Ask The Wizard #321

The key to problems like this is in setting them up. I recommend trying to boil the problem down to as few unknowns as possible. In this case, we can express the unknown distances on the square to just three, as follows:

It's easier to deal with rectangles than triangles. Given we know the area of three triangles, we can double the size and double the areas. That gives us:

• ab=10
• ac=16
• (a-b)(a-c)=14

Let's factor (a-b)(a-c):

a² - ab- ac + bc = 14

a² - 10 - 16 + bc = 14

(1) a² + bc = 40

Let's express b and c in terms of a, to get this down to a single variable:

b = 10/a

c = 16/a

Substituting those values for b and c in equation (1):

a² + (10/a)*(16/a) = 40

a² + 160/a² = 18

Next, let's get rid of that a² in the denominator by multiplying everything by a².

a⁴ + 160 = 40*a²

a⁴ - 40*a² + 160 = 0

Let's define a new variable y = a²

y² - 18y + 32 = 0

Next, let's solve for y using the quadratic formula:

y = (40 +/- sqrt(1600-640))/2

y = (40 +/- sqrt(960))/2

y = (40 +/- 8*sqrt(15))/2

y = 20 +/- 4*sqrt(15)

The area of the entire square is a², which conveniently is equal to y. From the equation above, if the +/- is negative, then y = apx 4.5081, which is obviously wrong, since we know the area is at least 20, not even including x. So that area of the square must be 20 + 4*sqrt(15).

The three triangles we're given have an area of 5+7+8=20. Subtracting that from the total area of the square leaves us the area of x: 20 + 4*sqrt(15) - 20 = 4*sqrt(15) = apx 15.4919.

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

Note my t-shirt in this picture. The movie theater cashier complimented it when I went to see Uncut Gems. I thanked her by torturing her with this problem, only with triangles of areas 2, 3 and 4. After the movie, I checked on her and she still hadn't solved it, but seemed to be trying. So I wrote out the following solution at the Suncoast bar for her. She actually seemed to appreciate it. I think that young woman will go far in life.

I won't go into all the math, but here are the jackpot points where multiple winners are equally likely as exactly one:

• PowerBall: $975 million
• Mega Millions: $1.65 billion

Not that you asked, but here are the jackpots where the probability of at least one winner is equal to the probability of none (50%).

• PowerBall: $704 million
• Mega Millions: $867 million

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

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

A quick answer is the value of the maximum you're allowed to bet. Just wait for a 22 and then play it, resulting in walking away from the table with an extra unit.

A more rigorous answer should look at the cost of waiting for that 22. This is something I don't know exactly, but can estimate.

The cost to the player of the dealer-push-22 rule is 6.91%. However, the dealer will get to 22 more than that, because the player won't care if he busted first. Depending on the rules, the player will bust about 15.7% of the time. If we assume this is uncorrelated with the dealer's probability of busting (which it's not), then the probability of a dealer bust is 6.91%/(1.0-0.157) = 8.2%.

I know this is rough, but let's assume the player's 22 probability is the same. I know the dealer will bust more than the player, but this coupon also doesn't count after a split, so let's just say those factors offset. So, if the player gets a 22 8.2% of hands, then he will get a 22 once every 1/0.082 = 12.2 hands on average.

Let's say the house edge in the base game is 0.75%. The cost of playing 12.2 hands at that house edge is 12.2*0.0075 = 0.0915. So, subtract that from the one unit value and you get a value of the coupon of 1.0 - 0.0915 about 91% of face value.

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