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Ask the Wizard #197
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
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.
James S. from Rock Island, IL
It will show the specific return of the game you played.
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.
Danny M. from Santa Barbara
No. I do not endorse any touts.
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.
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.
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.