TurboGenius Dice related comments - moved conversations

@turbo.
I gave this task in another forum, but because it has a big relationship with your claiming, post it also here.
It is a very old task. Two players decided to play dice game . They both throw dice and one wait dor sequence 6-6-6 , but another wait for exactly 4-5-6.
They do throws in order who will collect his sequence first - win.
Who will win ? Who have bigger chances to win?
Show how good you are in math

 

This is primary school maths. Both players have the exact same probability to complete their sequence.
1/(6x6x6) or (1/6)^3 = 1/216

 

No you are nor right, so you not know that primary school math . Not all is so simple

The answer is described in CC forum

 

"So as promised, I will explain this task.
Pitty but looks that peoples did not know that task and to solve by individual forces they cant...

So the task is who has better chances first to collect his sequence. Better chances to collect sequences have that, who collect sequences from three different numbers!
The easiest way to understand why that is, in my opinion, is this - say is starting position when both players hit his first number (666) hit 6 and (456) hit 4. Try to calculate how many numbers lead to success if say we have limit 4 throws at all.

This way (666) player after first hit 6 has only one successful next throw - that is 6, if he in the second throw hits another number he cant collect his winning combination till 4th throw.

Now, look at player (456) he after first hit 4 he has two numbers which still leed to possible win that is 5 and wining combination will be 456x, but also can win after hitting 4 this way wining combination will be 4456. So this player can win in two ways [456x;4456] but the player (666) only in one way [666x]. And player (456) in the next three throws after the first successful have better chances.
This way we can increase the number of throws and see that always player (456) have slight better chances to be first who collect his combination.

So here is some hint that nobody payss attention to the the task - that winner is that who is first .


TurboGenius infected your brain. "Can win after 4 this way winning combination will be 4456." And the other player who got his two 6 (66x) to complete his sequence?. He has the advantage compared to the other who got (44xx) to complete the sequence.

You are wrong, and the right answer to your problem is they have the exact same odds to complete their sequence, in other words in the long run, both will complete their sequence, either 6,6,6 or 4,5,6 on the 216th attempt.

 

I just got confirmation from a Maths teacher specialized in probability:


"If the order 4, 5, 6 must be respected, then the probability to obtain 6,6,6 is equal to the one to obtain 4,5,6 : (1/6)^3 .On the contrary if the order does not have to be respected, the probability is equal to (1/6)^3 x 3! let (1/6)^3 x 6 = (1/6)^2


Apologies accepted Bebedick.

 

if you at least can do elementary programming in excel do that and count how many times firstly will be reached sequence 456 and how many times and how many times first will reach 666 and you will see that 456 wins this competition.
But I see that your maths teacher is very weak. If he cants even write all variants in 4 hits and compare Nobody for him will help

 

This is a mathematical problem. The teacher gave you the mathematical probability formula that gives the answer: 4,5,6 or 6,6,6 sequences have the same probability to appear, therefore there is not a sequence that will appear before the other in the long run.

What is your mathematical probability formula that proves that sequence 4,5,6 will appear "slightly" before 6,6,6 sequence?. Prove it. We need something better that " I launched some sessions on excel ". You are a complete noob and amateur. And what is your story about 4 hits? It's just a cluster of the whole game, not the answer of your maths problem.

 

Dice question is a fool’s errand.

 

He wanted to make fun of Turbo but the boomerang effect occured: Bebedick showed he can't resolve his own primary school probability maths problem lol

 

I proved it if you can't read - that is your problem. By the way, this task is not mine it is the same task as problem Monty hall. That you do not know such tasks show only your weakness in math...Do all sailors are such weak in math or you are such alone?

 

I understand for you all who can solve equations in the second degree are fool's, but you all, who have problems with solving third-class tasks are smarts. Very easy to live

 

Yes, the effect is good - sailors not understand even what is written and can't even use Google for finding a solution ...That task has maybe a hundred years...not I am its creator .

 

Buddy, the sailor was traveling all over the world dicovering beautiful countries you've only seen on TV or on a postcard, while you were and still are in Croatia welding printed circuit resistors for selling your clocking device that never worked in the Casino lol

I have read your explanations which are invalid. You state 4,5,6 is more likely before 6,6,6 when there is a repeat. But it is the contrary, if I am the player who already got 6,6,x,x the odds I complete my sequence 6,6,6 is 1/6 because I am at one occurence to finish my sequence. While you, with your 4,4,x,x you still have TWO occurences hence the probability of 1/(6x6) to complete your sequence 4,5,6.

Go back to school and learn english, I suspect you re-wrote the maths problem very differently because your english is very bad, I believe the original problem was with 3 dices and the sequence to obtain in any order, in this case, yes, 4,5,6 is more likely to occur before 6,6,6.

 

Go back to school and learn some math because no matter what I wrote you not understand...
I do not have a chance to prove, of course, I have no wish to probe to ...ok will not name but if you think that you are better than me on match we can easily that solve

 

I registered but the school director emailed me that student Bebedick took the last maths class seat for the lesson: "Understanding probability of independant events: Dice and Roulette study".

 

Why its need the same repeat again and again?
Are two players one win when collecting 456, second when collecting 666. Say they decided to throw max 3 times after the first successful hit. Hit number must be always the same for both player - I think that is clear?
So now the first player hit 4 and second hit 6. Not need to explain that to reach this point they have equal probability? So they did the first step and both have made a first successful hit. Nimber of throw here is not important - simply 4 total hits are the minimal number to explain the situation.

Now left three throws. look to the second player his second throw can be one from 6 variants 123456. To wining combination leads only one hit 6. It is clear that if he will hit another number and have the combination for example 64 he can't win in the next two throws, or that is not clear?
So only hit if 6 keep him in a winning way! Does that clear ??? So he has chance 1/6 to stay in wining way and if next throw will be again 6 he has a chance to win if his opponent does not collect a winning combination. Does that clear? After first sucesseful hit and two more hits, he has 1/36 chance to stay in winning way!
P[2]=1/36 !!!

Now do the same procedure with the first player. He in the first throw hit 4. In next throw he can hit any number say he hit 5 and in third throw he can hit 6 chance of such event will 1/6 *1/6=1/36
So for the first player to stay in the fight are chance P[1]=1/36 +P[aditional]

You see that P[1]>P[2] if P[aditional] > 0 ?? Is this is clear ????

Now, what is P[aditional] ? In second throw first player not hit 5 as in the previous variant but hit 4 for that, he has the same chance 1/6, then in the third throw, he hit 5 and in fourth hit 6 and this way after 4 hits he collects his winning combination and wins !!!

So P[aditional] = 1/6*1/6*1/6=1/216 !!!

Ant total chance for the first player to be the winner after the fourth throw is :
P[1]=1/36+1/216=7/216 what is > than P[2]=1/36=6/216

If for you not clear that 7 is > than 6, so are no chances to explain better. If you cant read the task and read that in the text are no word about chances to hit for separate combinations again what I can do?

You simply read what you want, but in the text of task is clear wrote - the winner is that, who first collected, no word about chances to appear for one or for other combination...