**Motivation.** In a coin game, a player flips all their coins every turn, starting with just one coin. If the coins all land *heads* then the game stops; otherwise, the number of coins is doubled for the following turn. While the game may clearly terminate on any turn, there is in fact a positive probability that it will never terminate: this is the infinite product $$p = \prod\_{i=1}^\infty \left(1-\frac{1}{2^{(2^i)}}\right).$$


**Questions.** Do we have $p \in \mathbb{R}\setminus \mathbb{Q}$? Is $p$ transcendental?



Is $\prod_{i=1}^\infty (1-\frac{1}{2^{(2^i)}})$ transcendental?
I can prove that the number actually described by the word problem, which is $$ \prod\_{i=0}^{\infty} \left( 1- \frac{1}{2^{2^i}} \right),$$ is irrational, by a method similar to David Speyer's.


Expanding out the product, we get $$\sum\_{j=0}^{\infty} \frac{(-1)^{t\_j} }{2^j}= 1+\sum\_{j=1}^{\infty} \frac{(-1)^{t\_j} }{2^j} = \sum\_{j=1}^{\infty} \frac{1}{ 2^j} + \sum\_{j=1}^{\infty} \frac{(-1)^{t\_j} }{2^j} = \sum\_{j=1}^{\infty} \frac{1+ (-1)^{t\_j} }{2^j} = \sum\_{j=1}^{\infty} \frac{1+ (-1)^{t\_j} }{2} \cdot 2^{1-j} $$ where $t\_j$ is the number of $1$s in the binary expansion of $j$.


Thus the binary digit in the $j-1$th place is $1$ if $t\_j$ is even and $0$ if $t\_j$ is odd.


Since this sequence is not periodic, the number is irrational.


