Last week I discussed the probabilities of three effects: ACAAN, Triumph, and sympathetic Rubik’s cube. The Out of this World effect slipped my mind last week so I’ll mention it today.
We’ll use magicpedia for the definition of the effect.
A spectator separates a deck of cards into reds and blacks without looking at the faces of the cards.
Unlike facing, card color is not independent. There are only 52 cards in a deck so if you know the color of the first 51, you can determine the color of the last card with total accuracy.
If the spectator does not consider this fact and chooses between red and black independently, there is
\[\frac{1}{2^{52}} \approx 2.2 \times 10^{-16}\]of randomly choosing correctly. Note that this is the exact same number as the idealized Triumph from last week. This calculation also assumes an idealized OOTW which does not use any cards as indicators.
Now let’s consider if we require exactly 26 cards to be chosen to be of each color. For the first choice we can choose any of \(52\) cards. For the second choice, there are \(51\) cards remaining. For each successive choice, there is one less option to choose from.
Multiplying all the different choices together gives
\[52 \times 51 \times 50 \times ... \times 28 \times 27 \approx 2.0 \times 10^{41} \]However, this number isn’t correct because we have significantly overcounted. We don’t care about the order in which the cards were chosen. For example, the order A-K♠,A-K♣ is equivalent to A-K♣,A-K♠ which is also equivalent to K-A♣,K-A♠ and all other orders of these 26 cards. In total, there are
\[26! = 26 \times 25 \times 24 \times ... \times 3 \times 2 \times 1 = 403291461126605635584000000\]different orderings of 26 items. Then to get the true number of ways to choose \(26\) cards from a \(52\) card deck we simply divide the two numbers.
\[ \frac{52 \times 51 \times 50 \times ... \times 28 \times 27}{26!} = 495918532948104 \approx 5.0 \times 10^{14} \]The number of ways to choose \(k\) objects from \(n\) is known in math as a combination.
Based on the definition of the Out of this World effect we’re using, it doesn’t matter whether the specator chooses the reds or the blacks, so the probability of OOTW happening is
\[\frac{2}{495918532948104} \approx 4.0 \times 10^{-15}\]Bonus fact: above I wrote \(52 \times 51 \times ... \times 27\) is approximately \(2.0 \times 10^{41}\). This approximation is very, very close. The exact value is
\[199999709752403580401723552207732736000000\]which is a 3 millionth difference.