This isn't so much a question, but I'm just looking for some critique on my answer for this problem to confirm/deny if it's correct.
Drinking games are different everywhere, but the way I've been taught this game is that you draw cards one by one, trying to guess the following four questions all in a row and using a new card for every question:
- Color of the card (red/black)
- Higher/lower than the previous card
- In between or outside of the previous two cards
- Suit of the card
Every time you get a question wrong, you take a drink and start the process over. Once you answer all four questions correctly in a row, you're done.
For the sake of making the math simpler, let's assume that every card is drawn with replacement. Every card is put back into the deck immediately after it is drawn, and the deck is reshuffled. So there's no use in counting cards and, for example, determining that you should guess black for the first question because you've seen more red cards come out of the deck. This means that my answer will only be an approximation for the real thing.
Let $P_i$ be the probability of guessing correctly in game stage $i$, provided that you guess rationally.
Color
Since we're drawing with replacement, you always have a $0.5$ probability of getting this question correct. Pretty simple. $P_1 = 1/2$.
High or low
Let's establish that the order of card ranks is $A,2,3,4,5,6,7,8,9,10,J,Q,K$. The game is sometimes played with ace high, but the theory is the same; median card will just be shifted up by one.
Anyway, the median rank is $7$, with six cards lower than it and six cards higher. Since we can't count cards, we will pick lower if the previous card drawn is higher than $7$ and pick higher if it is $7$ or lower. So if the first card is an $A$, we have a $12/13$ chance of our higher guess being correct. We have a $11/13$ chance of being correct given a $2$, and so on. Drawing a $7$ first gives us a $6/13$ chance of correctness in this stage.
Therefore, the probability of guessing correctly here is
$$P_2 = 2\sum_{i=7}^{12}\frac{i}{13}\cdot\frac{1}{13} + \frac{6}{13}\cdot\frac{1}{13}$$
$$P_2 = \frac{2}{169}\sum_{i=7}^{12}i + \frac{6}{169}$$
$$P_2 = \frac{2}{169} \cdot 57 + \frac{6}{169}$$
$$P_2 = \frac{120}{169}$$
In or out
Our choice for this stage depends on the difference between the ranks of the two drawn cards. Assign ranks $A=1$, $J=11$, $Q=12$, and $K=13$. The difference between the two cards is in the range $[0,12]$. If the difference is $6$, there are $5$ ranks between the first two drawn cards and $6$ ranks "outside" of the two drawn cards. That is, if the first two cards are $3$ and $9$, we will be correct by guessing inside if the third card drawn is in $\{4,5,6,7,8\}$, and we will be correct by guessing outside if the third card is in $\{A,2,10,J,Q,K\}$. If the difference between the first two ranks is greater than $6.5$, we have our best chances of being correct by guessing inside. Otherwise, we have our best chances by guessing outside. It will always be imprudent to take the $2/13$ chance of guessing that the third card will be the same as one of the first two.
Given the above guessing scheme, here are the probabilities of guessing correctly for each possible difference between the first two ranks:
Difference$: P_{correct \: guess}$
$1: 11/13$
$2: 10/13$
$3: 9/13$
$4: 8/13$
$5: 7/13$
$6: 6/13$
$7: 6/13$
$8: 7/13$
$9: 8/13$
$10: 9/13$
$11: 10/13$
$12: 11/13$
$i < 6.5: \frac{12-i}{13}$
$i > 6.5: \frac{i-1}{13}$
Another thing to understand: Say we draw a $3$ first. If we made it this far, the other card will surely be higher than $3$ because we would have guessed higher in the second stage. This helps with the math below.
Let $C_n$ be the probability of correctly guessing inside or outside given that the first card drawn is $n$. Consider $i$ in the expressions below to be the difference between the second card and the first card.
$C_n = \sum_{i=1}^{13-n}(P[\textrm{Draw card }(i+n)] \cdot P[\textrm{Correct guess given the difference}])$ for $n \in [1,6]$
$C_n = \sum_{i=1}^{6}\frac{1}{13-n} \cdot \frac{12-i}{13} + \sum_{i=7}^{13-n}\frac{1}{13-n} \cdot \frac{i-1}{13}$
$C_n = \frac{1}{13(13-n)}(\sum_{i=1}^{6}(12-i) + \sum_{i=7}^{13-n}(i-1))$
$C_n = \frac{1}{13(13-n)}(12(6) - \sum_{i=1}^{6}(i) + \sum_{i=7}^{13-n}(i) - (13-n-7+1))$
$C_n = \frac{1}{13(13-n)}(72 - 21 + \sum_{i=1}^{7-n}(i+6) - (7-n))$
$C_n = \frac{1}{13(13-n)}(51 + \sum_{i=1}^{7-n}(i) + 6(7-n) - (7-n))$
$C_n = \frac{1}{13(13-n)}(51 + \frac{(7-n)(8-n)}{2} + 5(7-n))$
$C_n = \frac{1}{13(13-n)}(51 + \frac{n^2 - 15n + 56}{2} + 35-5n)$
$C_n = \frac{1}{13(13-n)}(86 + \frac{n^2 - 15n + 56}{2} - 5n)$
$C_n = \frac{1}{26(13-n)}(172 + n^2 - 15n + 56 - 10n)$
$C_n = \frac{1}{26(13-n)}(n^2 - 25n + 228)$
$C_n = \frac{n^2 - 25n + 228}{26(13-n)}$ for $n \in [1,6]$
$C_1 = 17/26$
$C_2 = 7/11$
$C_3 = 81/130$
$C_4 = 8/13$
$C_5 = 8/13$
$C_6 = 57/91$
The distribution is naturally mirrored for higher cards, where $C_1 = C_{13}, C_2 = C_{12}, ..., C_6 = C_8$.
The maximum difference between the first two cards if the first card is $7$ is only $6$, so the derived formula for $C_n$ cannot apply because we simplified the summation from $i=7$ to $13-n$ under the assumption that $13-n \geq 7$.
$C_7 = \frac{1}{13(13-7)}(\sum_{i=1}^{6}(12-i) + \sum_{i=7}^{13-7}(i-1))$
$C_7 = \frac{1}{13(13-7)}(\sum_{i=1}^{6}(12-i))$
$C_7 = \frac{1}{13(13-7)}(12(6) - 21)$
$C_7 = \frac{1}{13(13-7)}(51)$
$C_7 = 17/26$
Let $P_3$ be the probability of guessing correctly in stage $3$, given that stage $2$ has been completed.
$P_3 = \sum_{i=1}^{13} C_i P[\textrm{First card} = i]$
$P_3 = \sum_{i=1}^{13} C_i (\frac{1}{13})$
$P_3 = \frac{1}{13}(\frac{17}{26} + 2(\frac{17}{26} + \frac{7}{11} + \frac{81}{130} + \frac{8}{13} + \frac{8}{13} + \frac{57}{91}))$
$P_3 = \frac{82029}{130130}$
Suit
The probability of correctly guessing a suit is $P_4 = 1/4$.
Overall probability
When drawing with replacement, the probability of getting off the bus before taking your next drink is $P_1 P_2 P_3 P_4 = \frac{1}{2} \cdot \frac{120}{169} \cdot \frac{82029}{130130} \cdot \frac{1}{4} = \frac{246087}{4398394} \approx 0.055949 \approx \frac{1}{17.87}$.
Game strategy (for any readers looking for guidance)
If a player is not counting cards in any way, they should guess as such:
Color
Guess either red or black.
High or low
Guess higher if the first card is less than $7$. Guess lower if it is greater than $7$. Guess either one if the card is $7$. If playing with ace-high, use $8$ as the median instead of $7$.
In or out
Guess inside if the difference between the first two cards is greater than $6.5$. Guess outside if the difference is less than $6.5$. This does not change whether the game is ace-low or ace-high.
Suit
Guess any suit.