I’m not a statistician, but ISTM that calculating the odds of winning hands vs. losing hands is a lot more complicated than calculating the odds of a coin flip.
You would need to factor in all sorts of variables, such as the cards you were dealt, vs. the dealer’s cards, whether you choose to hit or stay, or double down, etc.
I don’t think you can assume that winning vs. losing is a 50/50 proposition.
We’re assuming that to simplify the question. Obviously, the house had a slight edge, but the overall long-term house advantage is pretty close to 50-50, so for a back-of-the-envelope calculation, it’s a valid approach to the problem and gives us an upper bound for the answer.
The actual chance of losing a hand is 48% in blackjack. Winning is 44%, tying is 8%. Source. This can vary a little if your strategy stinks to hell, but with perfect strategy, those are the numbers you should be looking at.
And to finally determine the win/loss odds in terms of MONEY, you need to consider how they pay out. Blackjack payout makes the final odds, in terms of money, a 1% house advantage provided you stay away from insurance, and double and split when you are supposed to according to standards (basic strategy).
Drink, have fun and linger at the table while you play smart basic strategy Blackjack…maybe catch a winning streak and call it a night.
Here’s how I look at it:
Let’s say the end condition is 2 failures in a row. On average, you’ll get your first failure after two rolls. There is then a 50% chance to immediately roll another failure, i.e. on average finishing on 3 rolls, but a 50% chance that you’ll start over. If you start over, there is then a 50% chance that you’ll end after 3 more rolls, on average; i.e., a 25% chance that you’ll end after 6 rolls, on average. And a 12.5% chance to average 9 rolls, etc.
If you’ve studied (and remember) infinite series, you’ll know that
1 * x[sup]0[/sup] + 2 * x[sup]1[/sup] + 3 * x[sup]2[/sup] + … = (1-x)[sup]-2[/sup]
And this series can be written as:
1.5 * (1 * 0.5[sup]0[/sup] + 2 * 0.5[sup]1[/sup] + 3 * 0.5[sup]2[/sup] + …)
So, it equals 1.5 * 0.5[sup]-2[/sup]
= 1.5 * 4 = 6
You can do a similar thing for an end condition of 3 failures in a row, but this time say that there’s a 50% chance to roll a failure after the first time you roll 2 failures in a row, which gives you a 50% chance to roll 3 failures in a row after 7 rolls, on average. Similarly, you’ve got a 25% chance to roll 3 failures in a row after 14 rolls, on average, etc. In a similar fashion to the above calculations, you can say that you’ll get 3 failures in a row, on average, after 14 rolls.
If you accept the above methodology, you can show by induction that the expected number of rolls to get f failures in a row for 50/50 odds is 2 * (2[sup]f[/sup] - 1).