So some friends and I were recently playing a slightly strange version of the drinking game 'ride the bus'. This involves setting up 9 cards in a diamond shape of 5 rows with 1, 2, 3, 2 ,1 cards in successive rows. So the idea is that you advance through successive rows picking one card to flip from each row and if at any point you flip over a face card or an ace your attempt is failed and you have to stop and take a drink with the aim being to get all the way without flipping a face card or ace. However you could only flip a card that was bordered by one you'd flipped in the previous row so this essentially left you with two strategies: you could pick an edge and follow that edge along to the end or in the middle row you could pick the middle card leaving you with a choice of 2 cards in the fourth row.
The argument that sprung up was whether one route had a higher probability of succeeding than the other. We were split into two camps, one group decided this was functionally the same as the Monty Hall problem and that having more choices on the fourth row increased your chance of getting through whilst the other group stuck to the idea that regardless of the route you took you were just drawing a random sample of 5/52 cards and that the probability would be the same regardless of route. I decided that my grasp of probability wasn't good enough and that I would resort to asking you guys. So Maths forum; is this an example of the Monty Hall problem meaning the probability depends on the route taken or is the probability the same regardless of route? (or is there some other option I'm completely missing?)