So in this instance I have a standard deck of 52 cards and am playing a high/low game with it (ie turn over the top card, guess if the next card is higher or lower) and maintain a record of all the cards used.
Each guess has an easily calculable chance of being either higher or lower. IE if you draw an 8 for a first cards it's 50% chance that it's higher, and 50% chance that it's lower and a ~5% chance it's another 8 which counts as a "win" regardless of your choice of high or low.
What I don't know how to calculate:
The odds that you will successfully win this game getting all 51 guesses correct strictly guessing the most probable choice.
Any solution I come up with would be different for every game and relies on the cards already drawn. I want to know this probability before even starting a game.
Edit:
Additional info:
- Suits don't matter
- Ties are wins.
- Aces are high
Best Answer
The number of ways, in a deck of $n$ cards, to win this game (out of the $n!$ total ways for the deck to have been shuffled) is given in OEIS as sequence A144188.
Playing with only 13 cards your chances of winning are about 5.246%.
Playing with half a deck, yet using suit order to break ties, your chances are down to 0.095%. What I find amazing is that this is much less than the square of the chances for a 13-card deck (which we would expect because it involves 25 guesses and the 13-card deck involves only 12) and is even 1.6 times less than the square of the winning chances with a 14 card deck and is less than the chances of winning two consecutive games, with a 15 card deck and a 14 card deck.
For large $n$ the chances of winning drop off by about a factor of 0.737 for each additional guess needed.