# How do you calculate the odds that the odds will be right?

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.

• Suits don't matter
• Ties are wins.
• Aces are high
• How do suits compare? Assuming you have some total ordering of the deck, each flip is, counterintuitively, independent of the others, in terms of what your chances of winning are (before seeing the card), because when there are $n$ cards left, they have some total order, which you know because you know the remaining cards. – aes Mar 5 '15 at 21:52
• Suits are not important, only card values. – leigero Mar 5 '15 at 21:53
• How do you break ties? – aes Mar 5 '15 at 21:54
• @aes a tie is a win, because you weren't "wrong" – leigero Mar 5 '15 at 21:54
• Let's specify what happens if there are two consecutive cards of the same rank. For example, if you have just seen a $9$ and guess "lower" but another $9$ comes out, do you lose or keep playing? For mathematical simplicity, I would suggest that you say the cards are ordered by suits if they tie in rank, so there is never a tie and we model the deck of cards by a permutation of the numbers 1 to 52. The prob;em is still quite difficult. – Mark Fischler Mar 5 '15 at 21:55

Assuming a total order (i.e. break ties with suits), because it's simpler, and also assuming that you reshuffle the cards every time, putting the last flipped card back in (so we get independence).

When $n$ cards are left, your chances of winning are $\sum_{k=0}^{n-1} \frac{\max(k,n-k-1)}{n-1}$. For $n = 2m$ even this is $\frac{1}{2m}\cdot2\sum_{k=0}^{m-1} \frac{2m-k-1}{2m-1} = \frac{3m-1}{4m-2}$. For $n = 2m+1$ odd, this is $\frac{1}{2m+1}(\frac{1}{2} + 2\sum_{k=0}^{m-1} \frac{2m-k}{2m}) = \frac{3m+2}{4m+2}$.

Therefore the answer is $\prod_{m=1}^{26} \frac{3m-1}{4m-2} \prod_{m=1}^{25} \frac{3m+2}{4m+2} \approx 1.6 \times 10^{-6}$.

The answer to yours will be lower, because outliers make it easier to keep winning, but you're less likely to have gotten one if you just won.

Your set up is complicated enough that I would just Monte Carlo it (i.e. run a computer simulation).

• That may be a good idea. I'm wondering if it can truly be calculated or if your odds of being able to win are just as random as the cards themselves. – leigero Mar 5 '15 at 22:29
• There is certainly an answer, it's just complicated to calculate. There are finitely many possibilities for how things will go, each of which have a well-defined probability assigned to them, so before you start you do in fact have a well-defined chance of winning before you know anything about the order of the deck. – aes Mar 5 '15 at 22:30

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.