SAT Math problem: Tarot probability

In honor of my close student friends, here’s the question:

“The tarot is an ancient oracle represented by 78 cards, divided into 2 groups: 22 of the Major Arcana (main cards), which show the generalized potential of a situation, and 56 Minor Arcana (auxiliary cards) that express the form in which the situation will take place. In an Astrological Mandala spread, 12 positions represent the astrological houses and one card sums up the querent’s moment. In each of these 13 positions, a Major Arcana card is combined with a Minor Arcana card to illustrate what is happening or will happen in each area of the querent’s life. What is the probability of a spread to repeat itself? “mandala_astrologica_taro (1)

a) I will not waste my time cracking up this numbers

b) Let the tarot reader crack this herself

c) It’s easier to win the Powerball

d) God willing, the spread may repeat itself because it’s a mystical matter

e) All the above

Mandala jogo ro

Answer:

a) Yes, don’t waste your precious time with this, because I don’t believe it’s gonna be on your test – unless your math teacher follows my blog and likes this question… So you’d better have a consultation with me to get advice on how to plan your life for the coming months!

b) I drafter this answer, but this time not on a paper napkin after 2 beers … Here is a little of that crazy calculation:

1st House: Choosing 1 of the 22 Major Arcana cards, combined with 1 of the 56 Minor Arcana cards = 22 choices x 56 choices = 1,232 possibilities

2nd House: Because you have already drawn a card from each arcana, there are still 21 Major Arcana cards remaining and 55 Minor Arcana cards: 21 x 55 = 1,155 possibilities

3rd House: following the same rationale: 20 x 54 = 1,080

4th House: 19 x 53 = 1.007

5th House: 18 x 52 = 936

6th House: 17 x 51 = 867

7th House: 16×50 = 800

8th House: 15 x 49 = 735

9th House: 14 x 48 = 672

10th House: 13 x 47 = 611

11th House: 12 x 46 = 552

12th House: 11 x 45 = 495

Querent’s Moment (13th position): 10 x 44 = 440

Considering that all houses interconnect to express the querent’s life, we have: 1,232 x 1,155 x 1,080 x 1,007 x 936 x 867 x 800 x 735 x 672 x 611 x 552 x 495 x 440 = 36,452,358 x 1,030, therefore the probability that a tarot spread will repeat itself is less than 1 in 1 trillion!

c) Yes, it’s easier to win the Powerball Lottery than to repeat a tarot mandala spread, because the odds of winning the Powerball are 1 in approximately 292 million *

d) What makes a person choose specific tarot cards and they (most of the times) magically match what is happening in his/her life? You may argue that it’s a matter of psychology. Or that the tarot reader adapts the readings to what she already knows about the person. Or that if the reader doesn’t know the querent, she manipulates the answer influencing the person’s future.

But how to explain, as in the case of one of my clients, that the tarot predicted that she was going to win a hefty sum of money, and a year later she comes back and tells me that a relative died and left her a significant life insurance payout?

I only read the cards, but who actually chooses and puts them in order? Chance?? After so many consultations, I don’t believe it’s just chance. So, with all the unlikelihood of a spread to repeat itself, since we are talking about “hidden forces / energies” who probably choose the cards – God willing, the spread may repeat itself because it’s a mystical matter!

e) All the above – yes !!! here’s our correct answer!

Answer: letter e.

* The original article in Portuguese had a Loto calculation. I borrowed the Powerball Odds calculation from http://www.durangobill.com/PowerballOdds.html

4 thoughts on “SAT Math problem: Tarot probability

  1. It’s funny you did this blog. I’m writing a short story where I tried to calculate the odds of someone drawing a single card, replacing that card, shuffling and drawing the same card again. (two times in a row) Turns out the odds are 1 in 78. The previous shuffle and draw have no impact on the current drawing. I wonder?

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    • Yes, first you have 78 choices.The second time, in your case, if you put that same card back into the deck, reshuffle and take one, you’ve got another 78 choices. So, the probability of being the same card is 1 in 78×78 = 1/6,084
      However, if you’ve already chosen the first card, then it’s already picked. Then chance of getting the same is 1/78 as you said, because now you have 78 choices to pick.

      Like

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