Probability of a simple card game.

[roffelmayo]

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?)

[jestingrabbit]

I believe that your game is functionally equivalent to choosing 5 cards and seeing if one is a face or ace. The "fourth row choice" is a choice between cards that we should be indifferent towards ie their distribution is identical.

[Dason]

I agree with jestingrabbit. I don't see any additional information that you could gain from flipping a card that would give you any information about which card in the next row you should choose.

[Xanthir]

Agreed with the others. There is no actual strategy here, as the cards have no relationship to each other and you have no information about any of them. Instead of making a diamond shape, just pull 5 cards out of the deck and drink if there are any faces or aces. Same exact probability.

[cowmandude]

These are functionally the same given that the deck was reshuffled and the board was rebuilt every time.

Not to steal you thread, but when I was discussing your thread I came up with another question regarding this game.

Say you didn't reshuffle every time and kept dealing off of the same deck. That means when a card is dealt you can assign it a probability of being a face card based on the cards that you've seen thus far. So say your strategy is to choose the path with the lowest probability of failure. If this strategy were successful it would imply that the game is not the same as drawling 5 cards from the deck. So I guess my question is, would this strategy be successful?

[Dopefish]

Surely all paths have the same probabilty of failure, even in that scenario?

[Ben-oni]

I think the supposition is that you leave any unturned cards where they are, so that not all paths are equal.

[Sizik]

Every unknown card, whether dealt or in the deck, has the same probability of being a face card.