Share this

Wizard Recommends

  • Up to $12,250 in Bonuses Play
  • $12,500 Welcome Bonus Play
  • $11000 Welcome Bonus Play

Ask the Wizard #320

Hello Wizard. In 100,000 hands of blackjack, how many losing streaks can I expect to see of 10 hands or more?

Michael

First we have to find the probability of a win for any given hand, which will depend on the rules, which weren't stated in your original question. In my page on variance in blackjack, I give the probability of a net win, push, and loss under "liberal Strip rules," which are: six decks, blackjack pays 3 to 2, dealer stands on soft 17, double after split allowed, surrender allowed, re-splitting aces allowed. Under those rules, here are the requisite probabilities:

  • Win: 42.43%
  • Push: 8.48%
  • Loss: 49.09%

Your question also didn't state how to treat pushes. I'll assume that a push counts as a hand played, but does not advance nor reset a series of losses. Taking out the pushes, the probability of a win and loss, given a resolved bet, are:

  • Win: 46.36%
  • Loss: 53.64%

That said, a very good approximation for questions like this is:

n × l × wm

Where:
n = number of hands played
l = probability of a loss
w = probability of a win
m = minimum number of hands in a losing streak

In this case, the expected number of losses is 100000 × 46.36% × 63.64%10 = 91.4. In other words, there will be a losing streak of at least 10 hands every 1,094 hands, on average. A random simulation backs this up.

At this point, I'm sure my perfectionist readers are preparing to Email me with intellectual whipping on Markov chains. I'd like to emphasize that my formula is an ESTIMATE and in fact quite a good one.

There’s a full 100-liter tank containing a water and 10 kg salt. If we add 10 liters pure water per minute, while simultaneously draining 10 liters of the solution per minute, how much salt will be left in the tank after 30 minutes?

Ace2

Let's start by defining a couple variables:

  • s = kg of salt in the tank
  • t = minutes since the salt was dumped in the tank

We're given that 10% of the salt is drained away per minute. To put that in mathematical language:

ds/dt = (-10/100) × s

Let rearrange that to:

ds = (-10/100) × s dt

-10/s ds = dt

Integrating both sides:

(1) -10×ln(s) = t + c

Next, let's find the dreaded constant of integration. To do that, we're given that s = 10 when t = 0. Plugging that into formula (1) above gives us:

-10 × ln(10) = 0 + c

So c = -10×ln(10)

Putting that into equation (1) gives us:

(2) -10×ln(s) = t -10×ln(10)

The question at hand is how much salt will be in the tank at t=30. Solving for s when t=30:

-10×ln(s) = 30 -10×ln(10). Next divide both sides by -10...

ln(s) = -3 + ln(10)

s = exp(-3 + ln(10))

s = exp(-3) × exp(ln(10))

s = exp(-3) × 10

s =~ 0.4979 kg of salt.

This question is asked and discussed in my forum at Wizard of Vegas.

I've often wondered about the statistical advantage of splitting aces when the dealer has a 10 showing. Is it really wise to equal the bet? Is it a hard rule that you must equal the bet? This inquiry is made assuming the player is not card counting.

Lee

The math never lies. According to my blackjack appendix 1, here are the expected values for all four ways of playing A,A vs. 10, assuming infinite decks, dealer stands on soft 17, and re-splitting aces is not allowed.

  • Stand = -0.540430

  • Hit = -0.070002

  • Double = -0.514028

  • Split = 0.179689

    So, this situation isn't even close, splitting is better by about 11% of the initial bet. It would be even more if re-splitting aces were allowed.

In the California Lottery there is a game called Hot Spot. It involves a randomly selected "Bulls Eye" ball drawn randomly from 1 to 80. There are 300 games per day. What are the chances that in five days the same Hot Spot number is drawn at the same daily game number in 3 out of the 5 days? For example, the number 23 being drawn in game number 134 (what is Biblical about that number?) on Monday, Wednesday, and Friday.

Centerflder

First, let's solve for the probability that any given game number has exactly 3 out of 5 days with the same number. That answer is COMBIN(5,3)*(1/80)^2*(79/80)^2 = 0.001523682. This is the number of ways to choose the 3 matching days out of 5 times the probability the second and third day match the first times the probability the other two days don't.

So, the probability there is not a 3 out of 5 day match for any given game number is 1 - 0.001523682 = 0.9984763.

The probability this does not happen for 300 days is 0.9984763300 = 63.29%.

Thus, the probability of the alternative, that there is at least one drawing number with 3 out of 5 days matching the same Bulls Eye number is 36.71%.

This question is asked and discussed in my forum at Wizard of Vegas.