QUIZ - Number of combinations in a 3 card spread?

Thread originally posted on the Aeclectic Tarot Forum on 25 Jul 2002, and now archived in the Forum Library.

divinerguy	25 Jul 2002
Quiz - How many possible card combinations are there in a three card Tarot spread? Assume a 78 card deck and reversals for all three positions. Just like in school, you have to show your work (your formula) to get full credit.

Mikael	25 Jul 2002
MM divinerguy, this was the same question i had a couple of days ago so here is my awnser: 78x77x76=456456 options Bb, Mik

fairyhedgehog	25 Jul 2002
Except that if you're allowing reversals, wouldn't that give you: (78 x 2) x (77 x 2) x (76 x 2) = 156 x 154 x 152 = 3,651,648 I think that the basic formula is right, cos I checked it using 3 cards (A,B,C) and 4 cards (A, B, C, D) and it worked :) Do you want all that set out here? EG, for 3 cards you can have ABC ACB BAC BCA CAB CBA =3 x 2 x1 = 6 options. For 4 cards, the combinations with A as the first card are 6, and there are 4 possible 1st cards, so there are 4 x 6 possible combinations, ie 4 x 3 x 2 = 24 I haven't checked my working with reversals - maybe someone else can tell me if I'm right or not :)

raeanne	25 Jul 2002
fairyhedgehog, I double checked with a college math instructor and you are absolutely correct with your formula: (78 x 2) x (77 x 2) x (76 x 2) = 156 x 154 x 152 = 3,651,648.

Sam	25 Jul 2002
c'mon! i'm on summer break! don't ruin it with this "math"! lol! j/k. very interesting!

Keslynn	25 Jul 2002
Now calculate the probability that a card will end up in the same position in a three-card spread two times in a row. *evil grin* ;) Kes

squigglywiggly	25 Jul 2002
Quote: Originally posted by Keslynn Now calculate the probability that a card will end up in the same position in a three-card spread two times in a row. *evil grin* ;) Kes Ugh to many smart people on this board. Come on y'all this is supposed to be fun and games don't make it into busywork. I always hated this stuff in high school anyways.

amyel	25 Jul 2002
EEEEEWWWWWWWWW!!!!! This post requires math!!! Divnerguy, don't mess with my poor lil' art history brain in this manner!!!! LOL

fairyhedgehog	26 Jul 2002
But amyel, I'm feeling so pleased with myself, cos it was a maths problem and I didn't freak out ;) When I was in primary school (I think that is elementary school in the States) I was so humiliated in maths lessons I felt terrified of maths altogether. It is a major thing for me to realise that I'm not petrified by maths any more. So, go divinerguy :D LOL Love and light, FH

divinerguy	26 Jul 2002
This illustrates how a book on Tarot card combinations would be a daunting task.

amyel	26 Jul 2002
Quote: Originally posted by fairyhedgehog But amyel, I'm feeling so pleased with myself, cos it was a maths problem and I didn't freak out ;) When I was in primary school (I think that is elementary school in the States) I was so humiliated in maths lessons I felt terrified of maths altogether. It is a major thing for me to realise that I'm not petrified by maths any more. So, go divinerguy :D LOL Love and light, FH Don't me wrong, all! I'm impressed with all the calculations...but now my head is spinning and I'm feeling like I'm falling in a dark well and my brain is trying to visually picture all these combinations and...ACK! I'm just gonna drink some more coffee now and try to find a nice book on the Fauve movement.....

Geenius at Wrok	26 Jul 2002
Quote: Originally posted by divinerguy This illustrates how a book on Tarot card combinations would be a daunting task. Indeed. Not counting reversals, there are 3,003 potential card pairs.

Phoenix	26 Jul 2002
Quote: Originally posted by Geenius at Wrok Indeed. Not counting reversals, there are 3,003 potential card pairs. And not only pairs, but there are also triplets and quads as well.

jade	26 Jul 2002
Quote: Originally posted by divinerguy This illustrates how a book on Tarot card combinations would be a daunting task. and impossible because depending on what the topic of the reading is.........would depend on the interpretation. *jade runs off for a cold compress*

Minderwiz	27 Jul 2002
Just to be really pedantic and mathematical during the summer break can I point out that there is a difference between Combinations and Permutations. The calculation originally given - 78 x 77 x 76 = 456,456 is a permutation. For a Tarot reading this is more appropriate than a combination because the sequence or placing of the cards determines the overall interpretation. A combination does not treat the sequence as being significant - that is the order in which the cards are drawn doesn't matter. The combination of three cards from 78 is 76,076 (you divide the first total by the result of 3x2x1). On reversals the correct answer was given above. Strictly speaking the draws are not independant because getting, say, the Emperor reversed precludes the possibility of drawing the Emperor in its upright position. but the 156 x 154 x 152 calculation does show the correct chances of drawing the cards. Getting the same three cards in the same three positions twice running is 3651648 squared Regards Minderwiz

Keslynn	27 Jul 2002
*stares at Minderwiz in awe* Wow! I was kidding. lol But you rock! :) Kes

squigglywiggly	27 Jul 2002
Quote: Originally posted by Minderwiz Getting the same three cards in the same three positions twice running is 3651648 squared Regards Minderwiz That's not what the girl asked, smartypants. Quote: Originally posted by Keslynn Now calculate the probability that a card will end up in the same position in a three-card spread two times in a row. What you calculated is easy as pie to calculate. Even I can do it in my peabrain. What she asked is hard! Do her thing if you can.

Minderwiz	28 Jul 2002
Well OK if you insist, but don't complain again! The probability of any card being drawn first in any spread is one in 78 or 0.12821. Therefore the a priori probability of the same card being drawn first in two consective spreads is 0.12821 squared or 0.000164 (just over 16 in 100,000) To be drawn second in two consecutive draws is conditional on the card not being drawn first in either draw. Which is 77/78 or 0.987179 chance each time. The probability of the card then being drawn second is one in 77 each time or 0.12987 So the probability of a card being drawn as the second card is 0.987179 x 0.12987 or 0.12821. This is the same probability as for a card to be drawn first and so the probability of it being in the same position twice running is also the same. For the card to appear as the third card in two consecutive draws is conditional on it not being drawn as the first or second card. That is 77/78 x 76/77 = 0.974359. The chances of the card then being drawn third are one in 76 or 0.013158. Thus the chances of a card being drawn as the third card is 0.12821. This again is the same probability as being drawn first. A priori the chances of any card being drawn in any position are the same because as more cards are drawn the probability that the card has not already been drawn goes down, and the probability of it being drawn next increases. Thus for any number of cards in a spread the probability of a card being drawn in the same position twice in a row is always the same, 0.000164 The probabilities above are a priori, that means calculated before the 'test' takes place. Obviously if we are looking at the position of, say the 10 of Swords, if it comes second on the first spread then the chances of it coming second in the second spread rise to nearly 13% however if it is then drawn first on the second spread, the chances of it being drawn second become zero, even before the second card is actually drawn. Well you asked for it, I normally woudn't have bothered with a 'correction' as most card users (apart from Poker and Bridge players) don't care for statistics. Best Wishes Minderwiz

jade	28 Jul 2002
i am sooooo way beyond impressed! excellent job! in light, jade

squigglywiggly	28 Jul 2002
You once again gave a correct solution to the wrong problem. You solved for a specific card coming up twice. Kes is asking for any card whatsoever coming up twice. Read her problem again, it's quite clear. Obviously I'm going to have to do this one myself. The answer to Kes's problem is: 1/78 + (77/78 x 1/77) + (77/78 x 76/77 x 1/76) = 3/78

Minderwiz	29 Jul 2002
I'm sorry but you really do need to learn something of the laws of probability as well as reading what Kes actually asked. All that you have calculated is the probability that a card (any particular card) will appear in a three card reading (in any position). That clearly is 3/78. What Kes said was 'Now calculate the probability that a card will end up in the same position in a three-card spread two times in a row' 'A card' is clearly any specific card from the 78 it can be any one of the pack but the inference is clearly that this 'chosen' card should occur in two consecutive readings (of a three card spread) in the same position. Only one outcome meets this criterion, i.e. the chosen card appears in the same position in both readings For this problem there are no alternative acceptable outcomes - i.e. the Emperor appearing second card in one reading and the 10 of Swords appearing second in the second reading is not an acceptable outcome. The Emperor (or ten of swords, or any other card) must appear in the same position in both readings. Therefore your use of the addition rule of probability is not valid - it is only used when assessing the probability of more than one outcome that meets the criteria set. All that you have calculated is the probability that any card will appear somewhere in a three card reading, either in the first, or the second or the third position - which is clearly 3/78. I teach this stuff, (for my sins) so I do know what I'm talking about. I think we should let the matter rest at this point Minderwiz

squigglywiggly	29 Jul 2002
Quote: Originally posted by Minderwiz I teach this stuff, (for my sins) so I do know what I'm talking about. Not very well, obviously. The original reading will produce a set of three cards, A-B-C. We don't really care what those three cards are. We just know that they are three different, unique cards. So what we want to calculate is 1) the probability that the first card of the second day's reading will be A, plus 2) the probability that the second card of the second day's reading will be B, given that the first card of the second day's reading is not A, plus 3) the probability that the third card of the second day's reading will be C, given that the second card of the second day's reading is not B & the first card of the second day's reading is not A. These three probabilities are completely independent of each other and so can be summed.

Minderwiz	29 Jul 2002
We seem to be interpreting Kes’ question in two different ways. My reading is that what is required is the probability of the same card occupying the same position in two consecutive spreads, irrespective of the other two cards in the spread. My answer to this interpretation was correct. Your latest post seems to suggest, that you interpret it to mean that only one card must be in the same position and the other two must be different, that is either A or B or C is in the same position but not two or more of them (not quite what Kes asked). However this interpretation does not quite fit your own explanation which has A appearing twice in the first position but apparently not bothering with whether B or C are in their relevant positions and which then goes on to ignore whether C is in the third position given that A is not first and B is second. I will proceed on the basis that you do want one and only one card to appear in the same position. Acceptable answers would be that A was first (B & C not recurring), or B was second (A and C not recurring) or C was third (A and B not recurring) and the addition rule would indeed apply. The probability of A being first in both spreads but B and C not being in their original positions is 1/78 x 1/78 x 76/77 x 75/76 or 0.0001601. This is the first of our three possible answers. The probability of B being second in both readings but A and C being different is 1/78 (think about it) x (77/78 + 77/78) [neither a nor b is first] x 1/77 [b is second] x 75/76 [c is not third] The probability of C being third in both readings but A and B are not in their original position is 1/78 x (77/78 + 77/78) [neither a nor c is first] x (76/77 + 76/77) [neither b nor c is second] * 1/75 [c is third] As each of these three outcomes satisfies the requirements then adding the three resulting probabilities together will give the overall probability – as you will no doubt realise it is nowhere near as high as 3/78. A patient Minderwiz

squigglywiggly	29 Jul 2002
Quote: Originally posted by Minderwiz I will proceed on the basis that you do want one and only one card to appear in the same position. Nope, you misinterpret. I solved for at least one card showing up again on two consecutive days. It's also ok if two or three show up again. 1) When I say 1/78, it covers all cases of card-matching that include A's. Some of these card-matching cases include B's and C's matching as well. 2) When I say 77/78 x 1/77, it covers all cases of card-matching that exclude A's but include B's. Some of these card-matching cases include C's matching as well. 3) Finally, when I say 77/78 x 76/77 x 1/76, it covers all cases of card-matching that exclude A's and B's but include C's. The reason for progressively eliminating each card in the series for the consecutive terms in the sum, is to make sure that we count each card-matching case once and only once.

Keslynn	29 Jul 2002
Wow, I didn't want to start an argument. I was just teasing! All this stuff is way over my head. However, I will clarify for the general peace. ;) I believe that Minderwiz is actually getting what I asked. I wanted to know the probability of a single card showing up in the exact same position 2 times in a row. For example, the King of Swords showing up in the Past position 2 times in a row. Does that help? :) (confuzzled) Kes

squigglywiggly	29 Jul 2002
You need to clarify a little bit more. Are you picking the card before you do the two readings. Or is it ok if any card that shows up in a first reading also shows up in the same position in the second reading. (Leave it up to fate to pick the cards for the first reading) Also, are you wanting one and only one repetition. In other words, is it ok if two or three cards repeat in the same positions.

Minderwiz	29 Jul 2002
OK Kes, I think that the discussion is getting a bit technical and we are in danger of overdoing it. I do now understand what Squiggly is trying to say and there is a genuine point to it - though if you meant one and only one card we've exhausted the 'argument' However if Squiggly is interested in continuing the discussion through the private message system, so as not to bore everyone else I'm quite willing to do so and there is more chance of both of us learning from each other. Best wishes to all Minderwiz

Ravenswing	21 Aug 2002
Quote: Originally posted by Keslynn Now calculate the probability that a card will end up in the same position in a three-card spread two times in a row. *evil grin* ;) Kes depends on what position. ravin'

Jewel	21 Aug 2002
Now you see, if statistics and probabilities had been taught to me using the tarot I bet I would have learned much quicker! *LOL*. Fascinating discussion ... I'm not bored!

Lightlike	21 Aug 2002
Quote: Originally posted by Jewel Now you see, if statistics and probabilities had been taught to me using the tarot I bet I would have learned much quicker! *LOL*. Fascinating discussion ... I'm not bored! I second that especially with Minderwiz teaching! :)

Minderwiz	21 Aug 2002
Thanks for your vote of confidence! Actually if people did use something they were interested in as the example in a statistics question then they really would do better. That's why gamblers and bridge players are very good at working out the probabilities. Incidently I learned my mental arithmetic playing darts and working out what I needed to win the game - though unfortunately whilst the calculations were correct they were not matched by the accuracy of my throws. Minderwiz

The QUIZ - Number of combinations in a 3 card spread? thread was originally posted on 25 Jul 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.