In a Zener card test using a 25 -card pack and no replacement, what is the expected score? 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:

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.)
 A: 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"!
A: 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.
