# Conditional Probability Question - Drawing Cards with Order

I am trying to calculate odds for a card game. Using a standard 52-card deck. Here is the basics of the game. You are given two cards, you add their values together. (Standard Blackjack values, suits don't matter) Once you know the total value of your two cards, you are given a third card face down and you then have to guess what the total value of your three cards are. What are the odds that you can guess your total?

As an example to illustrate. You draw two cards, let's say you get a 10 and a 5, what are the odds that you get a total of 25? (ie: Draw a 10). This example is a simple probability that I fully understand, however, the twist comes with calculating the odds of getting a total of 25 with any two-card combination with a value of 15? (ie: If you have 9 and 6, the odds of getting 25 are more than with a 10 and 5). Is there a way to generalize this into something like "Given all combinations of 15, what are the odds that your next card will be a 10?" That can be further generalized into "Given all combinations of X, what are you odds that you next card will be Y"?

The reason for this is because I want to have a list of odds to get any TOTAL value of 3 cards. For example, the odds of getting a total of 33 (Ace, Ace, Ace) is 1 in X. You'd have to calculate odds of getting an Ace when you already have 2. Simple enough, but gets very complication when trying to find the odds of getting 25 because your first two cards aren't as determined and there are only so many ways to get 25 from 3 cards, so the order in which you draw those cards matters as to the odds of getting the last one.

I hope this makes sense,

• Well... for any initial two cards and initial sum $x$ the most likely total will be $10+x$ since there are strictly more than $4$ cards with value $10$ remaining in the deck but at most $4$ cards with each other remaining value. As such the optimal strategy is to always guess $10$ more than what you started with. That being said, the probability of guessing correctly will vary depending on how many face cards or 10s are in your initial two cards. – JMoravitz Jun 19 '17 at 20:05
• With no $10$'s in your initial two cards, your probability will be $\frac{16}{50}$, with one $10$ in your initial two cards your probability will be $\frac{15}{50}$ and with two $10$'s in your initial two cards your probability will be $\frac{14}{50}$ of guessing correctly. If you insist on calling these odds instead, then that would be $16:34, 15:35, 14:36$ respectively. (remember, odds and probability are not the same, just related). If your question is "given your initial total is $x$, what is the probability of correctly guessing $x+10$" that will vary based on $x$. – JMoravitz Jun 19 '17 at 20:06
• Do you know only the sum of the first two cards, or the two cards themselves? Also, are you considering that an ace can be worth either $1$ or $11$? – John Jun 19 '17 at 20:14