Ask the Wizard #179
If two dice are rolled over and over, until either of the following events happen, then which is more likely to happen first:
- A total of six and eight is rolled, in either order, with duplicates allowed.
- A total of seven is rolled twice.
Somebody offered me a bet that that the six and eight would occur first. I accepted because seven is the most likely total. However, I lost $2,500 doing this over and over. What are the odds?
Anthony from Indiana
I’m afraid you had the square side of this bet. The probability of rolling two sevens before a six and eight is 45.44%. Here are all the possible outcomes. The first column is the order of petintent rolls to the outcome of the bet, ignoring all others.
Two Sevens before Six and Eight Bet
Basically, the reason the 6 and 8 is the better side is you can hit those numbers in either order: 6 then 8, or 8 then 6. With two sevens there is only one order, a 7 and then another 7.
I’m familiar with this promotion. When I was last at the South Point there was a leaflet for the promotion, but it hadn’t officially started yet. The leaflet didn’t mention anything about a limit. When I asked an employee he said he didn’t know. While I sympathize with your side, I think you would have a stronger case if you had verified before you started playing that the cards were unlimited. It is an unprofessional practice, in my opinion, to run a promotion with vague rules, letting the casinos interpret the details to their own advantage after the fact. That is why I like to ask questions before I play, rather than make assumptions.
If it were up to me to design a promotion then I would consider every possible question or situation that may occur, and write the rules to preempt such issues. Inconsistent application of the rules, I agree, is unfair. It is fine of them to put you on an undesirable list of players subject to a limit, but I think they should have reserved such a right in the fine print of the promotion, allowing you to inquire if you were on the list. This is all getting a bit out of my area of expertise, so please take my comments with a grain of salt.
Will from Rector
I didn’t know they had a buy bet in Crapless Craps. The following table shows the house edge of place and buy bets, assuming there were no rounding of winnings. In your example of a $30 buy bet on 2 or 12 the winnings would be 6*$30-$1=$179. So the expected return is [(1/7)*$179 + (6/7)*-$30] / $30 = -0.0048, so we’re very close.
Place and Buy Bets in Crapless Crapspass and buying oddsin Crapless Craps
|Bet||Pays||Prob. Win||House Edge|
|Place 2, 12||11 to 2||0.142857||0.071429|
|Place 3,11||11 to 4||0.25||0.0625|
|Buy 2, 12 (commision only on wins)||119 to 20||0.142857||0.007143|
|Buy 3,11 (commision only on wins)||59 to 20||0.25||0.0125|
|Buy 2, 12 (commision always)||119 to 21||0.142857||0.047619|
|Buy 3,11 (commision always)||59 to 21||0.25||0.047619|
Kevin from Long Island, New York
The Poisson distribution can be used to answer this kind of question. The general formula is e-m*mx/x!, where x is the number of the event you observed, and m is the expected number. In this case x is 3. The probability of a royal flush in "Not so Ugly Ducks deuces wild" is 0.000023. So the expected number in 10,000 hands would be 0.23. Thus the probability of hitting exactly three royals in 10,000 hands is e-0.23*0.233/3! = 0.161%. The formula in Excel for this is poisson(3,0.23,0).
John from Austin
Casinos don’t like to back up cards because it may cause some players who won because of the mistake to lose. The general policy is that if there is a mistake more than one card back then the hand is ruled dead. However often casinos will bend the rules a bit to keep the players happy. For example, last month I was playing blackjack at the Venetian by myself, when I doubled down. The dealer never saw me make the double down bet, thinking I stood, turned over his cards, and hit is own hand with a 4. I then brought the mistake to the dealer’s attention. The pit boss gave me the choice of accepting the 4 as my double down card or it could be burned for the next card in the deck. I chose to burn it and the next card was another 4, and I ended up losing. Although I was happy with how it was handled the pit boss told the dealer to push my bet anyway, which I thought was very nice and beyond the call of duty. To get back to the issue, as long as the floor gave you back your full bet then I think that was procedural.
Krisha from Bel Air, MD
Good question. In 9/6 jacks or better the probability of a royal flush is 22.65% of that of a straight flush, but a royal pays 16 times more. Overall the straight flush only contributes 0.55% to the return of the game. The straight flush is the Rodney Dangerfield of most forms of video poker, it gets no respect. I can only speculate that game makers wanted a big top prize. Nobody likes to come in second, so perhaps that is why the original game makers didn’t pay the straight flush much by comparison.
My question is, is there any way to calculate an optimum playing range of wins/losses? That if a player reaches X number of losses it is highly unlikely that they will recover and should just quit? Likewise, if a player wins X amount, then the player has achieved respectable winnings considering the probabilities of the game and should quit while ahead.
Chris from Tampa
I get asked variations of this question all the time. If you are playing a game with a negative expectation, which is almost always the case, the best strategy to preserve your money is to never play. However, if you are going to play anyway, for the sake of entertainment, there is no best quitting point. The more you play, the more you can expect to go down from wherever your bankroll is at the moment. As I have said many times before, a good time to quit is when you aren’t having fun any longer.