I ran accross this problem which I'm not able to solve (unless I do really huge calculations). I haven't read any probability yet, except for what we did in upper secondary - and that wasn't much.
The solitaire is played in a way that have your deck, and then you flip card after card. While doing this, you say "one, two, three, one, two, three" and so on. Notice that you may not choose between one, two and three, they just continue looping. If you flip an ace when you are saying "one", you lose. The same goes for saying "two" - getting a two, saying "three" - getting a three. All are losing conditions. If you manage to go through the whole deck without losing, you win (haha).
1. What are the chances of winning?
2. Imagine you take out all aces, twos and threes from the deck and play the solitaire with only these 12 cards. Are chances greater or smaller you'll win compared with playing with full deck?
I've tried a bit, but I don't manage to do anything smart. First I tried only with the 12-card deck, as that was more manageable. I found that it was 4/12 to lose on the first card. Imagine we don't hit an ace on the first card. It can of course not be fair to assume that there is 4/12 to lose on the second card as well, for two reasons: We only have 11 cards to play with, and the first card we picked might have been a two. So here the chances of losing must be either 3/11 or 4/11, depending on if we picked a two or a three on the first card. However, if we create a diagram to represent all possible alternatives, it soon gets out of hands. Are there perhaps some nice formulae to use here?
However, I have a thought.. Are the chances equal (playing with 12 or 52 cards)? May we as well discard the other 40 cards, as they have nothing to do with our solitaire? When playing with 52 cards, you'll end up with saying "one" 18 times, but saying "two" and "three" only seventeen times each. I supoose this should affect the probability in some way or another?