Ask the Wizard #146
Mike from Prescott, Arizona
You’re welcome. It’s rare for online casinos to intentionally let players win in free mode. I know the Elka casinos used to do this (for which I blacklisted them), but fortunately they seem to have vanished. If anyone can show me hard evidence that a casino is intentionally allowing players to win in fun mode I would be happy to investigate it. Hard evidence means, at a bare minimum, a record of hands won and lost in each mode, for several dozen hands. Simply telling me that you lost "a lot more" when in real mode is useless.
Sue from The Colony, Texas
Yes, if you give your e-mail address to a casino they will certainly send you e-mail. However the reputable ones will stop if you ask. The less reputable casinos will not only market themselves but also share your address with others. The NetGaming casino sold me out to pornography spammers. Bodog lets you play without surrendering your email address, as do the Wager Works casinos, such as at the Hard Rock Casino. Incidentally, Casino Meister has a new page about combatting casino spam. My webmaster tells of his own problems with casino spam at VegasClick.com. I've never seen Cleopatra online. It's rare for an online casino to have the same slots found in land casinos.
Annojh from Toronto
Let A = Choosing the normal die
Let B = Rolling a 6 with randomly chosen die
Answer = Pr(A given B) = Pr(A and B)/pr(B) = ((2/3)*(1/6))/((2/3)*(1/6)+(1/3)*1) = (2/18)/((2/18)+(6/18)) = 1/4.
Dan from Cairo, Egypt
The following table shows estimated probabilities that a pair will be beaten by at least one higher pair according to the number of players (including yourself). These probabilities are not exact because the hands are not independent. However to find the exact probabilities would get complicated and I think these are pretty close. My formula is 1-(1-r*combin(4,2)/combin(50,2))(n-1), where r=number of higher ranks than your pair, and n = total number of players. The table shows the probability of another player having a pair of aces, when you have a pair of kings, in a 10-player game, to be 4.323%.
Probability Pair Beaten by Higher Pair
|Pair||2 Pl.||3 Pl.||4 Pl.||5 Pl.||6 Pl.||7 Pl.||8 Pl.||9 Pl.||10 Pl.|
Ricardo from Malden, MA
When I find a casino is not playing fair I don't generally go back to check if they've stopped. Sometimes I do if requested by the casino and I feel the problem may have been accidental. Following is a basic strategy, based on infinite decks, where the dealer stands on soft 17 and deals seconds. What I mean by dealing seconds is that if the third card, and only the third card, would break the dealer it is skipped and the next card is played, whatever it is. Otherwise play continues normally. The house edge under this game would be 9.3%. I used to ask for donations but got so few I quit asking. Now the site is comfortably supported by advertising revenue anyway.
Karen from Eastchester, New York
No. Neither the amount you put in nor the denomination affects the odds. The same is true of slots.
Gerald from Alexandria
I lost the $1800 to another gambling writer, not Stanford. I would have preferred more rolls but there was an obvious time contraint. Assuming one throw per minute it would take 34.7 days to throw the dice 50,000 times. I wasn’t the one who decided on 500 but it seemed like a reasonable compromise between a large sample size and time. You are right that 500 is too few to make a good case for or against influencing the dice, but 500 throws is better than zero.
Aubrey from Kokomo
There are 6!/(4!*1!*1!) = 30 ways to arrange these numbers in any order. Another way to look at it is there are 6 positions to put the 1, and 5 left to put the 4, so 6*5=30. The probability of getting 666614 in exactly that order is 1 in 66 = 1 in 46656. Multiply that by 30 for the 30 possible orders and the answer is 30/46656 = 0.0643%, or 1 in 1552.2.