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?
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
| Relavant Rolls | Probability | Formula | Outcome |
| 6,8 | 0.142045 | (5/16)*(5/11) | Lose |
| 8,6 | 0.142045 | (5/16)*(5/11) | Lose |
| 6,7,8 | 0.077479 | (5/16)*(6/11)*(5/11) | Lose |
| 7,6,8 | 0.053267 | (6/16)*(5/16)*(5/11) | Lose |
| 8,7,6 | 0.077479 | (5/16)*(6/11)*(5/11) | Lose |
| 7,8,6 | 0.053267 | (6/16)*(5/16)*(5/11) | Lose |
| 7,7 | 0.140625 | (6/16)*(6/16) | Win |
| 6,7,7 | 0.092975 | (5/16)*(6/11)*(6/11) | Win |
| 8,7,7 | 0.092975 | (5/16)*(6/11)*(6/11) | Win |
| 7,6,7 | 0.06392 | (6/16)*(5/16)*(6/11) | Win |
| 7,8,7 | 0.06392 | (6/16)*(5/16)*(6/11) | Win |
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.
Dear Wizard, I played at South Point in Las Vegas under a promotion where I can get double the value of my cashback in the form of Chevron Gas cards. There was no limit on gas cards stated on the rules. However, I when I went to redeem, they told me I was limited to $500 at most in gas cards. I know that other players were allowed to get more than $1000, so I think that I am being treated unfairly. Do you think that this is fair thing for South Point to do? I lost a lot of money trying to earn those gas cards and expected to get a lot more than $500. Thanks.
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.
On a Crapless Craps table in Tunica, you can buy the 2, 3, 11, and 12. You listed the house edge when you place those numbers, but not when one is bought. What is the house edge on buying the 12 for $30 if you only pay the commission of $1 (rounded down from $1.50) when you win? According to my math, it’s about .47%, which would make it a VERY good bet. I got this by calculating the total money exchanged on all decision rolls ($211, including the vig) and the amount lost ($1). Am I doing this correctly? I want to make sure because this makes it a VERY appealing bet to make! Please detail how you arrived at the house edge as well, so I can make sure I am, in fact, doing it correctly. Thanks so much!
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 |
I recently had a tremendously lucky streak going with the deuces wild video poker game. I was in Las Vegas, and over the course of the weekend hit three natural royal flushes. I’m rounding here, so let’s say I played 10,000 hands during the weekend. What were/are my odds of hitting this/again? Thanks so much for all of your insight!
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).
I was playing one on one double-deck Black Jack in a casino in Louisiana. I had a large bet that I doubled down on a 4 and 7 against dealer’s 4. I opened my cards and place them infront of my chips and dobled down the bet. Dealer continued and opened his cards, 4 and two, then drew a ten and a three. We then noticed that the dealer had forgotten to deal me my double down card. The floor ruled that I lost the hand since I had eleven against nineteen. I asked the floor to give me the ten and draw a card on 4+2+3. Floor said that the cards cannot be backed up! and the best he can do is to return my bet. The next card in the deck was a ten, which would have been my 21 against dealer’s 19. I have decided to meet the casino manager for table games for compensation. I need advice on how to approach the manager on this matter. Is it common to get compensated for this type of casino errors?
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.
When the frequency of a Straight Flush is about four times that of A Royal Flush, how come it pays so low, about 16 times less? I concede that it is impractical. Yet, wouldn’t it be fair to set the payoffs of each hand in the inverse proportion to its frequency?
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.
I think one of the more important parts of gambling is knowing when to walk away. The most common stories I hear from gamblers is how much up they were playing a certain game, only to turn around and lose it all. And most players will simply keep playing a game until they lose their whole bankroll.
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.
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.