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OK Zamboe, lets settle this. New thread


emprworm
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i want a new thread as to not corrupt the other thread any more with this subject since we are on a new riddle over there

ya, i didnt mind the guessing answers on the hard riddles. that robot duel took up like 3 pages of math spam. what a headache!

When someone disagrees with you is spam ?.LOL. Ok. whatever, today more than ever I am sure that my answer are correct. Your answers are incorrect. period. You don't provide any source to backup your calculus, just your result as a benchmark to everyone else. It was not a headache for me, because it's just simple math. However I had to triple check just to make sure my affirmations were correct.

Sorry for the spam, it was unintentional, just an interest in finding a correct solution. Won't spam your posts anymore.

Ok zamboe. I proven this empirically. I asked you repeatedly to provide your equations and your proof, AND your iterations. You did nothing.

I spend 2 hours showing you MANUALLY the proof. I provide equations, iterations, AND a complimentary computer programming, all proving 100% accuracy of the results. I even do a manual count and prove it a fourth time.

It is spam and a waste of my time because you do nothing. I ask you to show me your equations and you show nothing and say nothing but be stubborn. If you show me your formulas, I will show you were your math is bad. I call it spam because I spend so much time with my post asking you to show me your proofs and you give me nothing, hence waste my time. and i dont like you wasting my time.

I am right, I am 100%, and as sure of it as the sun will rise tomorrow. I am as certain as the earth is round. This is basic math for me, Zamboe. For you it is obvious that you struggle with this math.

I will help you by showing you your error. Post your formulas, AND your iteration and I will expose your errors. If you cannot post a formula AND an iteration, then I will assume you have no idea how to solve this math and are just being stubborn.

Here is what I want you to prove and solve this problem:

If you cannot solve this problem then you do not understand the math. Here is your problem:

PROBLEM FOR ZAMBOE TO SOLVE

1 guy is flipping a 3 sided die, and the other a two sided coin.

COIN is labeled Side 1, Side 2.

DIE is numbered: 1, 2, 3

Player A has the coin

Player B has the die

goal: first person to throw a 1 wins.

Player A goes first, then Player B and repeat until a 1 is throwed. What are the odds that Player A will win this game?

I want you to post the ITERATIONS (probability trees), and the FORMULAS. Dont just say "My calculator, that can use the very same principles as any utility program, gives me a different result" I WANT TO KNOW WHAT YOU ARE EXACTLY TYPING INTO YOUR CALCUALTOR.

Post EVERyTHING you are doing to solve this problem

then i will expose your error and help you

i await for you to solve the problem.

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I am right, I am 100%, and as sure of it as the sun will rise tomorrow. I am as certain as the earth is round. This is basic math for me, Zamboe. For you it is obvious that you struggle with this math.

You might want to go over your basic math, then.

I keep my statment that his results are not correct, because among other reasons this problems is solved based on Bayes's Theorem and nothing else. Iterations are only used on distribution problems, and this problem is not about that. Very Basic.

You think your results are right, ok. I think they are wrong. Any problem with that ?.

After answering this post, closing this threath is a good idea, since I don't have to prove you something.

Last, If you keep thinking I spam your posts and give you headches, sorry about that, I repeat, it won't happen again.

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and, Zamboe, you dont know how to solve this do you?

If it is so basic, then lets hear the answer.

Or, are you too embarrassed, Zamboe?

I'm so glad you know all about Bayes theorem, however you do not need Bayes theorem to solve this (you could if you really wanted to, but why go through the hassle?)

Bayes theorem describes inverse probability based upon a known event trying to determine the probability of a specific cause.

This problem is NOT inverse probability, Zamboe.

Now I am fully convinced you really have no idea about this. Maybe you took a year of math in high school? I don't know.

What I do know is that you are too scared to solve that basic problem I asked you to solve and show the equation.

Actually you are not too scared...you just have no idea how to do it.

You can prove to everyone in this room right now that you are not a person that just blows smoke by stepping up and demonstrating to us all that you actually UNDERSTAND this math.

If you cannot demonstrate that you actually understand this math by SOLVING a very basic problem (and yes, it IS very basic, which is why it should be so easy for you)

If you cannot solve the basic problem, then all of us in this room and in this forum will know the truth: You dont know how.

Is ok Zamboe. Its ok to not know something.

There are many things I don't know in life. Its part of life. No one knows everything. You dont know this math....big deal. Theres nothing to be ashamed of. Just be honest and tell the truth. If you really know this math, then prove it to us.

we await you to show us the answer to the problem (which I will now repeat)

PROBLEM FOR ZAMBOE TO SOLVE

1 guy is flipping a 3 numbered die, and the other a two sided coin.

COIN is labeled Side 1, Side 2.

DIE is numbered: 1, 2, 3 (1/3 chance per each number)

Player A has the coin

Player B has the die

goal: first person to throw a 1 wins.

Player A goes first, then Player B and repeat until a 1 is throwed. What are the odds that Player A will win this game?

Once again, we await for Zamboe to show us he knows how to do this math, or just more meaningless talk.

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If you cannot solve the basic problem, then all of us in this room and in this forum will know the truth: You dont know how.

Is ok Zamboe. Its ok to not know something.

we await you to show us the answer to the problem (which I will now repeat)

You complain about spam, and you spam really good.

Stop behaving like you are superior and looking down all the rest, that wont work with me. Because you are not.

I know how to resolve it, I did it, if you don't understand when I posted it, well I won't teach you math and I won't be explaining my answer with a detail that a 10 year old child need, there is all the necessary information, that's why I DONT BELIEVE YOUR RESULT. Is it so difficult to understand ?.

Besides in the previous threat while the puzzle was being considered I asked you several times to give some partial results of your answer, you never did, instead you keept repeating the same thing, I asked you several times to provide that partial results.

How come you expect I answer your request, if you don't answer mines first ?

Last, don't talk for others, when you request something say "I wait" instead of "We wait".

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still no answer eh Zamboe?

Yea, like I thought.

The nice thing about math is that if you dont know it...you just dont know it. So me continually asking you to answer the question will do no good. We are talking about a new problem, Zamboe. One that is much easier than the riddle.

This is a very simple one. Should only take 2 minutes of your time.

That is, of course, if you know how to solve it.

If you don't know how to solve it, then I would expect your next post to be more of the same beating around the bush.

Now, you want to talk probabilities?

Ok, lets do it.

I predict that your next post will be one of 2 things:

1. I predict you will not solve the problem in your next post and provide formulas as I requested, and instead have more excuses

OR

2. YOu will not post again and run and hide instead.

ONE OF THOSE TWO THINGS WILL HAPPEN!

I predict this with 99.9% accuracy.

I am almost always right when I make predictions like this.

Don't you love it when I'm right?

But lets see what happens....

To prove how right I will be, I will not even edit this post so that future readers can see how you fufilled my prediction and did exactly what I knew you would do.

lol

:D

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emprworm you own words fall to you.

You haven't answered my question about partial results of the problem of the ROBOTS that I asked you, because you just don't know how to calculate them.

Don't try to deviate the issue, answer the questions of partial results , that I will repeat AGAIN :

A. ODDS OF R1 BEING KILLED

B. ODDS OF R1 SURVIVING

C. ODDS OF R2 BEING KILLED

D. ODDS OF R2 SURVIVING

E. ODDS OF R3 BEING KILLED

F. ODDS OF R3 SURVIVING

DONT DEVIATE THE PROBLEM. Don't run away.

Focus on the puzzle we disagree on.

If you don't post what I asked before you ever asked me anything , then you have no right to ask me something that you don't even do.

I am not only right, also now you make me laugh.

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lol! of course zamboe refuses to do a simple math problem. why am i not surprised. I am now talking to everyone else in here.

I want you all to see that I backup my claims with details and formulas. Watch zamboe's reply, will it contain his formulas or more cowardly running? Most likely Zamboe wont even read this post, but who knows...maybe he does have some honor. lets see how he responds.

R1 hits 30% of the time

R2 hits 50% of the time

R3 hits 100% of the time

Question: R1 goes first, what is his best move?

R1 has 3 possible first moves. Our task is to figure out which is the best according to the riddle. Therefore the only way to accomplish this task is to seperately calculate the odds of each move and then decide the best move based upon the best odds.

POSSIBLE FIRST MOVES FOR R1:

I: R1 could fire and hit R2

II: R1 could fire and hit R3

III: R1 could miss

In the case of I: Since the riddle specifically told us that all robots will maximize their odds, therefore, there is a 100% probability that R1 will not fire at R2, and a 0% probability that R1 would fire at R2. This decision is not subject to chance or randomness. The decision is fully predetermined, therefore to compute R1 firing at R2 and average it into our odds would be futile and errant. The probability of R1 firing at R2 as a first move is zero, therefore we do not average this decision tree into our probability computations.

In the case of II: R3 is dead, therefore it is a showdown with R2 vs. R1, with R2 taking the first shot.

A. ODDS OF R1 BEING KILLED = 77%

B. ODDS OF R1 SURVIVING

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R1 hits 30% of the time

R2 hits 50% of the time

R3 hits 100% of the time

R1 has 3 possible first moves. Our task is to figure out which is the best according to the riddle. Therefore the only way to accomplish this task is to seperately calculate the odds of each move and then decide the best move based upon the best odds.

POSSIBLE FIRST MOVES FOR R1:

I: R1 could fire and hit R2

II: R1 could fire and hit R3

III: R1 could miss

Information to begin with.

As I stated before, my comments were based on your math mistakes, I agree on the logical explanation.

In the case of I: Since the riddle specifically told us that all robots will maximize their odds, therefore, there is a 100% probability that R1 will not fire at R2, and a 0% probability that R1 would fire at R2. This decision is not subject to chance or randomness. The decision is fully predetermined, therefore to compute R1 firing at R2 and average it into our odds would be futile and errant. The probability of R1 firing at R2 as a first move is zero, therefore we do not average this decision tree into our probability computations.

Agree, no math needed to compute that option. Logic is enough.

In the case of II: R3 is dead, therefore it is a showdown with R2 vs. R1, with R2 taking the first shot.

A. ODDS OF R1 BEING KILLED = 77%

B. ODDS OF R1 SURVIVING

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As you know, all events MUST complement each other. It's a statics law that all possible events in concordance MUST be a total of 100%.

So, if R1 survives that means that R2 and R3 were killed in the process.

Based on you results : (S=survives, K=killed)

R1(S)+R2(K)+R3(K)=23+23+100=146 % mathematically incorrect. The mistake repeats with other combinations.

lol. My dear friend Zamboe. Where o where is your math professor? Ok, let me show you why this is wrong.

The ONLY thing that needs to add up to 100 is total SURVIVOR percentage. There will ALWAYS total 200% for the percentage for people dying.

I dont know who taught you this, but it is very wrong.

Illustration:

Imaging R2 and R3 have NO GUNS and LOSE THEIR TURNS. R1 gets to fire as long he wants.

New Odds Follows:

A. ODDS OF R1 BEING KILLED = 0%

B. ODDS OF R1 SURVIVING = 100%

C. ODDS OF R2 BEING KILLED = 100%

D. ODDS OF R2 SURVIVING = 0%

E. ODDS OF R3 BEING KILLED = 100%

F. ODDS OF R3 SURVIVING = 0%

So now, if we use your formula, we have R1(S) + R2(K) + R3(K) = 100 + 100 + 100 = 300%

LOL

Your formula is very very bad, zamboe.

So are you now going to say that the odds of R1 being killed if he loses his turns permanently is not 100?

Can you not see now, in light of this obvious blunder you have made, that this formula is wrong?

NEW RIDDLE! EVERYONE TRY TO SOLVE

Imagine 9 people getting thrown into the arctic ocean. Naked and no help in sight.

They are all tied up in chains and have 100 KG lead weights attached to their feet. Additionally, they have been poisoned with Arsenic- 100 Grams. A 10th person was dumped into a nice, warm tropical beach in Guam. He was given a million bucks and a suite at the Marriott hotel overlooking the bay. He has a nice cozy room, lots of food, music, and easy living, fullly healthy with any luxury he wants. 9 people must die, one must live. that is the rules. The game will continue until only one person is left alive.

QUESTION: Barring any unforseen event, and sticking stricly with the details of the story, what is the probability that person #10 will survive longer than the others?

Now, lets compute the odds of survival for these 10 people:

A. ODDS OF 1 BEING KILLED = 100%

B. ODDS OF 1 SURVIVING = 0%

C. ODDS OF 2 BEING KILLED = 100%

D. ODDS OF 2 SURVIVING = 0%

E. ODDS OF 3 BEING KILLED = 100%

F. ODDS OF 3 SURVIVING = 0%

G. ODDS OF 4 BEING KILLED = 100%

H. ODDS OF 4 SURVIVING = 0%

I. ODDS OF 5 BEING KILLED = 100%

J. ODDS OF 5 SURVIVING = 0%

K. ODDS OF 6 BEING KILLED =100%

L. ODDS OF 6 SURVIVING = 0%

M. ODDS OF 7 BEING KILLED = 100%

N. ODDS OF 7 SURVIVING = 0%

O. ODDS OF 8 BEING KILLED = 100%

P. ODDS OF 8 SURVIVING = 0%

Q. ODDS OF 9 BEING KILLED = 100%

R. ODDS OF 9 SURVIVING = 0%

S. ODDS OF 10 BEING KILLED = 0%

T. ODDS OF 10 SURVIVING = 100%

Do we all agree on those odds? Look carefully at them? You agree? Nema? Edric? Acriku? Gob? Anyone else reading this thread. Are these odds accurate?

Ok, now I'm glad we agree on that.

So now, lets quote Zamboe and use his formula and see where it takes us. First, Zamboe's quote:

As you know, all events MUST complement each other. It's a statics law that all possible events in concordance MUST be a total of 100%.

So, if R1 survives that means that R2 and R3 were killed in the process.

well in this case, we know that 10 survives, while all 9 were killed in the process. So, lets add them up and see if it totals 100.

10(S) + 1(K) + 2(K) + 3(K) + 4(K) + 5(K) + 6(K) + 7(K) + 8(K) + 9(K) = 100 + 100 + 100 + 100 + 100 + 100 + 100 + 100 + 100 + 100 = 1000%

Uh oh. We have a problem. Either the odds are wrong, or Zamboe's formula is wrong

so which is it?

Could Zamboe possibly be wrong?

Anyone...?

Anyone?

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The only necessary tool needed is Bayes's Theorem, in it's basics form.

Which is with n=3:

P = Probability

E = Event

P(E1 & E2 & E3) = P(E1) * P(E2/E1) * P(E3/(E1&E2))

For this particular case, GIVEN THAT THE EVENTS ARE SUCESSIVE, NOT AT THE SAME TIME, the formula symplifies to :

P(E1 & E2 & E3 ..... &E(n)) = P(E1) * P(E2) * P(E3)*......* P(En) (Formula 1)

So, getting in the particular problem.

R1 = 30% chance of killing

R2 = 50% chance of killing

R3 = 100% chance of killing

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

I. CASES THAT DON'T MAKE SENSE.

Cases that won't be calculed, because are not rational, since they *minimize* the odds of survival of each robot.

a. R1 hits R2 > R3 hits R1. Doesn't make sense for R1.

b. R1 fails > R2 hits R1 > R3 hits R2. Doesn't make sense for R2.

c. R1 fails > R2 fails > R3 hits R1. Doesn't make sense for R3 since he migth want to hit R2 because it's chances are 20% higher than R1.

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

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

II. CASES THAT MIGHT TAKE TO A ENDLESS SOLUTION

As I look for the max odd for R1 to survive, the endless situations don't provide the max values since the odds get smaller when the R1 and R2 don't hit.

a. R1 hits R3 > R2 fails > R1 fails > R2 fails ..... and so on.

b. R1 fails > R2 hits R3 > R1 fails > R2 fails ..... and so on.

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

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

III. CASES THAT GIVE ODDS FOR R1 TO SURVIVE

R1 can survive in the next cases:

a. R1 hits R3 > R2 fails > R1 hits R2

Based on the formula 1 : (h= hit, m= miss)

P( R1(h) & R2(m) & R3(h) ) = 0.3*0.5*0.3 = 4.5%

b. R1 fails > R2 fails > R3 hits R2 > R1 hits R3

P ( R1(m) & R2(m) & R3(h) & R1(h) ) = 0.7*0.5*1*0.3 = 10.5%

c. R1 fails > R2 hits R3 > R1 hits R2

P ( R1(m) & R2(h) & R1(h) ) = 0.3*0.5*1*0.7 = 10.5%

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

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

IV. REST OF POSSIBLE CASES

a. R1 hits R3 > R2 hits R1 : Not relevant because the problem is asking about R1's odds.

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

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

V. BEST ODDS FOR R1.

Getting the odds in the part III, a) b) and c)

a) 4.5%

b) 10.5%

c) 10.5%

The answer is :

Best thing to do for R1 is miss the first shoot and hit it's next shoot.

Max odd : 10.5%

I hope I've been clear enough.

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well finally. after 20 pages of my posts, i finally get something of substance. very very good indeed. now i can hopefully show you the problem with this.

The only necessary tool needed is Bayes's Theorem, in it's basics form.

Which is with n=3:

P = Probability

E = Event

P(E1 & E2 & E3) = P(E1) * P(E2/E1) * P(E3/(E1&E2))

For this particular case, GIVEN THAT THE EVENTS ARE SUCESSIVE, NOT AT THE SAME TIME, the formula symplifies to :

P(E1 & E2 & E3 ..... &E(n)) = P(E1) * P(E2) * P(E3)*......* P(En) (Formula 1)

100% correct. I am impressed, I seriously thought you had no understanding of this based upon your wierd answers about probabilities having to "add up" to 100.

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

III. CASES THAT GIVE ODDS FOR R1 TO SURVIVE

P( R1(h) & R2(m) & R3(h) ) = 0.3*0.5*0.3 = 4.5%

In this case we will calculate the Odds of R1 FIRING AT R3

I will assume you meant R1, therefore:

P( R1(h) & R2(m) & R1(h))= 0.3*0.5 *.3 = 4.5%

Sounds good. You are absolutely correct.

And what about all those times that R1 firing at and missing R3?

Well as you accurately compute:

P ( R1(m) & R2(m) & R3(h) & R1(h) ) = 0.7*0.5*1*0.3 = 10.5%

And remember this is only after 2 turns with R1.

Consider that R1 fires at R3 and:

P ( R1(m) & R2(m) & R3(h) & R1(m) & R2(m) & R1(h) ) has a probability of 3.675%

This gives R1 a win after 3 turns.

To be accurate, you must total these probabilities up. Remember that R1 can choose 2 possible paths: Aim at R3, or deliberately MISS. If he aims at R3, we MUST add up the probabilities of his misses IN ADDITION to his hits, becasue they are BOTH SUB-TREES BENEATH the action of "R1 firing at R3'

So lets do it:

In the case of R1 FIRING AT R3

Total Odds For R1 to WIN after just 3 turns when he takes aim and fires at R3:

4.5 + 10.5 + 3.675 = 18.675

If you take the limit of this, you will get precisely 3/13 or 23%, just like I said.

TOTAL ODDS FOR R1 TO WIN IF TRIES TO SHOOT R3: 23%

Now, lets examine the other scenario, and see where you went wrong:

b. R1 fails > R2 fails > R3 hits R2 > R1 hits R3

P ( R1(m) & R2(m) & R3(h) & R1(h) ) = 0.7*0.5*1*0.3 = 10.5%

False. You cannot multiply R1(M) in this equation. You are saying that R1(M) = .7 which is not true. R1 is deliberately missing. The probability is 100%. Therefore R1(M) = 1.00. The odds in this decision tree is 100% for a miss- because it is deliberate. The odds are not .7, but 1.00. R1 is not confined to a .7 per M. .7 is the LOWEST number for M which he can raise all the way to 1.00 if he wants. Only when R1 TRIES to hit R3 will he be forced with a .7M.

The Odds that R1 MISSES in his first turn is 1.00.

Your new, corrected odds now read:

P ( R1(m) & R2(m) & R3(h) & R1(h) ) = 1*0.5*1*0.3 = 15%

c. R1 fails > R2 hits R3 > R1 hits R2

P ( R1(m) & R2(h) & R1(h) ) = 0.3*0.5*1*0.7 = 10.5%

again, this is wrong. Same reasons. R1(m) = 1.00

Your new, corrected odds now read:

P ( R1(m) & R2(h) & R1(h) ) = 1 * .5 * .3 = 15%

Lets do it for yet another turn:

P ( R1(m) & R2(h) & R1(m) & R2(m) & R1(h) ) = 1 * .5 * .7 * .5 *.3 = 5.25%

and for another turn = 1 * .5 * .7 * .5 * .7 * .5 * .3 = 1.8375%

So for R1 to win after just 3 turns with this permeatation of deliberately missing his first shot (probability of 100% on the miss), his total odds are now:

15 + 15 + 5.25 + 1.8375 = 37.3625

After only 3 turns. Again, using limits, and taking an infinite amount of turns, this percentage approaches 38%.

TOTAL ODDS FOR R1 TO WIN IF DELIBERATELY MISSES: 38%

Now I requote my original post: (note the line in red- emphasis added)

TOTAL ODDS FOR III: R1 could miss

A. ODDS OF R1 BEING KILLED = (70% + 54%) / 2 = 62%

B. ODDS OF R1 SURVIVING = (30% + 46%) / 2 = 38%

C. ODDS OF R2 BEING KILLED = (100% + 46%) / 2 = 73%

D. ODDS OF R2 SURVIVING = (0% + 54%) / 2 = 27%

E. ODDS OF R3 BEING KILLED = (30% + 100% ) / 2 = 65%

F. ODDS OF R3 SURVIVING = (70% + 0% ) / 2 = 35%

I originally proved it just using basic algebra, but now here I proved it again using Bayes's Theorem. How wonderful they match perfectly. I have proven it manually in other posts. I have even proven it through use of computer iterations in the millions. Surely you wont still argue will you?

I hope I've been clear enough.

yes. You are clear that you did not fully think through this story problem. You really do understand that math as you eloquently shown. I now know you are very good at math. I wasnt too sure, but you have good understanding. However, in this case, you did not think through the logic of the story problem and mis-applied your math. Your math was great. Your application was faulty.

at least you posted your formulas so i still have respect for you.

i hope we can now lay this to rest.

peace.

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