Is this cool with you?

gAr · 2011-04-08 04:27:42

Wrong answer? I can't digest that!

Perhaps both are right and something took care of the many possible transitions. Maybe the errors got canceled in the markov chain!

Do you know of anyone with expertise in Markov chain? We can get so much information with a markov chain, that's why it's hard for me to let it go!

bobbym · 2011-04-08 04:32:42

Hi gAr;

Until you came here I was the only user of Markov chains I knew.

Whether the answers got cancelled or not, I do not know. But the fact that mine does not consider your possibility at all makes me think that maybe it does not.

We could try this:

1) A5 fires and kills B, game over.

2) A5 fires and misses so control passes to B5.

3) A5 has an empty chamber so he stays in A5. He spins the cylinder and repeats 3) until 1 or 2 occurs.

This complicates the analysis. Or does it? Maybe not!

gAr · 2011-04-08 04:41:20

Okay.

Yes, extra transitions complicates the analysis. I meant the same in #469.
Anyway, it's surprising that we got the same answer!

gAr · 2011-04-08 04:43:20

Hmmm, "Or does it? Maybe not!" - you added this when I was typing, you got a breakthrough?

bobbym · 2011-04-08 04:48:19

Hi gAr;

It seems that the way we solved the problem is like that. Blank cylinders do not count. Is it still correct to say 5 / 12 and 7 / 12 when we did that. The fact that A5 or any other state could stay in that state forever certainly affect the expected number of shots. But does it affect the probabilities?

gAr · 2011-04-08 05:21:29

Hi bobbym,

Looks like you have a point. But I'm not very sure.
In the state diagram, it only summarizes the probability of hitting the target and probability of not hitting the target.
Can it be some kind of transformation?!

bobbym · 2011-04-08 06:14:26

Hi gAr;

Why not just count the empty chamber as a miss. Then the empty chamber is absorbed into the 7 / 12. That makes the problem solvable.

gAr · 2011-04-08 14:35:02

Hi bobbym,

I still get confused how it solves the same problem!

bobbym · 2011-04-08 14:57:26

Hi gAr;

Say A5 for instance, He has 5 shots for 5 / 6 * 1 / 2 = 5 / 12 and B is dead as you know. Now everything else including the empty chamber counts as a miss so it now B's turn with 5 shots in his gun. The 5 / 12 is definitely his chances of hitting B, so 7 / 12 has to be everything else. If B is not dead the it his his turn and the game continues. Does not really matter whether it is an empty chamber or a live round that misses, B is still not dead ( it is not in the 5 / 12 ). This is the way it is starting to make sense to me.

gAr · 2011-04-08 15:23:58

Hmmm, still not very sure about the states, like how movement of A5->B5->A4 is justified when chamber is empty.
You may be tired explaining to me!

bobbym · 2011-04-08 16:11:42

Hi gAr;

The only thing I can think of is that we know for sure what the chances of hitting B is for the case of A5. It is definitely 5 / 12. The complement is everything else, we do not have to figure it out. There are only two results A kills B or it is B's turn. We are sure of the chances of B being dead, if he is not dead it is his turn.

If I were doing this for real and my opponent fired and he missed I would demand my turn. If my opponent squeezed the trigger and we just heard a click ( empty chamber ) I would laugh and demand my turn. I am either dead or it is my turn.

gAr · 2011-04-08 16:28:26

Hi bobbym,

Okay, I may need some time to understand.

I remembered seeing this question: http://www.mathisfunforum.com/viewtopic.php?id=12832&p=3 - #52
It is somewhat similar, except that adversary shoots back!

Can the problem we are discussing be done as a CCP too?

bobbym · 2011-04-08 16:41:13

Hi gAr;

I do not think that it can be modelled as a CCP. CCP is for expected number of throws or cards or something like that. Although the markov chain will get the expected value of this problem too, the ducks as you say do not shoot back. Also the duck problem can not end prematurely as this one does.

gAr · 2011-04-08 16:51:06

Okay, It would have been difficult if ducks shot back!

Anyway, have you tried to find the expected number of times the trigger is pulled before ending the game?

bobbym · 2011-04-08 17:01:59

Hi gAr;

That is a different problem. We would now have to consider an empty chamber as separate. There would be 3 options now

1) pull the trigger and B is dead
2) pull the trigger and the shot misses.
3) pull the trigger and it is empty.

This is because we need to count the pulls of the trigger not a probability, but an expected number. Now we have to make a decision as to what to do for each type of trigger pull. Where as before we come lump 2 and 3 together.

gAr · 2011-04-08 17:11:51

Oh, I see. Now it's difficult!

bobbym · 2011-04-08 17:14:22

Do you see now what is wrong with the way we did the other problem?

Also my arguments in #486 and #488 are now wrong, do you see why?

gAr · 2011-04-08 17:20:29

Hmmm, not sure, I cannot think of a reason beyond my #487.

bobbym · 2011-04-08 17:38:30

It is possible that the way we did the series solution and the markov chain solution may have gotten the correct probabilities. It is also possible that they did not. Also they definitely would not get the expected number of trigger pulls right. Here is why:

Using our familiar A5, A5 pulls the trigger and nothing happens ( empty chamber ) we took and moved it to B5 ( yes we did, it got lumped in with the misses! ). Now supposing B5 misses we move to A4. That is incorrect. A4 is not just a letter and a number. It means A has 4 bullets left. But that is wrong he has 5! He never fired his shot but he lost a bullet anyway!

For your problem there must be 3 conditions:

A shoots and kills B, game over!
A shoots and loses a bullet and misses now it is B's turn.
A pulls the trigger and nothing happens (empty chamber). We cannot move to B5. We must stay in A5 until an actual bullet is fired. That is what A5 -> A4 means!

gAr · 2011-04-08 17:53:28

That's exactly what I tried to tell in #463, #469 and #487, maybe I wasn't clear enough!

I have some confidence that following #467 will be right, which would contain 30 states out of which 3 are absorbing.
I calculated the probability of transitions for 3 states, would take much time I presume.

bobbym · 2011-04-08 18:03:26

Hi gAr;

That's exactly what I tried to tell in #463, #469 and #487, maybe I wasn't clear enough!

Sorry, I did not follow what you were saying.

I would just keep the 15 x 15 and fill in for A5 box with 1 / 6. Meaning 1 / 6 of the time you stay in A5 and do it again. the probability of going to B5 is now 1 - 1 / 6 - 5 / 12. Each of the others would be adjusted appropriately.

gAr · 2011-04-08 19:04:30

Hi bobbym,

It's ok.
Wouldn't filling A5 box with 1/6 mean that A gets the chance to pull the trigger until the bullet goes off?

bobbym · 2011-04-08 19:20:54

Hi gAr;

Yes, that is exactly what we want. He has 5 bullets, the chances of falling on an empty chamber is 1 / 6. His number of bullets cannot change until he fires one. Therefore he can not come out of A5 until that happens.

gAr · 2011-04-08 19:24:38

Hi bobbym,

I'm telling it may not be correct.
After A pulls the trigger, B should get the turn. If it remains in the same state, it means A keeps on pulling the trigger till the bullet goes off.

bobbym · 2011-04-08 19:34:09

But if you pass the turn to B without a bullet going off then he will pass the turn to A4 when he does not kill A, which is incorrect. A4 means A has 4 bullets that can only happen if he fires one.

Math Is Fun Forum

#476 2011-04-08 04:27:42

Re: Is this cool with you?

#477 2011-04-08 04:32:42

Re: Is this cool with you?

#478 2011-04-08 04:41:20

Re: Is this cool with you?

#479 2011-04-08 04:43:20

Re: Is this cool with you?

#480 2011-04-08 04:48:19

Re: Is this cool with you?

#481 2011-04-08 05:21:29

Re: Is this cool with you?

#482 2011-04-08 06:14:26

Re: Is this cool with you?

#483 2011-04-08 14:35:02

Re: Is this cool with you?

#484 2011-04-08 14:57:26

Re: Is this cool with you?

#485 2011-04-08 15:23:58

Re: Is this cool with you?

#486 2011-04-08 16:11:42

Re: Is this cool with you?

#487 2011-04-08 16:28:26

Re: Is this cool with you?

#488 2011-04-08 16:41:13

Re: Is this cool with you?

#489 2011-04-08 16:51:06

Re: Is this cool with you?

#490 2011-04-08 17:01:59

Re: Is this cool with you?

#491 2011-04-08 17:11:51

Re: Is this cool with you?

#492 2011-04-08 17:14:22

Re: Is this cool with you?

#493 2011-04-08 17:20:29

Re: Is this cool with you?

#494 2011-04-08 17:38:30

Re: Is this cool with you?

#495 2011-04-08 17:53:28

Re: Is this cool with you?

#496 2011-04-08 18:03:26

Re: Is this cool with you?

#497 2011-04-08 19:04:30

Re: Is this cool with you?

#498 2011-04-08 19:20:54

Re: Is this cool with you?

#499 2011-04-08 19:24:38

Re: Is this cool with you?

#500 2011-04-08 19:34:09

Re: Is this cool with you?

Board footer