteed. |
This was a question that confounded me in a dream. I mean, I know enough basic mathematics to conclude that the logical answer would be (1/52)×(0/51) which would be a clean zero.
But I was not entirely satisfied. Calculating for three aces to be drawn simultaneously under the same conditions gives us the following probability:
(1/52)×(0/51)×(-1/50) which also equals zero. [Since two aces of spades have hypothetically been drawn from a single pack, the pack now has -1 aces of spades in it.)
Now for the application of some skill.
What if you drew three cards, but placed the second card back in the deck without looking at any of them?
The probability of drawing three aces is still (1/52)×(0/51)×(-1/50).
However, you have placed the second card back into the deck.
So, the new probability becomes (1/52)×(-1/50), or (-1/2600), which is approximately -0.00038461538. A pretty small negative number, but a non-zero one.
Therefore, if you draw three random cards from a standard deck of cards, replace the second card you drew (possibly contingent on the second card being an ace of spades), then you have the probability of having two cards left in your hands being -0.00038461538. Unfortunately, this is a negative number. So, unless you use a mirror, the two cards are going to be two aces of spades in an alternate universe.
Of course, if you are going to use a mirror, there are several techniques that may be used to reach the same outcome at a much higher frequency. Some of these involve the use of highly advanced mathematical tools like using a trick deck and/or palming a card or two before you begin, which are beyond the scope of this exercise.
But I was not entirely satisfied. Calculating for three aces to be drawn simultaneously under the same conditions gives us the following probability:
(1/52)×(0/51)×(-1/50) which also equals zero. [Since two aces of spades have hypothetically been drawn from a single pack, the pack now has -1 aces of spades in it.)
Now for the application of some skill.
What if you drew three cards, but placed the second card back in the deck without looking at any of them?
The probability of drawing three aces is still (1/52)×(0/51)×(-1/50).
However, you have placed the second card back into the deck.
So, the new probability becomes (1/52)×(-1/50), or (-1/2600), which is approximately -0.00038461538. A pretty small negative number, but a non-zero one.
Therefore, if you draw three random cards from a standard deck of cards, replace the second card you drew (possibly contingent on the second card being an ace of spades), then you have the probability of having two cards left in your hands being -0.00038461538. Unfortunately, this is a negative number. So, unless you use a mirror, the two cards are going to be two aces of spades in an alternate universe.
Of course, if you are going to use a mirror, there are several techniques that may be used to reach the same outcome at a much higher frequency. Some of these involve the use of highly advanced mathematical tools like using a trick deck and/or palming a card or two before you begin, which are beyond the scope of this exercise.
No comments:
Post a Comment