Statistics on card frequency?

Thread originally posted on the Aeclectic Tarot Forum on 08 May 2004, and now archived in the Forum Library.

ana_ng	08 May 2004
I'd love to know if anyone has, or has seen, a chart or algorithm that shows the straight mathematical chance of drawing a given card in a spread. Or has any idea how to calculate it! For instance, if you drew 1 card, the card you drew would have a 1-in-78 chance of appearing in your hand. What about a 10-card spread? What about ratio of majors (only 22 cards in the deck) to minors? I've been pulling a LOT of majors in my spreads, and I want to calculate the odds of having five readings in a row contain over 50% majors.

Savoyali	09 May 2004
Hi ana_ng, welcome to AT! Hope you enjoy it. Maybe you'd like to hop on over to the New Members forum sometime and tell us a bit about yourself, we'd love to hear from you :) As for statistics, I'm sorry I don't have any, but there have been a couple of threads about the significance of many majors turning up in readings that might interest you, for example here and here. You may find more via the Search (the link is in the top right and corner of the forum pages). Hope that helps. :) Cheers, Savy

closrapexa	09 May 2004
Sorry, studied for my SATs a while ago but then promptly forgot everything I ever knew about statistics. However, alot of Majors in your readings may indicate that the cards are urgently trying to tell you something. The question is, are getting the message?

Alissa	09 May 2004
I'm not sure if I'm reading this right but I'll try to answer, in case I am.... (And I'm not a numbers person, so I may just shut up and go sit in the corner if I'm totally wrong in how I read this). You always have the same chance of pulling a card, 1 to 78. Each position in a spread has a 1 to 78 chance of receiving the same card that was there before. In psychology they taught us about "Gambler's fallacy," which is the belief that if you draw, say the Ace of Cups, you have "less" of a chance of drawing it again a second time - that it is "harder" to draw it twice in a row. But you always have a 1 to 78 chance, for every draw.

Osher	09 May 2004
Except, for every following card, the odds would reduce: It would be 1:78 for the first position, 1:77 for the second and so on. Or am I off a tangent?

hedgecub	09 May 2004
Quote: Originally posted by Happiness Except, for every following card, the odds would reduce: It would be 1:78 for the first position, 1:77 for the second and so on. Or am I off a tangent? You mean the odds would increase, since the sample set of possible cards is reduced :) It's been a couple years since I did Prob and Stat, but if nobody's gotten very far by week after next, then I'll have a go at a detailed answer with the help of my trusty textbooks ;) I've got exams next week, and it's Vector Calc, Differential Equations, and Complex Analysis I should be thinking about, so Probability will have to wait, though I much prefer it to all this math methods stuff :p

ana_ng	09 May 2004
While it is true that you have a 1 in 78 chance of drawing any one card, what I'm talking about is a little different. There are two types of cards in the deck: majors and minors. Since there are only 22 majors, your chances of that one card being a major versus a minor are (roughly) three to one. So in a 10-card spread, a "random" drawing should yield a ratio of 3 majors to 7 minors (here's where my math is shaky) over time. Given that the deck is more heavily weighted toward pips, the statistical likelihood of laying a spread that's 50% majors is much smaller than laying a spread that's 30% majors. What are those numbers and formulas? I want to do a graph of the last ten readings I've done, just to see how far off the "straight odds" I've been. What I want to generate is: The odds of laying one spread that's 50% majors are XX, The odds of laying three spreads in a row that are 50% majors are XXX (smaller number), The odds of laying ten spreads in a row that are 50% majors are XXXXX (etc). Yeah, that's a really small sample group, but it's just for my own amusement and edification.

ana_ng	09 May 2004
[quote]Originally posted by hedgecub [b]You mean the odds would increase, since the sample set of possible cards is reduced :) Oooh, that's right! So let me further elaborate: One deck of 78 cards, one-card draw. You have a one-in-three chance that the card you draw will be a major. You draw a second card from the same shuffle. Now the odds are slightly different based on your previous card: If you drew a pip, your odds of drawing a major are slightly higher (because the ratio of pips to majors has been reduced by one). If you drew a major, your odds of the next card being a pip are slightly higher (because your pool of potential majors has been reduced by one). How do those numbers march along in a 5-, 10-, 20-card draw? Am I strange in that I think this is FUN?

hedgecub	09 May 2004
Quote: Originally posted by ana_ng Am I strange in that I think this is FUN? Nope ;) I'm doing a pure maths degree for fun, and I intend to carry on doing maths for the rest of my life for fun... ...though it would help if I can get someone to pay me for it ;)

ana_ng	09 May 2004
[quote]Originally posted by Happiness Except, for every following card, the odds would reduce: It would be 1:78 for the first position, 1:77 for the second and so on. Or am I off a tangent? [/QUOTE Nope, you're totally right. I had amended that in a followup.

Arnaud	09 May 2004
How to compute the chance to have one reading with at least 50% majors out of ten, thus 5 majors out of ten. First step: How many different 10 cards spread there are ? Answer : TOT=78*77*76*75*74*73*72*71*70*69 =4566176969818464000 Second step: How many 10 cards spreads there are with five majors or more: A=22*21*20*19*18*73*72*71*70*69 =5695854848006400 (we force the 5 firsts to be major, and the others, we don t care, so we pick them among the remaining cards) Thus the ratio (hence the probability) is: P=A/TOT=0,00124740... (around 1 chance out of 1000) Now the chance to have five readings in a row, each one with the probability P, is P^5 =3.02e-15 Errr .. this is very, very, very low :-) (0.00000000000000302). Arnaud

Kittaine	09 May 2004
*cries* ;_; I'm gonna take up Statistics next term! I shouldn't have read this thread cos now I'm scared of Statistics! Oh well. Too late! >_< I'll save this thread and check it during the term, see if I'll understand it any better.

Savoyali	09 May 2004
::hands handkerchief to Kittaine:: don't cry! I barely scraped by high school maths, but then got straight twos (like American Bs) in uni psychology stats! It's all pretty logical and makes sense, don't worry :) Thanks, Arnaud, that was interesting!

ana_ng	09 May 2004
Quote: Now the chance to have five readings in a row, each one with the probability P, is P^5 =3.02e-15 Errr .. this is very, very, very low :-) (0.00000000000000302). *wild applause*

Inana	10 May 2004
Maths for fun? You are scary!! Have to applause indeed... before reading the posts again to understand it (sigh) :|

Arnaud	12 May 2004
Actually, there is one other thing to consider here, that is important: the intensity of the shuffling that was applied to the cards. Suppose we are in a situation where 5 Majors are present in a spread. We interpret them, then at the end we gather the spread in a stack, we put this stack back in the deck, and we shuffle the deck. IF the shuffling is not adequate (not intense enough), then the probability that these 5 cards will be next to each other in the deck (and thus the probability to have them again together in a spread in the future) is high. In this case, it would be *more* probable (not less) to have a second time 5 Majors in a row, than it was to have them the first time. So, the shuffling here is very important. The computations i suggested assumed there is a perfect shuffling between each spread (perfect in the sense that the order of the cards in the deck for each spread is completly independent from the order for the previous spread, which is may be difficult to achieve). Which leads to this question: how do you shuffle usually ? The classic way, by holding the deck in both hands ? The "poker" way by putting two stacks on the table and interleaving (?) them ? The "mud" way, by spreading all cards on table, face down, and mixing them around with both hands ? Arnaud Arnaud

Pook	12 May 2004
[quote]Originally posted by ana_ng [b] Quote: Originally posted by hedgecub Am I strange in that I think this is FUN? I think I'm gonna disagree with Hedgecub.....you're nuts! Or maybe it's that I would be if I tried 2 figure it out!! Wild Applause from me 2!! You have more patience than me Arnaud!!!

Savoyali	13 May 2004
Quote: Originally posted by Arnaud Actually, there is one other thing to consider here, that is important: the intensity of the shuffling that was applied to the cards. Suppose we are in a situation where 5 Majors are present in a spread. We interpret them, then at the end we gather the spread in a stack, we put this stack back in the deck, and we shuffle the deck. IF the shuffling is not adequate (not intense enough), then the probability that these 5 cards will be next to each other in the deck (and thus the probability to have them again together in a spread in the future) is high. Unless you pick the cards from a fan instead of dealing them from the top of the stack, because in that case it doesn't matter if they are technically next to each other since the order of picking is an additional randomizing factor.

Melpomone18	02 Jan 2005
I know that this is an older post, but when i read it I grew extremely curious and couldn't resist trying to figure it out for myself. So I fiddled with MS Excel for a bit and the answer that I got was different from the one that had been posted earlier. The previous answer neglected the orderings of the cards -- the majors may be at the beginning, the end, or mixed throughout the spread (only the first option was computed before). [the following is ugly, but i'm doing the same thing as before, just putting in the missing binomial coefficient, "10 choose n."] To find the answer, plug the following into Excel (the bracketed quantity is the cell number). [a1] 5 [a2] 6 [a3] 7 [a4] 8 [a5] 9 [a6] 10 [b1] =FACT(10)*(FACT(56)/FACT(56-(10-A1)))*(FACT(22)/FACT(22-A1))/(FACT(A1)*FACT(10-A1)*(FACT(78)/FACT(78-10))) [b2] =FACT(10)*(FACT(56)/FACT(56-(10-A2)))*(FACT(22)/FACT(22-A2))/(FACT(A2)*FACT(10-A2)*(FACT(78)/FACT(78-10))) ...and likewise for B3 to B6 -- make sure the number after "A" is the number after "B." [b7] =SUM(B1:B6) In words, B1 is the probability of getting exactly 5 majors in a 10-card spread (0.0799 ~ 8%); B2 6 majors (0.0218 ~ 2.2%); etc. B7 is the probability of getting at least 5 majors, .106 ~ 10%. The probability of getting at least 5 majors, three times in a row, is (.106)^3 ~ .001187, or roughly 1 in 1000. The probability of getting at least 5 majors, ten times in a row, is (.106)^10 ~ .000000000177 = 1.77E-10, or roughly one in ten billion. This was a pretty cool question. I had fun with it! :)

The Statistics on card frequency? thread was originally posted on 08 May 2004 in the Talking Tarot board, and is now archived in the Forum Library. Read the active threads in Talking Tarot, or read more archived threads.