Solution to the Gecko problem

[Skill Leverage]

Solution to the Gecko revolver problem from 'Think you know probability?'

Thought I would post this in a new thread, since it's an entirely different problem, and a brain-tingling one at that... Apologies to those of you who read it before I had realised my mistake; apologies to everyone else for being the world's most boring man.

For those of you not reading the other thread, Mr Gecko posed the following problem:

http://www.trade2win.com/boards/general-trading-chat/53502-think-you-know-probability.html

You have a 6-chamber revolver with multiple firing pins, containing 2 bullets in adjacent chambers. You put the gun to your head, pull the trigger once and get a click. If you are forced to pull the trigger again, should you re-spin the cylinder, or just fire again?

Now, I am going at a pretty fast rate with this, and it's been a while since I've attacked such a problem, so apologies for any glaring mistakes I make. I am going to tackle the situation for 2 firing pins in random positions, and have done so as follows:

At first I believed that for 2 firing pins we would need to know their configuration in order to achieve a solution; I have however come to the conclusion that for 2 firing pins we do not need to know the configuration of firing pins in order to be quantitative, since a) we know the bullets are adjacent, and b) we can visualise each scenario. The configuration does, however, affect our ultimate decision.

OK, here goes... In a 6-barrel gun, with 2 firing pins:

Initial conditions

Number the Firing Pins FP1 and FP2 (with FP1 always being the least-clockwise firing pin), number the chambers 1 - 6 (clockwise), with 2 bullets loaded in adjacent chambers. The firing pins hit 2 unknown chambers, which then rotate clockwise as part of the cylinder.

Calculation

We must first calculate the odds of dying should we re-spin the cylinder:

The odds of hitting a bullet, where there are 2 bullets and 2 firing pins in a 6-barrel gun:

P(hitting a bullet) = P(FP1 having a bullet, FP2 not) + P(FP1 not having a bullet, FP2 having one) + P(both having a bullet) = (2/6 * 4/5) + (4/6 * 2/5) + (2/6 * 1/5) = 0.6

We therefore have a 40% chance of survival, should we re-spin the cylinder.

Multiple Scenarios

Now, if the first pull of the trigger produces a blank, there are 3, rather than 1 ways of you dying - a bullet has moved into FP1's chamber, a bullet has moved into FP2's chamber, or a bullet has moved into both.

We know that two of the empty chambers have rotated out of play, leaving 4 chambers, 2 of which contain bullets which are adjacent. Since the bullets are adjacent, the probability of FP1 and FP2 now striking bullets is zero. The only way of us dying is if the two bullets immediately preceded either Firing Pin. This leaves us with three scenarios, assuming the first pull of the trigger produced a click:

Scenario 1 - The firing pins are adjacent - if the firing pins are adjacent, the only way we can die is if the two bullets were immediately behind FP1 - there cannot be 2 bullets immediately after FP2 however, since by definition there would have had to be a bullet in FP2 the previous pull, hence no click; therefore there is only two combinations out of three possibles (Thanks to Trendie here for correcting me) in which we survive - we have a 66.67% chance of survival should we pull the trigger again, compared with a 40% chance should we re-spin - we therefore pull the trigger again.

Scenario 2 - The pins are one chamber apart -if the pins are one chamber apart, there can be no bullet between the two FPs, leaving a 3-chamber 'half' of the cylinder in which the bullets remain.

There are 2 combinations of 2 adjacent bullets in 3 adjacent chambers, one of which will kill us, one of which won't. This gives us a 50/50 chance of survival should we pull the trigger again; we must, therefore, pull the trigger again.

Scenario 3 - The pins are more than one chamber apart - if the pins are more than one chamber apart, by definition in a 6-chamber cylinder they must be 2 chambers apart, and thus opposite one another.

Since the bullets are adjacent and a click was achieved on the last pull, it is certain death if we pull the trigger a second time - the bullets must have been in either of the 2-chamber gaps between the 2 firing pins, and therefore one must have rotated into a firing position. Therefore, to avoid certain death, we re-spin the cylinder.

Conclusion

In conclusion, when dealing with a gun with 2 randomly-located firing pins and 6 chambers containing 2 adjacent bullets, we must know the configuration of the firing pins in order to determine the correct action:

For all situations other than the firing pins being two chambers apart, we must pull the trigger again. If the pins are indeed two chambers apart, we die unless we re-spin the cylinder.

In either case, our chances of surviving both shots are slim to say the least.

That took waaaay longer than I thought it would; I hope that it's understandable, and indeed correct.

SL

 

[Skill Leverage]

Made a pretty major mistake which, upon correction, completely changed the outcome of this solution. Hopefully what remains is correct; if it's not then, again, apologies to those who actually care.

From here it's possible to work out your best course of action in terms of probability of survival, given that you don't know the config. of the pins... I believe from combination theory there are 15 different combos of 2 firing pins within 6 chambers; 3 of these are two chambers apart (meaning we re-spin), 12 of these are either adjacent or one chamber apart (meaning we pull the trigger).

From here we see that:

100% of the time, re-spinning the cylinder means 40% chance of survival.

80% of the time, pulling the trigger has a 66.67% chance of survival; 20% of the time, pulling the trigger means certain death. This means that by pulling the trigger, we survive 53.33% of the time, meaning that if placed in this situation, we must pull the trigger again to stand the best chance of survival

Thanks again to Trendie for correcting/verifying my analysis

 

[Skill Leverage]

Having just realised that I've spent a decent hour and a half working out a solution to a problem no-one cares about on Valentine's Day, I will now jump out of the nearest window.

 

[MrGecko]

"you must spread some around..."

Excellent. I am going to have to go away and think about this. The thing that strikes me as key is the number of "clicks" and the gap between the bullets... I think understanding the key relationships here is the source of the solution -

i.e.

If we know the bullets are x apart, we need f(x) "clicks" in order to determine something about the original conditions.

We need f "clicks" in order to determine the distance y between the two firing pins.

All assuming the number of Bullets and Chambers are known.

Once those are figured out, you can then make judgements given any scenario... Of course, there are quite a few adaptations you can make to liven things up - e.g. "16 chambers, 6 bullets, 8 "goes" and 4 bullets will kill you"...

Another thing that strikes me is that the Distance between the bullets and the position of the firing pins are infact the same function - or rather, they are substitutes for one another. Have to think about this more.

Good couple of threads Slev!

 

[Skill Leverage]

Cheers mate, glad to know someone got something out of my complete lack of a love life... your points are interesting, will have a think about them after I've watched Batman (the original, just about to start on C4) and had a little cry about being lonely.

SL

 

[MrGecko]

Having just realised that I've spent a decent hour and a half working out a solution to a problem no-one cares about on Valentine's Day, I will now jump out of the nearest window.

LOL

If it makes you feel any better, on Valentines day the most exciting thing that happened to me was noticing that sainsbury's had a "2 for 1" on their microwave Chilli con Carne and Scrumpy Jack.

There is a good puzzle thread here somewhere... I can recall working furiously at a problem FW posted so I could be the first to get the solution (WHICH IS ON A DIFFERNT THREAD). I can remember posting a Logic puzzle about the 3 gods, which I fear may have put the thread to rest... Infact, Slev, the "gameshow door scenario" problem (not wanting to give too much away) would be a good problem to post.

 

[Skill Leverage]

A gameshow consists of three closed doors; behind one of these doors is a car, behind the other two are turds. The contestant does not know where the car is, but the host of the show does.

The contestant picks a door and the host opens one of the remaining doors - one he knows doesn't hide the car. If the contestant has already chosen the correct door, the host is equally likely to open either of the two remaining doors.

After the host has shown a turd behind the door that he opens, the contestant is always given the option to switch doors. Should they take the switch, or stick with their original choice?

 

[MrGecko]

When nobody is in the house, I like to tuck my d!ck in between my legs and walk like a duck

Yes, But Edit out the name of it, otherwise (thanks to google) its a pointless exercise.

Puzzle thread

Answers Thread

Check it out.

 

[Skill Leverage]

Ha ok but now you need to edit your quote of mine otherwise people can still see it

 

[MrGecko]

Done

 

[Skill Leverage]

Ha now I'm the one who has to spread some around... will take a look at those puzzles.

In the meantime bitches, look at the problem mentioned above and tell us what's the best option - taking the swap or sticking with your original choice...

 

[JRP2891]

You need a girlfriend Skill.

 

[JRP2891]

As do i

 

[N Rothschild]

****ing dweebs

 

[Skill Leverage]

dweebs eh... I'll remember that one the next time I'm hanging out the back of yer ma

 

[arabianights]

tldr

 

[Skill Leverage]

You need a girlfriend Skill.

'need' is the wrong word... 'desperately want' would be more appropriate I feel.

 

[everyonerich]

'need' is the wrong word... 'desperately want' would be more appropriate I feel.

hope this help..
UK Escorts: Girl Directory

 

[trendie]

revolver, 2 chambers randomly distributed bullets.

you have a 2 in 6 chance of dying.
first pull of the trigger -click. empty chamber.

you have removed an empty chamber from the scenario.

if you pull the trigger again, the chances are 2 out of the remaining 5 chambers contains a bullet.

if you re-spin, you reset the chances, and have a 2 out of 6 again.

in the above scenario:
pulling the trigger again means a 3/5 = 60% survival rate.
re-spinning gives you a 4/6 = 66.67% survival rate.

 

[new_trader]

This is the way I figure it:

After the first spin you have a 2/6 chance of dying, also, you don’t know you have to pull the trigger twice, but I assume you are told there are two bullets?
I put the gun to my head and pull trigger- “click” - I don’t die.

Now I’m told I must pull the trigger again!

I would think to myself, - “If I spin the cylinder again then my chances of dying are going to be 2/6. But, because I now know that I’ve just fired an empty chamber, the chances of it being the empty one just before one with a live bullet is only 1/6. I’d be better of just pulling the trigger again.”

Am I dead?