Blackjack Probabilities

The calculations for blackjack probabilities can become lengthy even for the simplest cases. If a player chooses to stand it is a simpler problem since only the dealer is drawing cards. Even in this simple case there are 289 distinct possible outcomes. The values quoted are for a single deck where the dealer stands on a soft 17.

After the initial deal, the dealer is showing a 10 and the player has 68 for a total score of 14. If the player stands (unwisely) and the first card the dealer pulls from the shoe is a 10, the dealer now has a score of 20. At this point a showdown occurs since the dealer has a score between 17 and 21. The probability of that event occurring is ¹/₄₉​, since there are 49 cards left in the shoe after the dealer and player were both dealt cards. The dealer has a higher score (without busting) than the player, so the house wins the showdown. This event will contribute to the loss probability of the player, ¹/₄₉​, which works out to roughly 0.02. Since there are fourteen other 10's in the shoe, the total probability for this event is ¹⁵/₄₉​, which is about 0.31.

The probability for every event needs to be accounted for in this manner, so we reset the dealer hand and calculate the probability of the next showdown. If the next card in the shoe is a 5, the probability of this event is ⁴/₄₉​ since there are four 5's in the shoe. The dealer must draw another card, which is a 10, and busts. The probability of drawing the 10 was ¹⁵/₄₈​ because the dealer has the other 10. Since these events are independent they are multiplied together to get the final probability for this event, which is ⁶⁰/₂₃₅₂​. If we continue to calculate the probabilities for every possible event we would get a total approximate stand probability of 0.213, 0.787 and 0.000 for a player win, loss, and push, respectively.