Conditional probability in branching game where you have a chance to win more chances to win the game

Question

I've been stuck on this seemingly simple social media problem.

Here is the game:

1. You have one chest to open.
2. 50% of the time the chest is empty.
3. 30% of the time it contains a ticket. If you get a ticket, you win the game.
4. 20% of the time you open the chest to find two chests inside, each with odds identical to the original chest.
5. You can open all chests. You lose when you have not found a ticket and there are no more chests to open.
6. What is the chance that you will get at least one ticket and win the game?

The answers I've seen all seem to use the frequency of empty versus gold chests, but this seems wrong. The question appears to be: "in what share of possible iterations does a ticket appear at least once?"

The inverse would be, "in what proportion of iterations do we reach extinction before the condition T (ticket) occurs at least one time?"

What is the right method to use here?

It seems like a branching survival process with a 50% chance of death, 20% chance of reproduction, and 30% chance of condition T occuring at each iteration, and our question is what proportion of runs will T occur at least once (which is not the same as the frequencies, correct?).

Edit: follow up question - Suppose you open the chest and find two smaller chests. Then take all 9 possible combinations of the values of the two new chests and multiply them by 20% (the likelyhood of getting the two new chests in the first place) to determine the likelihood with which they occur, e.g., empty/empty = 50% × 50% × 20%, empty/ticket = 30% × 50% × 20%.

There is an obvious problem here. The probability of all possible results is greater than 20%. I'm assuming this is because the values of the two new boxes is dependant on the value of the initial one, i.e., they are not identically and independently distributed.

Does this mean the frequency of a distribution represented by the algorithm below vary from Empty = 50%, Ticket = 30%, More Chests = 20%?

"R = random number between 0 and 1 Distribution Function = If R ≤ 0.50 then "Empty" else If R ≤ 0.80 then "Ticket" else If R ≤ 1.00 then "Chest" and rerun Distribution Function 2 times

For the sake of conceptualization, let's say we ran the above a million times. Does our frequency vary from .5/.3/.2?

Extra detail - feel free to skip: The conditional probabilities become a bear to solve in a tree very quickly. In 20% of cases you get a box first. In 36% of those cases you get at least one more pair of boxes (7.2% of all cases); these would be a 4th and 5th box. In these cases, your chance of winning on either box 4 or box 5 isn't 51% (the chance that any two boxes have a ticket), but rather the chance that that box 4 or 5 exist, that neither box 2 or 3 had a ticket, and that either box 4 or 5 have a ticket.

However, in 0.8% of cases both boxes 2 and 3 have pairs of boxes inside, generating boxes 4,5,6, and 7. Obviously a ticket could not have been found yet. Going on from here the amount of conditions rapidly multiplies to an unfeasibly high number as you might generate 2-8 new boxes from this set. Likewise, even if only boxes 4 and 5 are generated, they might generate boxes 6 and 7, but now there are two places where a ticket could have occured previously to factor in.

Answer 1

In case anyone wants to know the answer to this, it is the quadratic solution mentioned in the comments. To intuitively visualize how this works, imagine the possibilities as a 10 x 10 square on a grid (so cut up into smaller squares, as if it was drawn on graphing paper). 50 of the squares are shaded red, those are empty. 30 squares are green and have a ticket. Then the remaining 20 squares are split into two new squares each 1/10th the size of the old one, filled with a smaller grid. These each have the same proportion of red to green, as well as two smaller squares in each of them. This pattern repeats infinitely, with the smaller grids each representing a smaller and smaller proportion of your 2D sample space.

The quadratic equation works perfectly for this sort of object that contains itself. I feel like it's quite intuitive if you consider the simple "complete the square" solution to quadratic equations: https://en.m.wikipedia.org/wiki/Completing_the_square, which Wikipedia has a nice video of.

What tripped me up was thinking about more complex variants of the problem and what all solutions would look like. Obviously if you change the winning conditions and just try to consider the number of wins before an iteration of the game ends, and up the reproduction of the chests to above 50%, make it conditional on the number of chests, etc. you need something more complex like the branching process models used for animal populations.