# 100 blank cards, minimize the EV

To solve this question, it comes down to finding the number of $$1$$s we should use, defined as $$x$$. We can then assign $$\lfloor\frac{x}{2}\rfloor$$ $$2$$s, $$\lfloor\frac{x}{3}\rfloor$$ $$3$$s and so on. We must now minimize $$x$$, given that the total sum must be greater than or equal to $$100$$:

$$min(x): \sum_{i=1}^{100}\left\lfloor\frac{x}{i}\right\rfloor \ge 100$$

The lowest integer $$x$$ for which this is true is $$28$$, resulting in a total sum of $$101$$. The expected value then equals $$0.28$$ when the person guesses $$1$$, and $$0.28$$ or lower for every other number.

Yes, you have the right idea: With $p_i$ the probability of drawing a card with number $i$, the expected value of choosing $i$ is $p_i \cdot i$, and you want to make this roughly equal for any $i$. Or, to be more exact: you want to find a value $E$ so that $p_i \cdot i$ is always smaller or equal to $E$ for all $i$.

Just by playing around a bit, I found that you can always get $p_i \cdot i$ at or below $0.28$:

$28$ cards with the number $1$

$14$ cards with the number $2$

$9$ cards with the number $3$

$7$ cards with number $4$

$5$ cards with number $5$

$4$ cards each for numbers $6$ and $7$

$3$ cards each for numbers $8$ and $9$

$2$ cards each for numbers $10$ through $14$

$1$ cards each for numbers $15$ through $27$

For a total of $28+14+9+7+5+2\cdot 4+2\cdot 3 + 5 \cdot 2 + 13=100$ cards, and the best the person can do here is to get an expected value of $0.28$ by picking any of the numbers $1$,$2$,$4$,$7$, or $14$.

Also, it is clear that this is the best you can do: to do better, you'd need to get $27$ cards of $1$, $13$ with $2$ ... and already you'd need a card with $29$. So, the best you can do is to write the numbers as spelled out above.