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Ask the Wizard #251
reno
From the Nevada Gaming Control Board 2009 revenue report, we see the win for "21" was indeed $1,008,525,000. That probably includes blackjack variants. According to my February 20, 2010 Ask the Wizard column, the cost of mistakes in blackjack is about 0.83%, according to gaming consultant Bill Zender.
The missing piece is what would be the house edge without the errors? I admit this is kind of crude, but the average of the house edge column in the April 2010 Current Blackjack Newsletter is 0.78%. So, the total house edge in blackjack, including errors is 0.78% + 0.83% = 1.61%. The portion of that due to errors is 0.83%/1.61% = 51.55%. So the 2009 profit from blackjack errors in Nevada could be roughly estimated as 1,008,525,000 × 0.5155 = $519 million.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
James from Spencer, MA
The following table shows the probability of each kind of royal, according to the number of cards held, given that there was a royal. It shows that 3.4% of royals are from holding one card. The probability of a royal to begin with is 1 in 40,391, so the unconditional probability of a royal holding one card is 1 in 1,186,106.
9/6 Jacks Royal Combinations
Cards Held  Combinations  Probability 

0  1,426,800  0.002891 
1  16,805,604  0.034053 
2  96,804,180  0.196154 
3  195,055,740  0.395240 
4  152,741,160  0.309498 
5  30,678,780  0.062164 
Total  493,512,264  1.000000 
AZDuffman
The formula is V = P × [(1(1+i)^{n})]/(i/(1+i)), where:
V = value of annuity
P = individual payment amount
i = interest rate
n = number of payments
Let’s say the jackpot was $15M. Using i = 4.66%, and n=25, the fair payment to keep up with inflation would be $982,525. You would actually get 15M/25 = $600,000. Actual payment/fair payment = 61.07%.
Not that you asked, but the formula if the payments are made at the end of each year is V = P × [(1(1+i)^{n})]/i.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
RbStimers
Let t be the number of turtles made, and x the number sold.
pr(x<=t)=0.9
pr(x14.29<=t14.29)=0.9
pr((x14.29)/3.5)<=(t14.29)/3.5))=0.9
The left side of the inequality follows a standard normal distribution (mean of 0, standard deviation of 1). This next step takes an introductory statistics course, or some faith, to accept.
(t14.29)/3.5 = normsinv(0.9) This is the Excel function.
(t14.29)/3.5 = 1.282
t14.29 = 4.4870
t = 18.77
Nobody is likely to buy 0.77 of a turtle statue, so I would round up to 19. According to the binomial distribution, the probability of selling 18 or less is 88.35%, and 19 or less is 92.74%. This question was raised and discussed in the forum of my companion site Wizard of Vegas.
pacomartin
The math works out more easily if he bet on the Player. I work out a similar problem in roulette at my mathproblems.info site, problem number 116. For even money bets, the general formula is ((q/p)^{b}1)/((q/p)^{g}1), where:
b = starting bankroll in units.
g = bankroll goal in units.
p = probability of winning any given bet, not counting ties.
q = probability of losing any given bet, not counting ties.
Here the player starts with $12 million, or 60 units of $200,000, and will play until reaches 120 units or goes bust. So in the case of the Player bet the equation values are:
b = 60
g = 120
p = 0.493175
q = 0.506825
So the answer is ((0.506825/0.493175)^{60}1)/(( 0.506825/0.493175)^{120}1) = 16.27%.
It is much more complicated on the Banker bet, because of the 5% commission. That would result in the distinct possibility of the player overshooting his goal. If we add a rule that if a winning bet would cause the player to achieve his goal, he could bet only what was needed to get to $12 million exactly, then I estimate his probability of success at 21.66%.
A simpler formula for the probability of doubling a bankroll is 1/[1+(q/p)^{b]. }
This question was raised and discussed in the forum of my companion site Wizard of Vegas.