In Blackjack, the player chooses whether to draw another card(s), or to stop drawing and make the dealer draw.

Some decisions are better than others. For example, if the player's cards add up to 8 or less, they should draw another card. If their cards add up to 17, then it's a better decision to stand. In between, it gets far more complicated. Rules like splitting, doubling, and surrendering also come into play.

The payout for winning is determined by the casino. Usually, it's 3:2. So if you place a \$100 bet and win, you'll win \$150 and end up with \$250 total. If you lose, you'll lose your \$100 and will end up with \$0.

The payout ratio can vary by casino, however. If the casino paid out 100:1, your odds of winning money over time would be very high. If the casino paid out 1:100, your odds of losing money over time would be very high. The question is, at what payout ratio would there be an even chance of winning and losing money?


1 Answer 1


More than just the payout ratio determines your overall odds of winning. How many decks are in the shoe? Does the dealer stand or hit soft 17? Can you result aces? Will you get even money on a blackjack after a split? etc...

Assuming the following: blackjack pays 3:2, dealer stands on a soft 17, an infinite deck, the player may double after a split, split up to three times except for aces, and draw only one card to split aces, it can be shown that the player's expectation is $-0.511734\%$.

Variations from the above assumptions will produce variations in the payout. I have seen calculations for games where blackjack only pays 6:5 and produces an expectation of about $-1.9\%$.

Based on this, I would imagine that only a slightly higher ratio would get you over the top. If you ever come across a game like this, there are probably other measures in place that prohibit a positive expectation.

  • $\begingroup$ How did you arrive at −0.511734%? $\endgroup$ Feb 22, 2015 at 0:11
  • $\begingroup$ These are calculations from wizardofodds.com/games/blackjack/appendix/1 $\endgroup$ Feb 22, 2015 at 0:16
  • $\begingroup$ I am not sure, but I don't think there is a good deterministic way to determine these odds short of Monte Carlo type methods. Most systems are tested on billions of simulated hands via computer rather than some deterministic probability formula. $\endgroup$ Feb 22, 2015 at 0:17
  • $\begingroup$ For an infinite deck, we can compute that blackjack happens $\frac {2 \cdot 4 \cdot 16}{52^2}=\frac 8{169}$ of the time, so adding $\frac {169}8\cdot .00511734\approx 0.108$ to the payout for blackjack makes the game even. That would make the payout for blackjack $1.608:1$ This assumes you only get the premium on an initial blackjack, not one after a split and that the $-0.511734\%$ was calculated that way. $\endgroup$ Feb 22, 2015 at 0:20
  • $\begingroup$ My calcs agree with the wizzardofodds's except for 10 and ace. Anybody know if these tables where calculated with blackjack peek? It's not clear from the description. $\endgroup$
    – plus1
    Jun 4, 2019 at 7:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .