Ask The Wizard #308
I won four jackpots in six bingo games. The requirement to hit the jackpot was a coverall within 50 balls. The casino then refused to pay, claiming there was a malfunction, and threatened to take my $100 deposit money too. This doesn't seem fair. What is your opinion?
The probability of a coverall within 50 balls on any given game is 1 in 212,085. The probability of getting one four out of six games is 1 in 134,882,670,482,530,000,000. That sounds like a malfunction if there ever was one. I think the casino has a legitimate case to decline the jackpots, as the games obviously didn't perform properly. However, I think it is just thievery to take your deposit money. I also have to question the integrity of the game, if it could gaff a win like this. Makes me suspect the draw may not be fully random.
This question is raised and discussed in my forum at Wizard of Vegas.
I was playing craps at one of your advertisers and got 38% too many sevens. I suspect they are cheating. Here is my full roll history: 7,5,7,2,4,6,8,7,9,4,9,6,6,6,5,12,7,11,8,4,7,7,9,5,12,5,11,5,8,1,7,7,6,6,6,5,5,9,8,10,9,7,7,11,8,9,3,7,6,10,6,7,8,7,8,6,6,5,5,9,6,7. I think you should quit endorsing this cheating casino!
In 61 rolls the expected number of sevens is 61×(1/6) = 10.17. You had 14. The probability of exactly 14 sevens is 7.96% and the probability of 14 or more is 12.77%. So, nothing unusual there. I also did a chi-squared test on every roll. I know that it isn't very kosher to do a chi-squared test on such a small sample, so take the results with a grain of salt. Here are the results:
Chi-Squared Test on 61 Dice Rolls.
| Dice total | Actual Observations |
Expected Observations |
Chi-Squared Statistic |
|
|---|---|---|---|---|
| 2 | 1 | 1.69 | 0.284608 | |
| 3 | 1 | 3.39 | 1.683971 | |
| 4 | 3 | 5.08 | 0.853825 | |
| 5 | 9 | 6.78 | 0.728597 | |
| 6 | 12 | 8.47 | 1.468944 | |
| 7 | 14 | 10.17 | 1.445355 | |
| 8 | 7 | 8.47 | 0.255829 | |
| 9 | 7 | 6.78 | 0.007286 | |
| 10 | 2 | 5.08 | 1.870219 | |
| 11 | 3 | 3.39 | 0.044627 | |
| 12 | 2 | 1.69 | 0.055100 | |
| Total | 61 | 61.00 | 8.698361 |
The bottom right cell shows a chi-squared statistic of 8.70. The probability of a statistic that high or higher with ten degrees of freedom is 56.09%. These results were close to the peak of the bell curve, so the casino easily passes the chi-squared randomness test.
On the game show Survivor there were two teams, one with nine players and other six. They then got randomly placed into three new teams of five people each. Each new team was composed of three members of the former nine-player team and two from the former six-player team. What are the odds of that?
Let's call the former team of nine player team 1 and the one of six player team 2. The number of ways you could pick three players from team 1 and two from team 2 is combin(9,3)×combin(6,2) = 1,260. The total number of ways to pick five out of 15 players is combin(15,5) = 3,003. So, the probability the first team is split 3/2 in favor of team 1 is 1,260/3,003 = 41.96%.
If that happened, then team 1 will have six players left and team 2 four players. The number of ways you could pick three players from team 1 and two from team 2 is combin(6,3)×combin(4,2) = 120. The total number of ways to pick five out of 10 players left is combin(10,5) = 252. So, the probability the second team is split 3/2 in of favor team 1, given that the first team is already split 3/2 that way, is 120/252 = 47.62%.
If the first two new teams are split 3/2, in favor of the former team 1, then the final team will be split 3/2 among the leftovers.
Thus, the answer to your question is 41.96% × 47.62% × 100% = 19.98%.
Formulas:
combin(x,y)=x!/((y!*(x-y)!)
x! = 1*2*3*...*x
This question is raised and discussed in my forum at Wizard of Vegas.
I think casinos that shuffle the cards early in a good count are cheating. I'm going to file a formal complaint with the Gaming Control Board against the Stratosphere for doing this to me. No particular question, I just wanted to vent.
Shuffling early, as a defense against card counters, has been part of game for 50 years. I would say that if casinos were using computers to tell the dealer when the count was good, as a hint to shuffle, that would be cheating. I also think if the dealer counted himself and shuffled early on recreational players, that too would be cheating. However, if the dealer is doing it when you raise your bets, well, that is just the way the game is played. If you won your case with Gaming, the casinos would ruin the game for counters, like they did in Atlantic City over the Ken Uston lawsuit. The next thing you would see is every game on a continuous shuffler. Both sides would be better off to leave the cat and mouse game as it is.
This question is raised and discussed in my forum at Wizard of Vegas.
What is the etiquette of tipping the shooter in craps?
There is absolutely no expectation of tipping the shooter ever! I would go as far as to ask you not to, lest it become a "thing," and leeches start hanging around the table, only betting on their turn, and shaking other players down for tips. This whole culture of tipping in casinos is getting completely out of hand.
This question is raised and discussed in my forum at Wizard of Vegas.