Okay, let’s ignore the other complication the other poster correctly mentioned for now to break it down to the basic problem.
We know that the probability of one die coming up 3 is 1/6. Because they’re independent events, the probability of two dice both coming up as 3 is 1/6 * 1/6 = 1/36. The temptation here is to say that two dice coming up with the same number is therefore 1/36, but the trick is that we don’t care what the number is. We only care that they’re the same.
So picture rolling them sequentially. The first die will come up with a number, let’s say 5. Since it already rolled, it has a 100% chance of being 5, so we just use a 1 in the probability equation. The chance of the second die also being 5 is again 1/6, so the probability of two dice matching will be 1/6 (1 * 1/6). Three dice would be 1 * 1/6 * 1/6, and so on.
It’s not for homework.
Fair enough. Did the reasoning make sense though?
Honestly, not really.
Okay, let’s ignore the other complication the other poster correctly mentioned for now to break it down to the basic problem.
We know that the probability of one die coming up 3 is 1/6. Because they’re independent events, the probability of two dice both coming up as 3 is 1/6 * 1/6 = 1/36. The temptation here is to say that two dice coming up with the same number is therefore 1/36, but the trick is that we don’t care what the number is. We only care that they’re the same.
So picture rolling them sequentially. The first die will come up with a number, let’s say 5. Since it already rolled, it has a 100% chance of being 5, so we just use a 1 in the probability equation. The chance of the second die also being 5 is again 1/6, so the probability of two dice matching will be 1/6 (1 * 1/6). Three dice would be 1 * 1/6 * 1/6, and so on.
Does that help?
Mostly.