This seems sort of similar to the famous "Quantum suicide" concept, although less elegant :-)

We can view Time 2 as the result of an experiment where 8 copies of you start the game, are assigned to two groups W and L (4 copies in each), then a coin is tossed, and if it comes up heads, than we kill 75% of the L group. After such an experiment, you definitely should expect yourself being in the W group given that you're alive with a 80% probability. Time 3 (not shown in your figure) corresponds to a second round of the experiment, when no coin is tossed, but simply 75% of the W group are exterminated. So if you're doing a single suicide, expect to be in W with 80% probability, but if you're doing a double-suicide experiment, then expect a 50% probability of being in W. If before exterminating the W group you dump all their memories into the one surviving copy, then expect a 50% chance of being in L and a 50% chance of being in W, but with four times as much memories of having been in W after the first stage of the experiment. Finally, you can also easily calculate all probabilities if you find yourself in group W after the first suicide, but are unsure as to which version of the game you're playing. I think that's what Bostrom's answer corresponds to.

The suicide experiment is designed more clearly, and it helps here. What would change if you had an exactly 100% probability of winning (the original Quantum suicide)? What if vice versa? And if you have a non-zero probability of either outcome, what if you just view it as the appropriately weighted sum of the two extremes?