Ask the Wizard #197
When using a five dollar match play coupon in Three Card Poker, would the optimal strategy change on the Ante/Play portion, based on the following assumptions? The $5 match play must be placed only on the ante bet. Also, in Washington State, the match play coupon has "no monetary value," and thus is not counted when determining how much must be bet on the play portion. Thus, a $5 bet on the ante with a match play requires a $5 bet on the play portion, not a $10 bet. Thank you.
Richard from Bremerton, WA
I assume if the dealer doesn’t qualify, then the player wins $10. In this case, the player should always raise. According to my calculations, the value of this coupon is $2.57.
I found these crappy roulette odds from the ferry between England and the Netherlands:
- 1 Number: 30 to 1
- 2 Numbers: 15 to 1
- 3 Numbers: 10 to 1
- 4 Numbers: 7 to 1
- 5 Numbers: 5 to 1
- 6 Numbers: 4 to 1
And it’s American-style (double-zero), despite the fact that the ferry is going between two European countries. What are the odds? — Spanky McBluejay
Shame on that ferry operator. The house edge varies from 13.16% to 21.05%, as follows.
Netherlands/England Ferry Roulette
While playing blackjack at a locals casino in Las Vegas, a dealer from another locals casino sat at my table. While making small talk, she told me that she could wipe out any player using what she called the "house shuffle." The lady dealing to us, who claimed to have been a dealer for 25 years agreed with her telling me that it’s "all about the shuffle." They were both referring to games dealt by hand as opposed to from a shoe. Is there a way to shuffle that lowers the players chances of winning, and if so wouldn’t this be a form of cheating? Have you ever heard of anything called the house shuffle?
Jim Y. from Downey, CA
I don’t believe it. Dealers are not the most skeptical group, often believing all the usual gambling myths. Usually the term "house shuffle" refers to the way the dealers are supposed to shuffle. For example, shuffle twice, riffle, and shuffle again. In this context, she seems to be saying she could alter the shuffle to the player’s disadvantage, which I doubt.
If a multi-game video poker machine is set up with 12 games having theoretical returns from 97% to 99.5%, and I only play the game with the best return, what will the casino’s player tracking system show for my play? Will it show the theoretical return for the specific game I play, or the average return of all games available on the machine?
James S. from Rock Island, IL
It will show the specific return of the game you played.
The Majestic Star in Gary, Indiana, offers double-deck blackjack, but you can’t split aces. How does that affect the house? The other rules are double on 10 and 11 only, no double after split, split other pairs once, and dealer stands on S17.
David T. from Chicago
Oy! Not being allowed to split aces costs the player 0.18%. Overall the house edge under these rules is 0.81%, based on total dependent basic strategy and a cut card game.
Other than the free sites, can you recommend a site where I can get professional sports handicapping for a monthly fee, or percent of win.
Danny M. from Santa Barbara
No. I do not endorse any touts.
In double-zero roulette, what is the probability that any number will not have hit by the 200th spin?
J.F.W. from Marshall
The probability that any given number will not have hit is (37/38)200 = 0.48%.
With 38 numbers, we could incorrectly say that the probability that any one of them would not be hit is 38 × (37/38)200 = 18.34%.
The reason this is incorrect is it double counts two numbers not being hit. So we need to subtract those probabilities out. There are combin(38,2) = 703 sets of 2 numbers out of 38. The probability of not hitting any two given numbers is (36/38)200 = 0.000020127. We need to subtract the probability of avoiding both numbers. So we are at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 = 16.9255%.
However, now we have canceled out the probability of three numbers not hitting. For any given group of three numbers, we triple counted the probability of any single number not being hit. We then triple subtracted for each way to choose two numbers out of the three, leaving with zero for the probability that all three numbers were not hit. There are combin(38,3)=8,436 such groups. Adding them back in we are now at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 + combin(38,3)×(35/38)200 = 16.9862%.
Yet, now we have over-counted the probability of four numbers not hitting. For each of the combin(38,4)=73,815 groups of four numbers, each was originally quadruple counted. Then we subtracted each of the combin(4,2)=6 groups of 2 out of the 4. Then we added back in the 4 groups of 3 out of the 4. So, for each union of four numbers, it was counted 4 − 6 + 4 = 2 times. To adjust for the double counting we must subtract for each group. Subtracting them out we are now at:
38×(37/38) 200 - combin(38,2)×(36/38) 200 + combin(38,3)×(35/38)200 - combin(38,4)×(34/38)200 = 16.9845%.
Continuing in the process we would keep alternating adding and subtracting, all the way until missing 37 numbers. Thus the probability of at least one number never being hit is:
Sum i=1 to 37 [(-1)(i+1) × combin(38,i) × ((38-i)/38)38] = 16.9845715651245%
Here are the results of a random simulation of 126,900,000 such 200-spin experiments.
Numbers Hit in 200 Roulette Spins
|31 or Less||0||0|
The ratio of times at least one number was not hit was 0.169833.
All Books I have seen regarding Baccarat say the banker wins more often than the player. None of them explain why this is. Common sense would make it seem that each side has an equal chance over the long run. Your explanation will be appreciated.
Al from Mississauga, Ontario Canada
Briefly, it is because the banker gets to act last. If the player got a third card that is likely to help, the banker will hit. If the player’s third card is likely to make the player hand worse, the banker will stand.
My wife and I bought a $20 raffle ticket for the Indiana lottery. My understanding of this game is that the drawing for the winning prizes (which number 777) will be held on August 16, 2007, regardless of the number of actual tickets sold and with the absolute maximum available number of tickets being 325,000. As of today, only 60,000 tickets have been sold. Would it be a good gamble to buy a few additional tickets? What would our odds of winning a prize be?
David B. from Evansville, IN
According to the Indiana Lottery web site, there is a total of $3,270,000 in prize money given to 325,000 ticket holders. That would make each ticket worth $10.615 each on average, assuming the series sold out. At a cost of $20 each, the return is 50.31%. If only 60,000 tickets were sold then each would be worth $54.50, for a return of 272.50%. The breakeven point is 163,500 tickets sold. If you believe that fewer than that will be sold, then buying tickets becomes a good bet, putting aside tax and the utility of money implications.