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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?

 Oct 29, 2020
 #1
avatar
+1

Chance that the dice get all sixes = 6^3, twice in a row = (6^3)^2 = 1/44656

 

Chance that the coins will get all heads = 2^3, twice in a row = (2^3)^2 = 1/16

 

So the three coins twice in a row is 2916 times more likely.

 

Sorry if I got this wrong, my probability isn't very strong.

 Oct 29, 2020
 #4
avatar+2092 
0

Your answer is more than wrong; it’s atrocious!

 

GA

--. .-

GingerAle  Oct 29, 2020
 #2
avatar+111600 
+1

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?

 

How can you roll 2 sixes when you only have one die?

 Oct 29, 2020
 #3
avatar+2092 
0

Hi Melody,

 

Two events are required to meet the criteria.

 

First event: roll a die and flip three coins at the same time.

Second event: roll a die and flip three coins at the same time.

 

At this point, it is now possible to have rolled two sixes twice in a row and/or flip a set of three heads twice on a set of coins two times in a row.

 

GA

 

Edit: strikeout of redundant word(s) that alters intended meaning of statement.  

 

--. .-

GingerAle  Oct 29, 2020
edited by Guest  Oct 29, 2020
edited by Guest  Oct 29, 2020
edited by Guest  Oct 29, 2020
edited by GingerAle  Nov 7, 2020
 #5
avatar+111600 
0

Hi Ginger,

 

"Two events are required to meet the criteria.

 

First event: roll a die and flip three coins at the same time.

Second event: roll a die and flip three coins at the same time.

"

At this point, it is now possible to have rolled two sixes twice in a row and/or flip three heads twice in a row."

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

 

At this point that is not possible.

Two sixes twice in a row = 4 sixes

 

just as,

3 heads twice in a row = 6 heads.

Melody  Oct 29, 2020
edited by Melody  Oct 29, 2020
 #6
avatar+111600 
0

Anyway,  I am being pedantic with the wording, which is appropriate in mathematics.

 

I can never remember how to do these. Would you like to answer it for us?

Melody  Oct 29, 2020
 #7
avatar+2092 
+1

For the die, think of it as a single six occurring in event one, then a single six occurring in event two.

 

For the coins, think of it as a set of three heads occurring in event one, then a set of three heads occurring in event two.

 

The in a row element exists only after two events. 

 

------

 I am being pedantic with the wording, which is appropriate in mathematics.

 

Yes, it is. Pedantic wording reduces ambiguity.LOL

My description above provides an interpretation for a solution.

 

-----

Yes. I’ll post a solution

 

GA

 

--. .-

GingerAle  Oct 29, 2020
 #8
avatar+2092 
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
 #9
avatar+111600 
0

Thanks Ginger,

 

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

 

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?

 

Here is my extremely dubious attempt.

 

For any individual turn 

   P(HHH)=1/8      P(not HHH)= 7/8

   P(6)=1/6            P(not 6) = 5/6

   

For any 2 consecutive turns

   P(HHH any number, HHH not 6 (or vise versa)) = 1/8 * 1* 1/8 * 5/6 *2 = 5/192

 

    P(6  any number of heads, 6 not all heads, (or vise versa)) = 1/6 *1 * 1/6 * 7/8 *2 = 7/144

 

I am only comparing these two things so the total for comparison is  5/192 + 7/144 = 43/576

 

Prob of favourable outcomes / prob or relevant outcome = 5/192  divided by  43/576 = 15/43

 

15/43 is approximately 0.3488

 

I doubt very much that mine is correct, Ginger seems more confident with her answer.

 Oct 29, 2020
edited by Melody  Oct 29, 2020
 #10
avatar+2092 
+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
 #11
avatar+260 
0

Thanks!

Pangolin14  Nov 8, 2020
 #12
avatar+111600 
+1

Nope, I still do not follow your explanation Ginger.  Thanks for your effort though.

 

Your table of what can happen with any one turn is helpful and easy to understand.

 

After that I am lost.

 

I did go off on a tangent and found that in any 2 consecutive goes:

the prob of throwing 6 heads and NOT 2 sixes is  (50+5)/(48*48) = 55/2304

And the prob of throwing 2 sixes and NOT 6 heads is  (56+7)/(48*48) = 63/2304

I do not know what to do with that ...

I am just trying to show you my train of thought.

 

I do not care if you give up but if you want to explain what you are saying clearly I will try and understand.


Perhaps Pangolin14 has a class solution that can be shared with us both?

 


Melody  Nov 12, 2020
 #13
avatar+2092 
+1

Hi Melody,

 

I figured out why you couldn’t follow my logic.  My solution has a shovel-full of bullshit 🐃💩 mixed in with the logic and math.  I didn’t see or smell this until I started constructing another solution;  I’m still not sure where it came from.  It’s Spooky!  (This may be an example of Nauseated’s Spooky Dumbness at a Distance.) Christ! I’m Doomed!

 

My new solution gives (36.364%) as the probability that a set of 3 coins has three heads twice in sequence before a die has two sixes in sequence. 

 

I will present the solution with coherent details after I triple check it for logic and counting errors.

 

GA

 Nov 13, 2020
edited by GingerAle  Nov 13, 2020
 #14
avatar+111600 
0

Ok Ginger, not a problem.   smiley

Melody  Nov 13, 2020
 #15
avatar+111600 
0

I have been thinking about what I did here:   

 

I found that in any 2 consecutive goes:

the prob of throwing 6 heads and NOT 2 sixes is  (50+5)/(48*48) = 55/2304

And the prob of throwing 2 sixes and NOT 6 heads is  (56+7)/(48*48) = 63/2304

 

I am not going to redo that, so I will assume it is correct so far.

 

The prob of either of these two things happening in any 2 consecutive turns is

 

(55+63)/2304 = 118/2304   

 

so maybe the prob that 6 heads comes up before 2 sixes (in a row) is

 

\(\frac{(55)}{2304} \div \frac{(118)}{2304}\\ =\frac{55}{118}\)

 

I don't know if it is right but it sounds logical to me.

 Nov 15, 2020
 #16
avatar+2092 
+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?

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

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%

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

 

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
 #17
avatar+111600 
+2

Thanks Ginger,

NOW our answers almost match.

 

So there are 2304 possible outcomes (for 2 consecutive throws) but we only care about the ones inside this Venn diagram

So there are 36 ways to get six heads 

and 64 ways to get two 6s.

and one way to get 2 sixes AND six heads.

So there is 35 ways out of  99 relevant outcomes where 6 heads is thrown BEFORE 2 sixes.

 

P(HHH,HHH before 6,6) =   \(\frac{35}{99} = 0.\bar{35}\)

 

 

The logic I used last time was more complicated but similar. I think I just made careless errors.

 Nov 20, 2020
 #18
avatar+2092 
+1

NOW our answers almost match.

 

Actually, they match exactly:

\(\text {Odds against } \mathbf {64:35} = \dfrac{35}{99} \; \mathbf \approx \; 35.356\%\)

 

It's very cool  when numbers meet on a high hill Very Cool

 

 

GA

GingerAle  Nov 21, 2020
edited by GingerAle  Nov 21, 2020
 #19
avatar+111600 
+1

YES  IT IS  !    laugh

 

The approximate answer is   35.354%    wink

Melody  Nov 21, 2020
 #20
avatar+2092 
0

OMG! You’re right!

I can’t completely untangle myself from this Quantum Entangled Spooky Dumbness at a Distance BS.  I need a doctor!

 

The only doctors who could fix Quantum Entanglement problems are the time-traveling doctors: Dr. Tud and Dr. Who.

Dr. Tud is lost in time, and Dr. Who is on First and seems stuck there.

 

I may be doomed ... I wonder how long it takes for this to wear off....Maybe never

 

GA

GingerAle  Nov 21, 2020
 #21
avatar+111600 
+1

Don't know.... maybe it depends on what universe reality you are in   wink

Melody  Nov 21, 2020

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