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The Zener cards were invented by Karl Zener and used by J B Rhine in his experiments on extrasensory perception (ESP) at Duke University in the 1930s. They comprise cards of each of five types, showing a square, circle, star, cross, or wavy lines:

cards

A standard pack contains 25 cards, five of each type.

A run consists of a subject trying to guess each card in a pack in turn. If we replace and shuffle after each guess, his expected score is 5. But what if we don't replace, and we allow our subject to keep a record (or he just remembers) how many cards of each type have already come up. What is the expected score then?

The answer is greater than 5. The probability of getting the 25th card right must always be 1, and that of getting the 24th card right must be either 0.5 or 1, depending on what has gone before. Similarly the probability of getting the 23rd card right cannot be smaller than 0.33. So the expected contribution to the score from the last three guesses alone is greater than 1.83. So the expected score after a run is greater than $\frac{22}{5}+1.83=6.23$. How large actually is it?

(I've now asked the misère version of this question - what's the expected score if we try to minimise it - here.)

  • What is his method of guessing after the first guess ? – true blue anil Mar 11 '16 at 03:46
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    @trueblueanil - The method is as follows: always guess that the next card will be of the most numerous type remaining. If two or more types are equally most numerous, choose one of them at random. So if the first card is star, then guess that the second card is one of the other types. –  Mar 11 '16 at 11:27

2 Answers2

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The expected score is 8.65, to 2 decimal places.

I ran a Monte Carlo simulation with a million runs, getting an average score of 8.649, and after searching the web for "8.65" and "Zener cards" I soon found Ronald Read's article in the American Mathematical Monthly 69(6), 1962, in which he gets 8.65 by using an exact method that I do not fully understand.

The guessing technique I used was a variation on the one I gave in the comment in answer to @trueblueanil's question. I wrote the card types in the order (square, circle, star, cross, wavy) and always guessed that the next card would be of one of the most numerous types remaining. But when two or more types were equally most numerous, rather than choosing among them at random I chose the rightmost one on the above list. Over a million runs, the average number of hits was 8.648557, the largest number was 18, and the smallest was 5.

An example of when 5 occurred was the following ordering:

(cross, star, cross, star, star, cross, star, circle, square, wavy, cross, star, square, circle, square, circle, square, circle, square, cross, wavy, wavy, wavy, wavy)

when every guess is "wavy"!

  • Just for fun, here is the exact number: ${23148348523\over 2677114440}= 8.646753451$. –  Mar 13 '16 at 00:33
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I suspect the lowest expected score comes if all the stars are last, and the total is $1/5+1/4+1/3+1/2+5=6\frac{17}{60}$.
I suspect the largest expected score comes if you have one of each, then another one of each, and the score is $5(1/5+1/4+1/3+1/2+1)=11 \frac5{12}$.
I don't know how to average over the actual sequence.
As I claim in comments, we average $1+1/2+1/3+1/4+1/5$ for the first cards of each suit.
The average score for the second cards of each suit depends how many four-card suits remain once the first cards of all suits have appeared. For example, if three suits still have four cards each when the last suit has appeared, we expect a score of $1+1/2+1/3$ from the second-cards.
I think we have a score $$\sum_{n=1}^5\sum_{m=1}^{a_n}\frac1m$$ where $a_5=5$ and $a_n$ is the number of $n$-card suits that remain when the last $n+1$-card suit has gone.

Empy2
  • 50,853
  • +1. Nice. But how do you get the second to fourth terms in the sum for the first case? Say the first card is circle. Then we choose our guess for the second card at random among star, square, cross and wavy. The probabilities that we're right are 0 if we choose star and $\frac{5}{20}$ otherwise, giving an expected score for the second card of $\frac{1}{4}\frac{5}{20}3=\frac{3}{16}$. –  Mar 11 '16 at 12:04
  • We get $1/5$ chance of picking the first suit to appear. We get $1/4$ of picking the second suit to appear. Then $1/3$ chance of picking the third suit; $1/2$ of picking the fourth suit, and $1$ chance of picking the last suit to appear. – Empy2 Mar 11 '16 at 12:07
  • Thanks - I understand the train of thought now. But what assumptions are you making in the first case about the ordering of the cards before the stars at the end? If cards 16-20 are all cross, isn't our expected score for cards 16-25 equal to $5*\frac{1}{2} +5=7.5$? –  Mar 11 '16 at 12:27
  • Oops - my 7.5 there is wrong. We would guess either star or cross for card 16, with probability $\frac{1}{2}$ of being right, and then we would guess star for each of cards 17-25, being wrong the first 4 times and right the last 5, giving an expected score of 5.5 for cards 16-25: $\frac{1}{2}$ for the cross and 5 for the stars, which supports what you said. –  Mar 11 '16 at 13:07