Is there a statistician in the house?

Thread originally posted on the Aeclectic Tarot Forum on 20 Nov 2002, and now archived in the Forum Library.

dangerdork	20 Nov 2002
I was just thinking, it would be really interesting to know the odds of certain things happening with the tarot. Since the Celtic Cross is so common as to be almost a standard (like the RWS deck), let's assume a "spread" to be 10 cards. What are the odds of getting 5 Trumps in a spread? 6? 7? What are the odds of getting any particular card in a spread (easy, 10 chances, 78 cards - about 1 in 8 - i think). What are the odds of the same card turning up in 3 spreads in a row? 4? 5? What are the odds of getting multiple Court cards, etc. I'm sure most of you have seen things that statistically just seem outrageously unlikely to turn up by coincidence. Has anyone with an interest in statistics ever bothered to figure this stuff out? (I'm too lazy to drag out the math books for the formulae and do the work myself). Is there anything like this on the web? Any good stories of outrageous coincidences?

Linda698	20 Nov 2002
Hi, i did a spread of 7 cards a while ago and 5 of the cards were court cards. I did wonder what the chances of this happening are! I hope someone can come up with some info. Take care increase the peace.

raeanne	20 Nov 2002
Hi all, OK, I hope I can explain this well enough so that I don’t totally confuse everyone. Let’s say you want to figure the probability of 5 out of 10 cards being a court card. You first write down what the probability is that one card will be a court card. This is a fraction of the number of court cards divided by the total number of cards in the deck. For a regular tarot deck this would be 16/78. Then you write down the probability that it won’t be court card, which would be the number of non-court cards divided by the total number of cards in a deck. For a regular tarot deck this would be 62/78. Now, we want the probability that 5 (let’s call this number “matches”) out of 10 (call this number “total) cards court cards. Take the fraction 16/78 and raise it to the power of “matches” and multiply that by the fraction 62/78 raise to the power of “total” minus “matches”. Are you totally confused yet? So, 16/78 to the power of 5 times 62/78 to the power of 5 equals 0.000115 or slightly greater than 1 in 10,000. Now, to see the odds that 3 of the 10 cards would be court cards you would have 16/78 to the power of 3 times 62/78 to the power of 7. This comes out to be 0.00173 or 1 in 1730. In figuring the odds for a Major arcane card, the fractions would be 22/78 for a major and 56/78 for it not to be a major. So if you have a 15 card spread and you want to know the odds of 7 of them being majors, you would have 22/78 to the power of 7 times 56/78 to the power of 8 which is 0.00001 or one in 100,000! Linda698, For 5 out of 7 to be court cards, you would have (16/78)^5 times (62/78)^2 which equal 0.000229 or slightly better than 2 in 10,000 or 1 in 5,000!

ihcoyc	20 Nov 2002
This is where the math gets tricky. First, to get it exactly right, you need to subtract one from the deck after each card is drawn. Most readers do not deal from a shoe into which several different decks have been shuffled. The odds for the next card are changed, also, by whether the first card drawn meets the criteria you are testing for. For example: Odds that the first card drawn is a major: 22/78, or about 28.2% Odds that the next card will be a major: If the first card was not a major, 22/77, or about 28.6% If the first card was a major, 21/77, or about 27.3% And so forth for each consecutive card drawn. Likewise, the odds of drawing five consecutive court cards is not: (16/78)^5, but rather (16/78)*(15/77)*(14/76)*(13/75)*(12/74) Since you will run out of courts well before you run out of deck, the odds of the next card being a court diminish with each successive court dealt.

raeanne	20 Nov 2002
ihcoyc, You are figuring different odds than I am. If you want to figure the odds of five cards in a row being court cards, then yes, your method is the correct one. If, however, you want the odds of five out of ten cards being court cards then you have to use the formula I presented. I'm not asking that the first five cards drawn be court cards. You could have a pip card, then a couple of court cards, a couple more pips then a court, etc. This changes the odds from the simple card reduction method to an exponential one.

dangerdork	20 Nov 2002
this is fun! how about seeing the same card in consecutive SPREADS? or the same 2 cards, or 3... or the same card in the same position in consecutive spreads? surely somebody somewhere has already geeked on this and there's a nice chart of it on the web, so we can be lazy instead of doing all the math ourselves...

zander770	20 Nov 2002
great topic-- i was just thinking about this (in a way), too! i was going to merely take all the cards i had drawn--say two-hunded, to make the math simple. math is NOT my Strong Point--and just "average them out" on a calculator to see which card appeared most, for me (and, this thought came about because i hadn't drawn ANY MAJ KEY over #16--the tower, in what seemed like WEEKS). "weird spreads", for me, last week? 1) i read a 7 card spread were every key was ill-dignified! 2) i read a 7 card spread where the maj keys of #'s 5, 6, 7, 8, 9, 10 AND 15 appeared! (and, YES; i DO "shuffle"! ha! i mean, even if i DIDN'T "shuffle" the cards couldn't appear like that! or, i guess, they COULD, for they did.), and; 3) i read THREE 7 card spreads, regarding the same person, and only ONE CUP appeared, and that one was ill-dignified! over-all, lately, my readings for people have been "negative" (here's one: over the summer i read for a woman i hardly knew at all and w/in 3 minutes she begins "shaking," then: "crying"!), but, i do, however ("mostly") "tell it as it appears," for we ARE "ethical," and, moreover: "they asked for it..." right? right! ~Z~770 })

Mystica	21 Nov 2002
I'm sure that I'm the only one, and I will be sorry I posted this.... But...my eyes are rolling back in my head reading this!!!ROFLMAO

WillieHewes	21 Nov 2002
Remember that event that have incredible odds happen to millions of people every day. I don't think that calculating the statistics of certain particular combinations of cards adds to one's understanding of tarot. All combinations are meaningful, that's the whole point.

divinerguy	21 Nov 2002
In a related thread - the question was, How many card conmbinations in a 3 card spread (with reversals)? Take a look http://www.tarotforum.net/showthread.php?threadid=5834&highlight=statistics

Laurel	21 Nov 2002
Vianne- don't feel bad. Me too. But nonetheless its been an interesting thread! Laurel

dangerdork	21 Nov 2002
Well, to those of you rolling your eyes, here's one of the reasons I posted this: I think we all have a logical side and an intuitive side, and like a yin and yang we need to keep them in balance. Maybe it's the right brain / left brain thing, maybe something else. But there's a duality at work between the "scientific" thinker that wants to see everything in terms of cause and effect, logic and evidence; and the intuitive, spiritual, "go with the flow" side. There can be conflict between the two, and I believe that, like the Temperance card, one's outlook and philosophy should be a blending of both sides... One way that I am able to reconcile my scientific, logical side with the part of me who attaches so much significance and emotional impact to some little pieces of cardboard with pictures on them is by sheer weight of evidence. My rational persona simply cannot deny that something happens with the tarot that is absolutely inexplicable in terms of probability. I don't try to reason HOW they work, but my reasoning side can't deny THAT they work.

Mystica	21 Nov 2002
Dangerdork, no offense was meant. Maybe it was the way I wrote it, sometimes that's a problem online, we get misunderstood....I didn't mean rolling eyes in a sarcastic way, I meant all that math is way to much for me! I think it's very interesting, but I do get lost in all those calculations. I was laughing at myself because I couldn't keep up.

dangerdork	21 Nov 2002
no offense taken vianne - I keep telling myself I *could* look up the formulas and do the math if i really wanted to - I just want to talk somebody else into doing the hard part! LOL

juice	22 Nov 2002
Before I forget, what do you mean by ill-dignified? I was trying to compare the feel of the odds with something well known like how often you see a pair of aces. A straight flush in poker is so rare that I have never drawn one. Everybody gets a pair of aces regularly. So in Tarot a 10 card spread with 3 majors or courts might not be all that rare, so as to be worthy of a note beyond a simple comparison instead of big deal overall patern. There might have to be 5 or 7 to be significant enough claim that a life event is doing a sewer snake through the plumbing of your life. Most people only keep track of a card being seen again and not where it is seen, so we will ignore position for now. I just ran ran off fairly quickly the number of different combinations you can have based off of number of cards in a Tarot spread. 6006 2 card 456,456 3 card 34 million 4 card 2.5 billion 5 card 184 billion 6 card 13 trillion 7 card 945 trillion 8 card 66 quadrillion 9 card 4.5 quintillion 10 card 310 quintillion 11 card 20 sextillion 12 card 1.3 septillion 13 card 89 septillion 14 card 5.7 octillion 15 card that's 57 with 26 zeros after it. I did the same for a regular poker deck. 2652 2 card draw 132 thousand 3 card 6.4 million 4 card 311 million 5 card Then I ran into a problem. I'm currently stumped for formula. I did a simpler version of Raeanne's suggestion to help me get it down. What are the chances of seeing a 3 card spread with exactly 2 aces in it? By the given formula... (4/78)^2 X (72/78)^1 = 0.0024 The Harder Way. There are only 6 ways for 2 out of 4 cards to show up. There are only 74 cards that can sit inthat 3rd position. That's 444 matches. There are 456456 total possible combinations of 3 cards. Number of matches divided by total possible gives us, oops. (6x74)/(78x77x76) = 0.0009727 I ran some other math that would take a while to explain so I'll just short cut. I suspect the problem with the suggested formula from earlier in this thread is it is telling you at least so many and not exactly so many. Better, (example) you're finding out the odds that 5 or 6 or 7 or 8 or 9 or 10 out of 10 cards are courts. Next post...

juice	22 Nov 2002
Here are the odds that X out of 10 cards will be courts. All the numbers in the left column should add up to 1. 10 1.319e-6 9 3.792e-6 8 1.469e-5 7 5.694e-5 6 2.207e-4 5 8.55e-4 ...5 or more 0.00115 4 0.003313 .4 or more 0.004466 3 0.012.....3 or more 0.017 2 0.05.......2 or more 0.067 1 0.19.......1 or mroe 0.26 0 0.75.......none 100.6 ain't 'puters kewl Well that took a while to do. For anybody who's ever played dice, anything that happens as often as 5% of the time is going to feel like a common event. So 3 or 4 courts is noticeable but to a tarotholic who reads fairly often that would only be a road sign. I wouldn't start qawking til you hit 5 or more. May do some more charts or not. Are there any gee wiz I really need those?

raeanne	22 Nov 2002
Hi Juice, I think the problem you are having is in mixing combination mathematics with probability. These are two very different concepts and you will have major problems if you try to mix them together. Your statement of there being 6 ways for 2 out of 4 cards to show up is combination math for specific aces in combination. This is a very different concept than looking for the probability of two out of three cards being aces. If you are looking for two aces in a three-card spread, you can end up with one of the following three states: Ace, Ace, Non-ace OR Ace, Non-ace, Ace OR Non-ace, Ace, Ace For the different states list, the probability would be: 4/78 for the first ace, 3/77 for the second ace, and 74/76 for the last non-ace card OR 4/78 for the first ace, 74/77 for the non-ace, and 3/76 for the second ace OR 74/78 for the non-ace, 4/77 for the first ace, and 3/76 for the second ace Multiply these out and you will see that the probability is 0.002 for each state. It’s fairly easy to see where the formula (4/78)^2 times (74/78)^1 comes from by looking at these three states. I hope this helps clarify things a bit more.

dangerdork	22 Nov 2002
I really need those :) That is, if you really need an excuse to do them! This post turned out way more interesting than I thought. While everyone still seems interested... we have the chances of N court cards showing up in a 10-card spread... how would you apply the formula for the Trumps? 22 cards, not 16... Also, is it probability or combination math to ask... this happened specifically to me on Halloween... I did 8-10 readings , for strangers, with different decks, and the Hanged Man showed up every time. so what about probability of the same card showing up at all in multiple consecutive spreads? or in say, 7 out of 8? I'll do the math, but I don't know the algorithm? Or 2 or 3 or 4 cards showing up in consecutive spreads? It's fun geeking on this stuff...

raeanne	22 Nov 2002
Dangerdork, For the major cards, the formula would be (22/78)^N times (56/78)^Draw-N, with N being the number of majors that show up and Draw being the number of cards in your draw. So the probability of 6 majors showing up in a 10 card draw would be (22/78)^6 times (56/78)^10-6. For 7 majors to show up in a 15 card draw, it would be (22/78)^7 times (56/78)^15-7. There is a 10/78 probability that any specific card will show up in a ten-card draw. For the same card to show up in two consecutive spreads you would have (10/78)^2. For it to show up in three consecutive spreads, it would be (10/78)^3, and so on. For a fifteen card draw there would be a 15/78 probability that a specific card would show up. For the same card to show up in two consecutive 15 card draws it would be (15/78)^2. I hope this gives you the info you need.

SlyR	23 Nov 2002
I posted your question at rec.gambling.poker, and Barbara Yoon, the resident authority on probability, answered as follows: Example.....there are 52/3 x 51/2 x 50/1 = 22,100 'ways' to have 3 cards out of a 52-card deck (say as hold'em or Omaha flops, or seven-card stud starting hands)... So all you need to do to answer your question here is to similarly calculate the number of 'ways' to have 5 out of 16 "court cards," times the number of 'ways' to have 5 out of 62 'non-court cards,' out of the total number of 'ways' to have any 10 out of the 78 cards... P.S. This is for EXACTLY "five court cards" (that is, NOT for 'five OR MORE court cards' Bottom line answer is that there will be exactly five 'court cards' about 2.25% of the time, or once about every 44.5 times.

violinlily	23 Nov 2002
no offense intended, you like totally lost me!! lol, I have a little (verrrry little) background in statistics, but, whoa..... what I did read (and could understand, that was about 5 lines), was really interesting. great thread!!

juice	26 Nov 2002
Here is the chart if I do the numbers exactly as you listed Raeanne. No offense intended and maybe I just misunderstood. This can't be right since those probablilities do not equal 1 or 100 or any number that stands for all possible combinations. All of the percentages have to add up to 100% of the time. Truncating... 0 0.1006 1 0.0259 2 0.0067 3 0.0017 4 0.000465 5 0.0001152 6 0.00002973 7 0.00007674 8 0.000001980 9 0.0000005111 10 0.0000001319 My math wasn't quite right and I assumed the computer was rounding for me. Sorry about that. I had wondered why the formula was giving fits for 3 card percentages. Let's do it Barbara Yoon's way. First we need the giant divisor in her equation which is the total possible ways X number draws out of 78 cards can be. If you will fogive me the formula I used for how many possible combinations was dependant on which way the cards were drawn and not independant. So by Barbara Yoons formula. 1. 78 trivial :) 2. 3003 3. 76076 4. 1426425 5. 21111090 6. 256851595 7. 2641902120 8. 23446881315 9. 182364632450 10 1258315963905 11 7778680504140 12 43430966148115 13 220495674290430 14 1023729916348425 15 4367914309753280 16 17198662594653540 17 62724534168736440 18 212566476905162380 19 671262558647881200 20 1980224548011249540 If I didn't mess up the math wich is a possiblity. I did double check. :) I got carried away once I reallized it was a simple step from each entry the list to the next. When you get to 36 draws, that will be the biggest number, and then you go back through the list in reverse order. At least my mixed up way from before told you how many different celtic cross readings you could have. Now let's do a chart for X number of major card combinations. 1. 22 OK that one was easy! :) 2. 231 3. 1540 4. 7315 5. 26334 6. 74613 7. 170544 8. 319770 9. 497420 10 646646 11 705432 12 646646 13 497420 14 319770 15 170544 16 74613 17 26334 18 7315 19 1540 20 231 21 22 Now we need a chart for possible non-major card combos. 1. 56 I like these entries :) 2. 1540 3. 27720 4. 367290 5. 3819816 6. 32468436 7. 231917400 8. 1420494075 9. 7575968400 10 35607051480 11 148902215280 12 558383307300 13 1889912732400 14 5804731963800 15 16253249498640 The formula runs: (# of combos) x (# of non-major) / (total possible) out of 3 cards 0 1 x 27720 / 76076 .. 0.364 1 22 x 1540 / 76076 .. 0.445 2 231 x 56 / 76076 ... 0.17 3 1540 x 1 / 76076 ... 0.02 The numbers in the right column add up to 1 minus rounding error. The only time a major showing up in a 3 card spread ?might? count is if they're all majors. Correcting the court card possibilities gives us. 1. 16 2. 120 3. 560 4. 1820 5. 4368 6. 8008 7. 11440 8. 12870 9. 11440 10 so just go back up the chart non-courts 1. 62 :) OK so I can enjoy silly pleasures 2. 1891 3. 37820 4. 557845 5. 6471002 6. 61474519 7. 491796152 8. 3381098545 9. 20286591270 10 107518933731 11 508271323092 12 2160153123141 13 8308281242850 14 29078984349975 15 93052749919920 That should cover a lot of territory. Let me know if there are any important ones I missed. Just for entertainment the next line is the type of line I'de write off to the side to keep track of where I was in the math. 62 61/2 60/3 59/4 58/5 57/6 56/7 55/8 54/9 53/10 52/11 51/12 50/13 49/14 48/15 Just because the celtic cross is as common as dirt. And, we started with this question. 0. 1 x 107518933731 / 1258315963905 = 0.085 1. 16 x 20286591270 / 1258315963905 = 0.258 2. 120 x 3381098545 / 1258315963905 = 0.322 3. 560 x 491796152 / 1258315963905 = 00.219 4. 1820 x 61474519 / 1258315963905 = 00.089 5. 4368 x 6471002 / 1258315963905 = . 0.022 6. 8008 x 557845 / 1258315963905 = .. 0.0035 7. 11440 x 37820 / 1258315963905 = .. 0.00034 8. 12870 x 1891 / 1258315963905 = ... 0.000019 9. 11440 x 62 / 1258315963905 = ..... 0.00000056 10 8008 x 1 / 1258315963905 = ....... 0.00000000636 In short 5 courts isn't all that rare. Now 6 or more could be notable.

dangerdork	26 Nov 2002
JUICE! You RULE! Now if I can just figure out what exactly those are lists OF.... ;)

raeanne	26 Nov 2002
Juice, If we go back to the problem of two aces showing up in a three-card spread, I think I can explain why we are getting different numbers. The way I look at the problem, there are only three ways for this to happen: Ace + Ace + Non-ace Ace + Non-ace + Ace Non-ace + Ace + Ace. I don’t care which ace it is, just any old ace will do. The probability of any of these three events happening is the same, which is 0.002 (after rounding). Using your calculations there are six possible COMBINATIONS of two aces in a three-card spread: Ace of Wand + Ace of Cups Ace of Wands + Ace of Swords Ace of Wands + Ace of Pentacles Ace of Cups + Ace of Swords Ace of Cups + Ace of Pentacles Ace of Swords + Ace of Pentacles You might be figuring the probability for two SPECIFIC aces to show up but not the probability that ANY two aces will show up. With a 10 card spread for 6 Majors to show up, you are calculating the number of ways 22 cards can combine as six cards multiplied by the number of ways 56 cards can combine as the remaining four cards and dividing everything by the number of ways 78 cards can combine in a ten-card spread. Is this really going to give you the information you need if you want to know the probability of 6 Majors showing up in a ten-card spread?

juice	27 Nov 2002
Danger dork - Those list from 1 to 9 (or 15 or 20) were the numbers you plug into Yoon's equation without doing the bigger chunk of the math again and again. Yoon's example of (52/3 x 51/2 50/1) was to show how to get the number possible combos of 2 cards out of a 52 card deck and not care which card was drawn when. I translated that into: size of deck ..... deck size -1 .... deck size -2 ....... ------------ . x . ------------- .x. ------------- ...etc # of draws ....... # of draws -1 ... # of draws -2 ...... until (# of draws -N) = 1. If you are drawing for majors, the size of your "deck" is 22. If you are counting the part of the deck or possible draws that are not majors, then the size of your "deck" is 56. When you're trying to think of 5 out of 7 cards are majors and 2 out of 7 are not, it is like you have 2 "decks" shuffled together. That was only part of the formula, the part inside one set of parantheses in the following equation. (# of Y card combos in your majors "deck" you could draw) times (# of Z-Y card combos in your non-majors "deck" you could draw) divided by (# of Z card combos from the combined whole deck you could draw) = chance that Y out of Z cards will be majors If you want to find out what the odds that 5 out 7 cards will be pips, you do the formula once and go on about your business. But if you want to ask a whole string of questions like how many minors need to turn up before the odds of "that" happening go below 1%, you'll end up doing some of the math over. Lets do it with a bit more detail. Most of those lists were the number of cards of type "whatever mentioned" and the possible number of combinations for that number of cards of that type. Using the second version of the math, the one suggested by Barbara Yoon, lets do the simplified case of 2 aces in a 3 card spread. From the list posted yesterday the number of possible combinations of 3 of the 78 cards no matter what order they are drawn in is 76076. 76076 = (78/3 x 77 / 2 x 76 /1) We can skip that last /1 part because anything divided by 1 equals whatever it was before. That isn't that much figuring on a calculator. A 15 card spread may make you do it again and again to be sure you didn't lose track. (Eample 78/15 x 77/14 x 76/13 x 75/12 x 74/11 x 73/10 x 72/9 71/8 x 70/7 x 69/6 x 68/5 x 67/4 x 66/3 x 65/2 x 64/1 :) ) Back to the shorter case. The number of ways to have 2 out of 4 cards would be 4/2 x 3/1 which equals 6. The 3rd number we need for the formula is to have number of ways to have 1 non-ace card which is 74. 6 x 74 / 76076 = 0.0058 (getting out calculator again) Now we can verify the math by calculating 0, 1, and 3 aces to complete the question of how many aces can come up in a 3 card spread and what are the possibilities of each way? no ace (# of no ace combos: 1 way = none) x (# of 3 non-ace card combos) / (# of 3 card combos) 1 x (74 / 3 x 73 / 2 x 72) / 76076 = 0.8521 or 85.21 % (rounded) which makes sense since we rarely see aces. 1 ace (4/1) x (74 / 2 x 73) / (76076) = 0.1420 We see 1 ace a little less than 15% of the time. 3 aces (4/3 x 3/2 x 2/1) x (1) / (76076) = 0.00005 or 0.005% All of the time you draw 3 normal tarot cards you will get one of the four conditions. You'll get no aces 85.21% of the time. You'll get 1 ace 14.20% of the time. You'll 2 aces 0.58% of the time annd 3 aces 0.005% of the time. Now did that make it clearer? Clear as mudd? Raeanne - I'm afraid this formula may approximate the odds part of the time but it just doesn't add up to all the chances. Some of the chances are missing. (4/78)^0 x (74/78)^3 = 0.853908 (4/78)^1 x (74/78)^2 = 0.04615 (4/78)^2 x (74/78)^1 = 0.00249 (4/78)^3 x (74/78)^0 = 0.00013486 Yes I do want to know the combinations. Lets make the case even simpler with 4 aces and 4 twos and no other cards. If you drew 1 card at a time, in a perfect world, all 8 of the cards would be drawn once each in 8 1 card spreads. That is 4 times you'll draw an ace no mater what ace it is. Not just once. Trying to say it another way. If I were being picky about what order the cards fell in, ace-wand-ace-cup-other would not be the same as ace-cup-ace-wand-other or other-ace-cup-ace-wand or ace-wand-other-ace-cup. Whoever that was= Somebody will ned to go back to the poker forum for card in particular position or card showing up again which is like having 2 or more decks. This formula doesn't handle those variables.

dangerdork	27 Nov 2002
Juice - well it seems at least the question I asked as the title of this post has been answered. ;) I'm gonna have fun with this!

raeanne	27 Nov 2002
Juice, I have the Schaum’s Outline series Statistics book. When I work the examples with your formula, I get the wrong answer. When I work them with mine, I get the right answer. So, I don’t know what to say. This is tarot! Do it however you want. Have a good holiday.

The Is there a statistician in the house? thread was originally posted on 20 Nov 2002 in the Using Tarot Cards board, and is now archived in the Forum Library. Read the active threads in Using Tarot Cards, or read more archived threads.