Ask The Wizard #320

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 × w^m

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.

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.

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.

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.