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Ask the Wizard #311

If you want to enclose 355 milliliters in a can, what should the dimensions be to minimize the surface area?

"Anonymous" .

Good question! I was just wondering this when I saw some skinny soda cans at a gaming show, which held the usual 355 milliliters that the standard size does. Surely both can't be right (and don't call me Shirley).

Let:
r = radius of the can
h = height of the can
v = volume of the can
s= surface area of can

We know from simple geometry that the surface area = 2*pi*r^2 + 2*pi*r*h.

Likewise, we also know the volume is pi*r^2*h, which we're given equals 355.

So, 355=pi*r^2*h.

Let's rearrange that to:

(1) h = 355/(pi*r^2)

We know:

(2) s = 2*pi*r^2 + 2*pi*r*h.

Let's get that to a function of just one variable by substitute our expression for h in equation (1) into (2):

s = 2*pi*r^2 + + 2*pi*r*(355/(pi*r^2))) = 2*pi*r^2 + 710/r.

Let's take the derivative of s and set it equal to zero, to solve for the optimal r.

ds/dr = 4*pi*r - 710/(r^2 ) = 0

4*pi*r = 710/(r^2)

Multiplying both sides by r^2:

4*pi*r^3 = 710

r^3 = 177.5/pi.

r = (177.5/pi)^(1/3) = 3.837215248.

Plug that value into equation (1) to get h = 7.674430496.

two cans -- rotated

This question is raised and discussed in my forum at .

Just got back from Poker Night at the VFW. Got 6-6 in the hole three times in a row! Never had that happen before. What are the odds of getting a pocket pair of the same rank three times in a row in an evening? You may assume that an evening consists of 120 total rounds.

AZDuffman

The answer and solution appear in the following spoiler tag.

There are four possible states you can be in at any given time:

  • State 1: The first hand or any hand where the last hand was not a pocket pair.
  • State 2: Last hand was a pocket pair.
  • State 3: Last two hands were the same pocket pair.
  • State 4: Three of the same pocket pairs in a row has already been achieved.

If you're in state 1, you can advance to state 2 with a probability of 3/51. Otherwise, you stay in state 1.

If you're in state 2, you can advance to state 3 with a probability of (4/52)×(3/51). Otherwise, you go back to state 1.

If you're in state 3, you can advance to state 4 with a probability of (4/52)×(3/51). Otherwise, you go back to state 1.

If you're in state 4, you stay there.

That said, you can create your transition matrix, T, as follows:

0.941176 0.058824 0.000000 0.000000
0.941176 0.054299 0.004525 0.000000
0.941176 0.054299 0.000000 0.004525
0.000000 0.000000 0.000000 1.000000

There are 120 total hands played, so find T^120.

0.941044 0.058549 0.000265 0.000141
0.941025 0.058548 0.000265 0.000162
0.936786 0.058284 0.000264 0.004666
0.000000 0.000000 0.000000 1.000000

The upper right cell shows us the probability that starting at state 1 will lead us to state 4 after 120 starting hands in a three-hand sequence, which is 0.000141471.

Take the inverse of that number, the probability is 1 in 7068.605131.

This question is raised and discussed in my forum at .

In your explanation of a video poker pull-tab machine, you give the example: "Even if the game looks like a five card draw video poker game your outcome is predestined. For example if you get a royal flush on the deal and throw all of it away you would get another royal flush on the draw." My question is, what happens if you throw-away cards that make a predestined outcome impossible (like a deuce in Deuces Wild for a predestined 4-Deuces or an Ace in a Double Bonus game for 4 Aces?) Maybe these types of games aren't offered but just those, like Jacks or Better, where no such situation is possible?

"Anonymous" .

What I hear happens is a fairy comes along and changes your hand on the draw to whatever it was predestined to be. For example, if you were predestined to get two deuces on the deal and improve to four deuces after the draw, if you threw away the deuces, you would probably get the other two naturally on the draw, and then the fairy would change two junk cards to the two deuces you threw away.

It seems to me that most gambling professionals I know prefer to know a game’s volatility expressed as variance rather than as standard deviation. Of course, the former is just the square of the latter. However, I prefer the standard deviation since it is in the same units as the bet and win/loss. Perhaps they like a bigger number to make a bigger volatility stand out? What is your take – is there a preference for using "variance" by gamblers and, if so, why?

Gary J. Koehler

I agree that you hear the variance of a game bandied around more than it's standard deviation, which I have always found a little annoying. The reason I think gamblers should care about the volatility of a game is to associate a win or loss to a probability for a session of plays. For example, what would be a 1% bad loss after 200 hands of blackjack. To answer that, you would use the standard deviation of blackjack, which is about 1.15, depending on the rules.

The specific answer to that question is 1.15 × 200^0.5 × -2.32635 (which is the 1% point on the Gaussian curve) = -37.83 units south of expectations. Don't forget that due to the house edge you can expect to lose something. If we assume a house edge of 0.3%, then after 200 hands you could expect to lose 0.003*200 = 0.6 hands. So a 1% bad loss would be 0.6 + 37.83 = 38.43 hands.

The casino in Milwaukee, which started as a bingo hall, had a record 290 bingos in one game this week. The pattern was the letter I, either up and down (3 on top and bottom and all the Ns) or sideways (3 Bs and Os with the middle). It took 43 calls for the first G ball to be called, resulting in mass winners. Each person got $25.

Here is an article about it: .

My question is what are the odds of going 43 calls without calling any numbers of a particular letter?

PlayYourCardsRight

I've been in similar situations where most people were waiting on a particular letter, but the most winners I've ever seen at once is around 25.

I show the probability of going 44 calls and avoiding any one letter (not just G) is 1 in 1,517,276. Here is a formula to that probability: 5*combin(60,44)/combin(75,44) - combin(5,2)*combin(45,44)/combin(75,44)

How do I convert odds in sports betting back and forth between the American and European ways of expressing them?

Teddys

Let's let a be the odds expressed the American way and e the European way.

To go from American to European:

If a>0, then e=1+(a/100).
If a<0, then e=(a-100)/a.

To go from European to American:

If e>=2, then a=100×(e-1).
If e<2, then a=100/(1-e).