Ask the Wizard #192
I was at Foxwoods the other day, watching the final two tables of the Foxwoods Poker Classic. When Vince Van Patten (one of the World Poker Tour hosts) came in to watch, he started making all kinds of prop bets with some of the poker pros who were hanging around. He was offering anyone 20 to 1 if they could flip through an entire deck of cards cycling through the ranks and saying aloud while peeling each card Ace, 2, 3, 4, and so on up to King and starting again at Ace without ever having the card they’re announcing come up. Nobody made it all the way through and Vince won a few hundred dollars in about 10 minutes before everybody gave up. I know this has to be possible, but I have a suspicion Vince has quite the hustle going offering only 20 to 1 on this. What are the odds of actually getting through the whole deck?
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.
You have the odds and combinations listed for five-card stud with one joker fully wild. Would you also post the same for two jokers fully wild, as all decks come with two jokers (1-red, 1-black) and many people play with both used as wilds.
Dave K. from Ohio
Follow this link.
I recently made a trip out to Vegas, where I came upon a game called the "World’s Most Liberal Blackjack" at the Las Vegas Club. In this game you are allowed to: double down with any 2, 3 or 4 card combination, split & re-split aces as often as you choose, split & re-split any pair as often as you like, surrender your first two cards for half of your original bet and any hand with six cards automatically wins. The caveat is blackjack pays even money unless it’s suited in which case it pays 2 to 1. Is this a better game than a 3 to 2 BJ with 6 decks and the dealer standing on a soft 17? Also, in this case, would it be beneficial to double down since the BJ only pays even money?
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.
I went to Vegas last month and played Three Card Poker for the first time. I got a straight flush, and was so excited to win that I didn’t notice that the dealer only paid out 20 to 1 instead of 40 to 1. I lost a few hands and left the table to go cash in, then realized what had happened. My question is, if I notice an error in the future, what should I do? I’m assuming after I left the table it was way too late, but what is the rule if I’m still at the table? If I don’t point out an error before the next hand begins, is it too late?
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.
For baccarat, if you found a casino that allowed a player to bet both player and banker at the same time, is there any advantage to do that? How about if they rated you for the total of both bets? (I.E. - bet $25 on Banker, $25 on Player and be rated for $50)
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.
Many casinos allow bets behind another player’s on the blackjack table. Can you please tell us the proper pair-splitting strategy when the "behind" wager far exceeds the regular wager, assuming the two bettors are working together?
Jim from Brick, NJ
I hope you’re happy, I added a new page to answer this question. Please see my Blackjack Appendix 19.
This is about controlling the dice at Craps. You previously discussed the Stanford Wong Experiment, stating, "The terms of the bet were whether precision shooters could roll fewer than 79.5 sevens in 500 rolls of the dice. The expected number in a random game would be 83.33. The probability of rolling 79 or fewer sevens in 500 random rolls is 32.66%.... The probability of rolling 74 or fewer sevens in 500 random rolls is 14.41%."
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%.
What is the expected number of rolls needed to get a Yahtzee?
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.