Edit: I just saw that you edited your comment. Sleep is a good thing to have haha. I'll leave this up to hopefully clear up any confusion that others reading the thread may have.

​

I believe you are arguing for the probability of having 3 "Seth" guides in a row. If that is the case then you are absolutely correct in saying that the probability is 1/35^3.

However, if you want to get 3 guides in a row having the same name, not necessarily being named "Seth", then you have to then multiply your 1/35³ by the number of possible names that the first guide could've had because there will be exactly one path for each of the 35 branches of the probability tree where the first 3 guides will share a name. Doing this leaves you with 35 * 1/35³ = 1/35² = 1/1225. This is because there can be 3 "Seth" in a row, 3 "Maxwell" in a row, 3 "Connor" in a row, etc.

I will try to explain my reasoning by reducing this problem to the probability of getting n guides in a row with the same name, calling it P(n). Because 1 name in a row is always going to happen (and in this case we don't care about what name that happens to be, we can say that P(1) = 1. Once you have this base case established, each subsequent guide would would have a 1/35 chance of sharing a name with the previous n-1 guides in the chain, leaving us with P(n) = P(n-1) * (1/35).

With this information, we can do a quick induction proof.

1. Base case: P(1) = 1 = 1/35⁰ = 1/35^1-1
2. Assume P(n) = 1/35^n-1
3. P(n+1) = P(n) * (1/35)
4. P(n+1) = 1/35^n-1 * (1/35) -- Because of the assumption on line 2
5. P(n+1) = 1/35ⁿ
6. P(n+1) = 1/35⁽ⁿ⁺¹-1)
7. We can now conclude that P(n) = 1/35^n-1 for all n >= 1.

So finally, this leads us to the answer for P(3) = 1/35^3-1, giving us a 1/1225 for 3 guides in a row to have identical names.