# Optimal strategy in a coin game has unexpected symmetry?

The game:

I am going to toss a fair coin and you are trying to determine if I tossed a Head or Tail.
You do this using the rule I follow when I toss my coin:

I have before me $$2$$ $$\color{red}{\text{red}}$$ boxes each with a $$\color{red}{\text{red}}$$ marble inside them, I also have $$2$$ $$\color{blue}{\text{blue}}$$ boxes each with a $$\color{blue}{\text{blue}}$$ marble inside of them. Finally, there are two empty white boxes.

If I toss a Head I must move a $$\color{red}{\text{red}}$$ marble from a $$\color{red}{\text{red}}$$ box into an empty white box. Similarly, if I toss a Tail I must move a $$\color{blue}{\text{blue}}$$ marble from a $$\color{blue}{\text{blue}}$$ box into an empty white box.

To aid you in your guess of my toss, once I have moved a marble, you are then permitted to open and examine the contents of a single $$\color{red}{\text{red}}$$, $$\color{blue}{\text{blue}}$$ or white box.

The strategy:

In this primitive instance of two boxes of each color, we are actually indifferent between what box we peek into. We have 75% probability of correctly guessing the toss.

Consider now we have $$R$$, $$\color{red}{\text{red}}$$ boxes, $$B$$ $$\color{blue}{\text{blue}}$$ boxes and $$W$$ white boxes. The optimal strategy is to peak into min{ $$R,B,W$$ }.

I was curious as to why we have symmetry between the colored and white boxes and simply follow the heuristic of the minimum number of boxes, is there a nice transformation of the game that makes this symmetry more obvious? Thanks

Here’s a transformation that clearly doesn’t change the game, but makes the analysis simpler: have the first player privately distinguish three boxes, one of each color, in advance before flipping the coin. They then proceed as before, but must move the marble from the distinguished red or blue box to the distinguished white box.

We now reason as follows.

• If the second player happens to examine a non-distinguished box, they gain no information and every strategy will win 50% of the time.
• If the second player happens to examine a distinguished box, they will win 100% of the time with optimal play:
• If they examine a white box and find a red or blue marble, they know that marble was moved.
• If they examine a red or blue box and find it empty, they know a marble of that color was moved.
• If they examine a red or blue box and find it full, and they assume it was one of the distinguished boxes, they will correctly guess that a marble of the other color was moved. They can and should make this assumption, since we established above that it will not hurt.

Therefore, the second player should maximize their chance of examining a distinguished box, which means they should examine a box with the fewest number of boxes of the same color.

• I don't see this addresses the symmetry question. Forget about boxes R G W with different values, assume R=G=W. The OP asserts (I think that rightly) that examining the R (or G) box is equally optimal (gives the same probability of right guess) as choosing the W box. This is not obvious. And I don't think this helps to see that. Commented Feb 27, 2022 at 18:38
• @leonbloy I've reduced this to the game where the first player privately picks three distinguished boxes, one of each of the three colors, and the second player tries to find a distinguished box, winning 100% of the time on success and 50% of the time on failure. This is symmetric in the three colors. Commented Feb 27, 2022 at 19:32
• I upvoted because this is a neat trick. But IMHO there is a tiny gap. In the last bullet, the player certainly can make the assumption, but you need a bit of math to show that the player should make the assumption. Or perhaps you consider it "obvious" that if you see a full R/B box you "obviously" should bet the other color. :) Commented Mar 4, 2022 at 20:28
• @antkam That's not a gap; the problem explicitly asks about the optimal strategy, and I was clear about where I used this. Without that provision, there's no reason to expect that the win probabilities will be symmetric at all (just like there's nothing to stop a non-optimal rock-paper-scissors player from asymmetrically committing to always choose paper). Commented Mar 4, 2022 at 20:43
• My point is that "optimal" = "make that assumption" needs to be proven. But perhaps you consider that obvious. Commented Mar 4, 2022 at 21:02