A Quick Puzzle,

there is a six chamber revolver with 2 bullets in it,

the bullets are next to each other, on a random spin you pull the trigger, click! an empty chamber, now what are the odds of it firing if you pulled the trigger again?

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25% as it happens, the key is that this means one of the bullets cannot be fired without the other being fired first, the first trigger is fundamentally different from the second trigger because the first is completely random, while the second is a sequential shift of the chambers after the first. The answer would be completely different if the second trigger was randomized. The chance of the gun firing at least once from being triggered twice is 50% (it goes up as expected) compared to the 33% of it firing if triggered once,

Let’s break down how this percentage was obtained by steps.

First trigger: Let’s label the chambers 1–6 in a way that bullet 1 is in chamber 1 and bullet 2 is in chamber 2 (it doesn’t matter because the order of the chambers in the gun is random). Illustrated with bullets bolded 1 and 2 and empty chambers as 3–6

(1 2) 3 4 5 6

There were 2 bullets and 6 chambers, so the chances of it firing are 2/6 or 33%.

However, you know that it didn’t fire so this probability is not used for any calculations. It does provide insight into which chamber was triggered. Neither of the bullet chambers could’ve been triggered, so one of the 4 empty chambers was triggered (chambers 3–6 in X below). For the sake of the mathematics, let’s say the next chamber is +1 of the previous. This sets up the base for the second trigger.

(1 2) X X X X

Second trigger: We know one of the 4 empty chambers was triggered by the first shot, and now the gun is on the next chamber. In terms of numbers, this means the first shot was in chambers 3–6, which also means this limits the placement of the second shot. The current chamber must be 4, 5, 6, or 1, marked by X below.

(X 2) 3 X X X

As seen by the way the numbers are set up, only chamber 1 out of the 4 chambers that may be triggered contains a bullet, so the chances of the gun firing are 1/4 or 25%,

please keep in mind that the way the chambers and bullets is arbitrary, for example, the bullets could’ve been placed in chambers 4 and 5, as long as they are beside one another,

 just to clarify that the 25% chance refers to the second trigger only ASSUMING the bullet didn’t fire on the first trigger. In other words, if the gun was pointed at you, triggered once (didn’t fire), then triggered a second time, and this whole sequence was done an infinite number of times, one quarter (1/4) of these sequences would result in the second trigger causing a bullet to fire. The scenario is completely irrelevant if the first trigger causes the bullet to fire, which is why the 33% chance of the bullet firing from the first trigger is not directly relevant to the percentage calculation, so now I know!