Ask the Wizard #192
Matt from New Britain
A simple way to estimate the probabiity of winning is to assume that every card has a 12/13 probability of not matching the stated rank. To win this bet, the victim would have to do this successfully 52 times. The probability of 52 wins is (12/13)52 = 1.56%. A fair price to pay would be 63.2 to 1. At 20 to 1 Vince had a 67.3% advantage (ouch!).
Accoring to G.M., who is a better mathematician than I am, the actual probabiliy is 1.6232727%. The reason for the difference is the outcome of each pick is positively correlated to previous picks.
Dave K. from Ohio
Follow this link.
James from Chicago
The house edge of this game is 1.30% or 1.33%, as shown in my survey of Las Vegas blackjack rules, depending on whether the number of decks is five or eight. The odds are better in ANY game where blackjack pays 3 to 2. If you were to play this game, which you shouldn’t, you should still always stand on blackjack. Personally, I think the "World’s Most Liberal Blackjack" claim on the marquee is false advertising.
Scott from San Diego
Ideally, you should challenge the hand before it is over, while it is still easy to run back the cards. It doesn’t hurt to ask later than that, but you are not entitled to anything. This is getting outside my area of expertise, but the decision whether or not to review the tape would likely depend on the amount of money involved and your value as a player.
William R. from Las Vegas
I asked Barney Vinson this question, author of Ask Barney: An Insider’s Guide to Las Vegas. He said the casino would likely only rate one of the bets, in your case $25. An advantage to doing that is that it certainly lowers risk. This might be a good play if you needed to put in a lot of action, for example to qualify for an event you were invited to, and didn’t have much money to lose. However I think that if large bets were involved ($100 or over) it would set off a red flag, and you probably wouldn’t be invited to the next event.
Jim from Brick, NJ
I hope you’re happy, I added a new page to answer this question. Please see my Blackjack Appendix 19.
The question I have about this bet is that 14.41% still isn’t "statistically significant" [ i.e. p < 0.05 ] , which is usually taken to mean greater than two Standard Deviations from the Mean -- or a probability of less than a *combined* 5% of the event happening randomly on EITHER end of the series.
How many Sevens would have to be rolled in 500 rolls before you could say that there is a less than 2.5% chance that the outcome was entirely random (i.e. that the outcome was statistically significant) ?
Many Thanks & BTW , yours is ABSOLUTELY the BEST web site on the subject of gambling odds & probabilities that I’ve found .... keep up the good work !!!
Plexus from Warwick, Rhode Island
Thank you for the kind words. You should not state the probability that the throws were non-random is p. The way it should be phrased is the probability that a random game would produce such a result is p. Nobody expected 500 rolls to prove or disprove anything. It wasn’t I who set the line at 79.5 sevens, but I doubt it was chosen to be statistically significant; but rather, I suspect the it was a point at which both parties would agree to the bet.
The 2.5% level of significance is 1.96 standard deviations from expectations. This can be found with the formula =normsinv(0.025) in Excel. The standard deviation of 500 rolls is sqr(500*(1/6)*(5/6)) = 8.333. So 1.96 standard deviations is 1.96 * 8.333 = 16.333 rolls south of expectations. The expected number of sevens in 500 throws is 500*(1/6) = 83.333. So 1.96 standard deviations south of that is 83.333 − 16.333 = 67. Checking this using the binomial distribution, the exact probability of 67 or fewer sevens is 2.627%.
Ian F. from Provo
Assuming the player always holds the most represented number, the average is 11.09. Here is a table showing the distribution of the number of rolls over a random simulation of 82.6 million trials.