Ask the Wizard #268

That would be the Casino di Venezia in Venice, established in 1638.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

My initial answer was the Golden Palace 20% monthly bonuses, up to bonuses of $2,000, in 1999 and 2000. There were no restricted games and the play requirement was the amount of the bonus only. That came to a sudden stop, unfortunately.

However, after some discussion on my forum, I think that the award has to go to the Casino on Net single-zero roulette promotion, in which they paid 70 to 1 for bets on 0 and 7. That has a player advantage of 92%! I’m told that the casino lost $4 million on that two-hour promotion but paid everybody.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

The six center faces of the cube are fixed. By turning the faces all you can do is rearrange the corners and edges. If you took the cube apart, then there would be 8!=40,320 ways to arrange the eight corners, without respect to the orientation of each piece. Likewise, there are 12!=479,001,600 ways to arrange the 12 edges without regard to orientation.

There are 3 ways each corner can be oriented, for a total of 3⁸=6,561 corner orientations. Likewise there are two ways each edge piece can be oriented, for a total of 2¹²=4,096 edge orientations.

So, if we could take the cube apart, and rearrange the edge and corner groups, then there would be 8! × 12! × 3⁸ × 2¹² = 519,024,039,293,878,000,000 possible permutations. However, not all of these permutations can be arrived at from the starting position by rotating the faces.

First, it is impossible to rotate just one corner and leave everything else the same. No combination of turns will achieve that. Basically, every action has to have a reaction. If you wish to rotate one corner, it would disturb the other pieces somehow. Likewise, it is impossible to flop just one edge piece. For these reasons, we have to divide the number of permutations by 3 × 2 = 6.

Second, it is impossible to switch two edge pieces without disturbing the rest of the cube. This is the hardest part of this answer to explain. All you can do with a Rubik's Cube is rotate one face at a time. Each movement rotates four edge pieces and four corner pieces for a total of eight pieces moved. A sequence of rotations can be represented by a number of piece movements divisible by 8. Often a sequence of moves will result in two movements canceling each other out. However, there will always be an even number of pieces moved with any sequence of rotations. To swap two edge pieces would be one movement, an odd number, which can not be achieved with the sum of any set of even numbers. Mathematicians would call this a parity problem. So we have to divide by another 2 because two edge pieces cannot be swapped without other pieces being disturbed.

So there are 3 × 2 × 2 = 12 possible groups of Rubik's Cube permutations. If you disassembled a Rubik's Cube and put it back together randomly, there is a 1 in 12 chance that it would be solvable. So the total number of permutations in a Rubik's Cube is 8! × 12! × 3¹² × 2¹² / 12 = 43,252,003,274,489,900,000. If you had seven billion monkeys, about the human world population, playing randomly with the Rubik's cube, at a rate of one rotation per second, a cube will pass through the solved position on average once every 196 years.

Links

• Group Theory via Rubik's Cube by Tom Davis (PDF)
• Rubik's Cube entry at Wikipedia

The worst hand is the dreaded 1-2. It is composed of a high 6, low 6, low 4, and any 7. The probability of this hand is 2×2×2×4/combin(32,4) = 32/35,960 = 0.09%, or 1 in 1,124.



Interestingly, if the player played 0-3, the high hand would be a high-3, which is normally the minimum low hand the house way tries to achieve. I don’t know if that is a coincidence or not.

I like creative bets like this. It can be found at the Boyd casinos, as well as the Palms, El Cortez, and South Point. To answer the question I looked at another set of bets on the specific player to score the first touchdown. Those odds are shown in the second column of the table below. For the sake of simplicity, I’m ignoring the field at 5-1, and no touchdown scored at 100-1. I then converted those wins to the &lquo;fair probability” in the third column, meaning the probability of winning you would need for the bet to be exactly fair. These probabilities are inflated, due to depressing what each win pays, which is why the sum is 166%. The “adjusted probability” in the fourth column shows the fair odds divided by 1.660842, so that the total probability is 100%. The fifth column shows the number of Scrabble points in each player’s name. The “expected Scrabble points” in the sixth column is the product of the probability and Scrabble points. The lower right cell shows the mean Scrabble points is 14.18521.

Going by the mean, the over looks to be the right side. Going player by player, the probability of 11 or more Scrabble points is 0.641894, which converts to a fair line of -179. So that makes the over at -115 an outstanding bet. Laying 115, the player has a 20% advantage on the over.

Unfortunately, by the time I went to bet it the line had moved to -180.