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Should I hit or stay on 15?

Question: “When should I hit on my 15?” Sam says: “Always hit a hard 15 when the dealer's up-card is a 7 through an ace. If you stand on that 15, you will win 25.36 percent of the time and lose 74.64 percent of the time. That makes a net loss of 49.28 percent.

aymag.com - Ask Dr. Blackjack: Hitting on a Hard 15 and 16 - AY Magazine
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With casino gaming being one of the most attractive recreational activities in the United States and over 500 casinos to choose from, AY About You Magazine would like to help you enjoy your gaming experience when playing blackjack by letting you ask professional player and author of “Ask Dr Blackjack”, Sam Barrington, your puzzling questions. If you have a question about blackjack, please contact AY Magazine or Sam Barrington directly at sbarrington21@aol.com. Your question and the answer will be shared with our readers (anonymously, of course, if desired). For an autographed book just go to www.AskDrBlackjack.com and find out my best tactics for playing. Follow AY About You Magazine to find out the dates and times of free blackjack seminars by Sam Barrington. My last post with AY Magazine had me giving seminars at the Tunica casinos. I have since pulled out of there with the dying casino trend there. At one time they had ten casinos operating there at full capacity. Now, they have six casinos. Increased competition is killing them. When the completion of Oaklawn and Southland is finished, it is rumored another two Tunica casinos will close.

So, let’s get to the questions:

From George in Memphis, Tenn.:

Question: “When should I hit on my 15?”

Sam says: “Always hit a hard 15 when the dealer’s up-card is a 7 through an ace. If you stand on that 15, you will win 25.36 percent of the time and lose 74.64 percent of the time. That makes a net loss of 49.28 percent. Now, by hitting that 15, you will win 29.02 percent of the time and lose 66.46 percent of the time and push 4.52 percent of the time. All totaled up, you are 11.84 percent better off hitting the hard 15.

From Rufus in Little Rock:

Question: “What are the rules on hitting a “hard 16?”

Sam says: “This is a tricky play you have to watch out for or it can cost you a little money. To sum it up, never do it. Just sit there like a lump. The “hard 16” is the worst hand you can get, so as the saying goes. You work with what you got when facing the dealer’s up-card of 7 through an ace. If you hit the 16, you will win 25.23 percent of the time, bust out 69.31 percent of the time and push 5.46 percent of the time for a net loss of 44.08 percent of the time. By standing on the “hard 16” you will win 29.01 percent of the time and lose 70.99 percent of the time for a loss of 41.98 percent of the time.

You end up 2.10 percent better off by “standing”.

That’s the same as winning two more hands out of every 100 times you face that scenario. That’s better than the annual return on a CD, and you can do this hundreds of times a year.

READ MORE: Ask Dr. Blackjack: Doubling Down and Insurance

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What are the odds of getting 20% 3 times in a row?

Assuming that the dice is fair and that the rolls are independent, which is what one would expect, the probabilities multiply. So the probability is indeed (1/20)3=1/8000.

math.stackexchange.com - Chance of getting 20 three times in a row on a 20-sided die.

$egingroup$

The probability of rolling a $20$-sided die three times and getting three $20$s is $$ left(frac1{20} ight)^3=frac1{8000} $$ If we were to roll the die three times and compare the triple rolled with $(20,20,20)$, it would, on average, take $8000$ of these triples to get the triple $(20,20,20)$. However, if we were to simply roll the die until three $20$s occurred in succession, it would take, on average, $8420$ rolls to get three $20$s in a row. Here are a couple of ways to compute this expected duration.

Generating Functions

The generating function for the probability of getting three $20$s in a row in exactly $n$ rolls is $$ egin{align} f(x) &=sum_{k=0}^inftyleft(vphantom{frac{19}{20}}

ight.overbrace{ frac{19}{20}x }^ ext{non-$20$}+overbrace{ frac{19}{400}x^2 }^{substack{ ext{non-$20$} ext{one $20$}}}+overbrace{frac{19}{8000}x^3}^{substack{ ext{non-$20$} ext{two $20$s}}}left.vphantom{frac{19}{20}}

ight)^k overbrace{ frac{x^3}{8000} }^{ ext{three $20$s}}[6pt] &=frac{x^3}{8000-7600x-380x^2-19x^3} end{align} $$ The probability of getting three $20$s in a row eventually is $$ f(1)=1 $$ The expected number of rolls to get three $20$s in a row is the derivative at $1$. $$ f'(x)=frac{24000x^2-15200x^3-380x^4}{left(8000-7600x-380x^2-19x^3 ight)^2} $$ The expected number of rolls to get three $20$s in a row is $$ f'(1)=8420 $$

Recursion

Let $E$ be the expected number of rolls to get three $20$s in a row. $$ egin{align} E &=overbrace{frac1{8000}cdot3}^{ ext{three $20$s}}+overbrace{frac{19}{8000}(3+E)}^{substack{ ext{two $20$s} ext{non-$20$}}}+overbrace{frac{19}{400}(2+E)}^{substack{ ext{one $20$} ext{non $20$}}}+overbrace{frac{19}{20}(1+E)}^ ext{non $20$} ag1 &=frac3{8000}+frac{57}{8000}+frac{38}{400}+frac{19}{20}+left(frac{19}{8000}+frac{19}{400}+frac{19}{20}

ight)E ag2 &=frac{421}{400}+frac{7999}{8000}E ag3[6pt] &=8420 ag4 end{align} $$ Explanation:

$(1)$: collect the contributions to the expected duration

$(2)$: separate constants and coefficients of $E$

$(3)$: simplify

$(4)$: $8000cdot(3)-7999cdot(1)_ ext{left}$

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