The Math: Rolling Doubles with 3 six-sided dice. probability

Hello, I'm new here. I enjoy reading through the different discussions on probability. I did a search with "roll 3 dice" "roll doubles" and couldn't find what I was looking for.

I play a gambling game where if you roll 3 dice and get doubles (ie: 1 and 1, 2 and 2, 3 and 3) or triples (1, 1 and 1, etc). You double your bet. it doesn't matter what the 3rd number is (although getting triples wins just as easily).

So if I bet 1 dollar and roll the three dice and get 1, 6, and 1. I'll get my dollar back plus 1 more dollar.

Likewise if I bet 100 dollars and roll three dice and get 3, 3, and 3. I'll get my 100 back plus 100 more.

But if I roll 1, 2, 3 or 3, 6, 1, etc. I lose my bet.

Does the house have an edge?

What's the probability of getting doubles with 3 dice?

Could there be any strategy to the game or is it pure chance?

What are my odds of winning?

Thanks! I appreciate your response.

216 possible results on 3 dice.

When rolling the chance of all three dice being different is 6/6*5/6*4/6 = 120. Therefore 96 sets have a double or triple (6 triples and 90 doubles).

Chance of winning therefore is 44.4%. As the bets are evens, it's a losing game, with a house edge of 11.1%.

If triples paid six times your stake (dollar bet, plus $6 winning), edge would be 0%.

There is no skill in this game, as there is no decision to make apart from to roll or not roll. Assuming all rolls are random and independent, does not make any difference when you bet.