GingerAle

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 #6
avatar+2099 
0

Ron, if you hadn’t responded to this, I’d never have known it was you. LOL

 

The post I trolled is definitely BB esque –it’s exactly the kind of post that at least two of the forum’s BBs would make (I can give examples).  Even though you’ve claimed ownership, the only corroborating evidence is that you included a copy of the question in your post. The BBs rarely do this. 

 

I’ve made mistakes before, believing a guest poster to be a BB. It usually happens when the post gobbles like a turkey, walks like a turkey, and poops like a turkey; but it’s really a duck playing a turkey on the net. Before this, the only time I know for sure that the BB I trolled was an impostor is here: https://web2.0calc.com/questions/kim-has-10-identical-lamps-and-3-identical-tables#r9  EP was the imposter.  EP is no BB, but he did an excellent job of playing one on the forum.  And you did too.

 

 While I usually enjoy busting rumps with an occasional troll post, it’s especially good when the target post is a gold nugget to a troll. This is the first time I’ve trolled you. Truly, it was a wonderful first time, but I do wish it was as good for you as it was for me.  Perhaps the next time, when I know it’s you, it will be good for both of us, and we’ll respect each other in the morning.    

 

My favorite target is EP. Busting EP’s rump with an occasional troll post is wonderfully enjoyable. Here’s a recent one: https://web2.0calc.com/questions/helllllppppp_2#r3.  This didn’t start out as a troll post.  I just made a terse comment to Asinus.

Asinus, this is NOT the correct equation for this question.  

 

Then EP, [Hello EP], walks half way across the Troll Bridge and jumps in the river, without any help from me. Then he swims to the river bank, dries himself off, and then walks right into my troll cave to deliver a message...  God, I laughed for ten minutes. This was two gold nuggets. 

 

I’ve never trolled Asinus, except for a few light teases.  He’s worthy of more, but Asinus is very polite, so I’m reluctant, but still I’m tempted.  

 

I remember when I first saw Asinus log on the forum.  When I saw his name, I laughed for an hour.  I thought, “Someone has a wonderful sense of humor or is clueless...”  Here’s why:

 

Sinus translates from the Latin as “half chord [of a circle]” The well known trigonometric function (that is not usually thought of in terms of half chords).  Arcsinus (also Latin), translates as “the arc that makes this half chord.” Asinus is the abbreviated form of Arcsinus. This is (obviously) Asinus’ intended meaning for his user name. 

 

It’s relatively rare to hear or read the Latin word for “sine,” and it’s very rare to hear or read the abbreviated form “asinus,” but asinus is actually a Latin word, unrelated to the words above.  It translates to “ass” as in donkey; it also means “fool” as in jackass.  Even after six years, I still find this very funny. [LOL]

 

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Here, I speculate if you might descend into the BB realm. I didn’t think you would, but if in fact you made that post with the intent of anything other than the objective of imitating the humor of Gracie Allen, then you have.  Even so, I do not think I need to write a Requiem for Ron just yet. 

 

Visiting the Shrine of Organized Stupidity and Perpetual Quantum Dumbness doesn’t mean you have joined The Congregation of the Brain-Dead and Damned.

 

Though you may have descend into the BB posting zone, I do not think that will be your new residence.  Now, the question is, “Why did you, after posting continuously for more than eighteen months, create a post that seems like you were blessed by the God of Stupid (instead of Gracie Allen)?” 

 

I can speculate on this:

... A door to another universe is now open. Enter at your own risk... and read on to find out why and what may have compelled you to visit the BB zone. 

 

Continued...

 

GA

Nov 21, 2020
 #3
avatar+2099 
+1

Solution:

 

\(\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{ }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} \left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \approx 31.64%\\ \)

\(\text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \dfrac{81}{256} \approx 31.64 \% \\ \text{ }\\\)

\( \text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \dfrac{405}{2048} \approx 19.78\% \\ \text{ }\\\)

\( \text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games in six trials and then win the seventh game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 6})* \rho \text{(Cbs win 7th game}) =\\ \text{ } \hspace{34 mm} \dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\dfrac{405}{4096} \approx 9.88\% \\ \text{ }\\\)

\( \text{ The sum of the individual probabilities gives }\\ \text{ the overall probability of the Cubs winning the series.}\\ \text {Sum of individual probabilities: } \\ \left(\dfrac{81}{256}\right)+ \left(\dfrac{81}{256}\right) + \left(\dfrac{405}{2048}\right)+ \left(\dfrac{405}{4096}\right) = \dfrac{3807}{4096} \bf \approx 92.94\% \)

 

 

 

GA

Nov 16, 2020
 #16
avatar+2099 
+2

I roll a die and flip three coins at the same time, and repeat this multiple times. What is the probability that I flip three heads twice in a row before I roll two 6's twice in a row?

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Solution:

 

The probability, in two (2) sequential events, that a set of three coins presents three heads twice before a die presents two (2) sixes, is equal to their respective ratios when calculated as odds against in the (fully populated) sample space.    

 

The three coins and six-sided die have (8*6) 48 unique arrangements.

The two (2) events have (48^2) 2304 unique pairs of these arrangements.

Of the (2304) pairs of arrangements ONLY the pairs that have either two sixes or two three-head coin sets are of interest. The other 2205 pairs are irrelevant.

 

Sixty-four (8^2) of these pairs will have two (2) sixes and thirty-six (6^2) of these pairs will have two (2) sets of three heads. Note that one (1) pair will have both a pair of sixes and a pair of three-head sets that is shared by the (64) and (36) pairs. It’s necessary to subtract (1) from the (36) pairs of three-head sets, because this simultaneous appearance fails the requirement that the two sets of three-heads appear before the two sixes. (This does not affect the success of the die pair.)

 

The new sample space is now populated with 99 elements with the ratio of (64) pairs of die with a six to (36-1) pairs of three-head coin sets. When sampled randomly the odds against the coins are 64:35. This gives a probability of 35.356%

 

So the probability that a set of 3 coins has three heads twice in sequence before a die has two sixes in sequence is 35.356%

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Descriptive Arrays

The two arrays list the pairs of index numbers (from the array in post #10) that gives success to the Die or the Coins. 

To use: choose a number on the left and then one of the corresponding numbers on the right.

For example (7, 3):  Number 7 corresponds to (6-TTH) and the 3 corresponds to (6-HTH).

 

\(\begin{array}{|c|c|} \hline \text {Die} & \text { successes }\\ \hline \text {First selection} & \text{Second selection}\\ \hline 1 & \;\;\;2,3,4,5,6,7,8 \\ 2 & 1,2,3,4,5,6,7,8 \\ 3 & 1,2,3,4,5,6,7,8 \\ 4 & 1,2,3,4,5,6,7,8 \\ 5 & 1,2,3,4,5,6,7,8 \\ 6 & 1,2,3,4,5,6,7,8 \\ 7 & 1,2,3,4,5,6,7,8 \\ 8 & 1,2,3,4,5,6,7,8 \\ \hline \end{array} \text { } \begin{array}{|c|c|} \hline \text {Coin} & \text {successes }\\ \hline \text {First selection} & \text{Second selection }\\ \hline 1 & 1, 9, 17, 25, 33, 41 \\ 9 & 1, 9, 17, 25, 33, 41 \\ 17 & 1, 9, 17, 25, 33, 41 \\ 25 & 1, 9, 17, 25, 33, 41 \\ 33 & 1, 9, 17, 25, 33, 41 \\ 41 & 1, 9, 17, 25, 33, 41 \\ & \\ & \\ \hline \end{array} \)

 

-------------------

The title says this is tricky.  There is an illusion when solving this: The die has six sides and eight selections, and the three coins have eight arraignments and six selections.   Because of this cross match, I inverted the logic in the first solution –subtracting one from the die sets instead of the coin sets. I also doubled counted the success selections. 

 

My quantum state of dumbness opened a portal and transported a shovelful of BS into my solution.  This is definitely Spooky Dumbness at a Distance.  The spookiest part is I probably would not have noticed this for months, or ever, if not for the parallel solution I constructed at your behest.  Then the contamination became instantly obvious. 

 

Now that I’ve extracted the bovine excrement, I’ll use it to fertilize my quantum flower garden.  Flower equations thrive on this.LOL

 

 

 

GA

Nov 15, 2020
 #10
avatar+2099 
+2

Unfortunately, I really do not understand the logic behind what you have done.

 

That’s interesting ...if not off-putting. W T F?

 

What part of my logic do you not understand? You’ve demonstrated logic and mathematics for probability questions far more advanced and complex than this one.  Investing a few minutes of time in analyzing the steps should bring brilliant illumination in understanding the logic.  If there is a flaw in my logic, it should be discernable in one of the steps outlined above. 


Here’s a list of all possible die rolls with all possible combinations of H/T for the three coins. 

\(\begin{array}{|c|c|c|} \hline 1& 6 & HHH\\ 2& 6 & HHT\\ 3& 6 & HTH\\ 4& 6 & HTT\\ 5& 6 &THH\\ 6& 6 &THT\\ 7& 6 &TTH\\ 8& 6 &TTT\\ \hline \end{array} \)  \(\begin{array}{|c|c|c|} \hline 9& 5 & HHH\\ 10& 5 & HHT\\ 11& 5 & HTH\\ 12& 5 & HTT\\ 13& 5 &THH\\ 14& 5 &THT\\ 15& 5 &TTH\\ 16& 5 &TTT\\ \hline \end{array} \)   \(\begin{array}{|c|c|c|} \hline 17& 4 & HHH\\ 18& 4 & HHT\\ 19& 4 & HTH\\ 20& 4 & HTT\\ 21& 4 &THH\\ 22& 4 &THT\\ 23& 4 &TTH\\ 24& 4 &TTT\\ \hline \end{array} \)   \(\begin{array}{|c|c|c|} \hline 25& 3 & HHH\\ 26& 3 & HHT\\ 27& 3 & HTH\\ 28& 3 & HTT\\ 29& 3 &THH\\ 30& 3 &THT\\ 31& 3 &TTH\\ 32& 3 &TTT\\ \hline \end{array} \)   \(\begin{array}{|c|c|c|} \hline 33& 2 & HHH\\ 34& 2 & HHT\\ 35& 2 & HTH\\ 36& 2 & HTT\\ 37& 2 &THH\\ 38& 2 &THT\\ 39& 2 &TTH\\ 40& 2 &TTT\\ \hline \end{array} \)   \(\begin{array}{|c|c|c|} \hline 41& 1 & HHH\\ 42& 1 & HHT\\ 43& 1 & HTH\\ 44& 1 & HTT\\ 45& 1 &THH\\ 46& 1 &THT\\ 47& 1 &TTH\\ 48& 1 &TTT\\ \hline \end{array} \)

Randomly pick a number from 1 to 48, and then do it again. 

 

To me, the logic and math used for the tabulation of these simultaneous events seems simple and straightforward.  The only complexity, which seems trivial after a full view, comes from the relative time (dimension) introduced in the question by use of the word “before,” What is the probability that I flip three heads twice in a row before I roll two 6's twice in a row?

 

This necessitates setting up an equation to calculate a probability that one event (two sequential sixes) occurs BEFORE another event (two sequential sets of three-heads) occurs.  This requires factoring out non successes –where neither a six nor a set of three heads occurs, leaving only the successes of the die or the coins. 

 

After factoring, (28.125%) of the twin events in the sample set will have a success for either the die or the set of coins. There are no ties in this sample set: if the three heads appear simultaneously with a six on the die, twice in two events then the coins have the success because the six has to appear twice BEFORE the set of three-heads occurs. In this space, the coins are successful (44.44%) of the time and the die is successful (66.66%) time. This means randomly sampling from this set will produce a selection where the coins are successful (44.44%) of the time time when compared to the die.

 

So the probability that a set of 3 coins has three heads twice in sequence before a die has two sixes in sequence is (44.44%).

 

 

GA

Nov 7, 2020
 #8
avatar+2099 
0

Solution 1:

 

Probability of two (2) sixes on a single die occurring before the occurrence of two (2) sets of three heads, in two sequential events.

 

The first roll: the probability of a six is (1/6). At this point, the outcome of the coins does not matter. Probability: (1/6) * 1 0.16667

 

The second roll: the probability of a six is (1/6). At this point, the outcome of the coins cannot be three heads. Probability: (1/6) * (1-(1/8)) 0.14583

 

The two events can occur in any order.  Mean probability (0.16667 + 0.14583)/2   (0.15625)

----

Probability of two (2) sets of three heads occurring before or in concurrence with two (2) sixes occurring on a single die, in two sequential events.

 

 

The first roll: the probability of three heads is (1/8). The outcome of the coins does not matter. Probability: (1/8) * 1 0.125

 

The second roll: the probability of three heads is (1/8). The outcome of the coins does not matter. Probability: (1/8) * 1 0.125

 

The two events are not distinguishable.  Mean probability: (0.125 + 0.125)/2 = 0.125

----------------------------

Solution 2:

 

Each event can have 48 unique arrangements of die number and ordered (H/T) coin sets. Any of these 48 arrangements can occur twice in the two events.

 

In the two (2) events, there are 96 total arrangements, and there are 15 arrangements that give success to the die. Probability of success for the die in two events: (15/96) = (0.15625).

 

In the two (2) events, there are 96 total arrangements, and there are 12 arrangements that give success to the set of coins. Probability of success for coins in two events (12/96) = (0.125)

 

The question asks:

What is the probability that I flip three heads twice in a row before I roll two 6's twice in a row?

 

Rewording this to clarify:

What is the probability that I flip a set of three coins to three heads twice in sequence before I roll a 6 on a die twice in sequence?

 

In a large sample of twin events 71.875% of the time there is no success for either the die or coins. For 28.125% of the either the coins or the die have a success.   In this space, the coins are successful 44.44% of the time and the die is successful 66.66% of the time. So the probability of set 3 coins has three heads twice in sequence before a die has two sixes in sequence is 44.44%  Who the heII knows?It seems logical though.

 

Ambiguous wording aside, this is an interesting question.  

 

GA

 

--. .-

Oct 29, 2020