Gambling Guide
Ask the Wizard #311
"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).
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.
This question is raised and discussed in my forum at Wizard of Vegas.
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 Wizard of Vegas.
"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.
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: Bingo! Record set at Potawatomi for number of winners in single game.
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)
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).
