Brain Teaser No 3

Thanks Both!

My bedtime reading is hair raisingly exciting! ATB from Fredrik

Paul's logic is (methinks) impeccable, but a cute, short way to come up with the numbers for each still escapes me.

Ian

3am in the morning .... thanks Don!

I get F surviving 52.22% of the time, Tarquin next with 30% and finally poor old Tsunami on 17.77%.

If Don tells me these numbers are right, I'll try to outline how I got to them but it aint pretty so I'll only do it if anyone cares.

Ian

quote:

If Don tells me these numbers are right, I'll try to outline how I got to them but it aint pretty so I'll only do it if anyone cares

Your figures are spot-on Ian. Well done. And Pual's logic is indeed, impeccable.

Let's see how pretty your solution is. I think mine is very satisfying, even if it doesn't look very pretty (I did it two ways - one without a "fault tree"; one with a fault tree)

3.00 am ????????????? OOOhhhh, the joys of this forum! We definitely care Ian, give it a go, but not at 3.00 am again.

Cheers

Don

Given the optimal strategies are optimal,

Round 1. Who gets to face F in the final shootout...

Ta goes first half the time, Ta wins.
Ts goes first half the time, Ts wins 80%, Ta 20%, if Ts misses Ta wins.

So Ta 0.5+0.2*0.5=0.6 of seeing F, Ts 0.5*0.8=0.4

Round 2.

F and Ta, probability 0.6.

F wins 50%, Ta 50%
Ta then has an overall probability of winning of 0.6*0.5=0.3
F gets the same, 0.3.

F and Ts, probability 0.4.
This is the complicated one.
If F hits, he wins, 0.5*0.4.
But if he misses he may also win on a second shot if Ts misses,
0.5*0.4*0.2*0.5=0.2*0.1, or on a third shot, 0.2*0.1*0.1. This could go on forever...

So the total probability of F winning against Ts is,

0.2 + 0.2(0.1+0.1^2+0.1^3....0.1^n)

If we start to write out that infinite sum it gets much easier...

0.1+0.01+0.001+....= 0.11111...

which is exactly 1/9.

So F wins 0.3 + 0.2 + 0.2/9=0.5222...
From earlier Ta wins 0.3
So Ts must win 0.1777...

Paul

Oh bugger I just typed this out to find Paul had beaten me to it ...

Here it is anyhow.


I did it using a fault/decision/outcome tree applying the same logic that Paul describes above.


There are three possibilties for who shoots first so each starting point has a probability of 1/3

(1) Tarquin to shoot first

He takes out Tsunami, then has 50/50 chance of surviving the shot from F. If he does then he takes out F. The out comes of this branch are then that both F. and Tarquin have a (1/3 x 1/2) = 1/6 chance of surviving it and tsunami a zero chance.

(note we'll use this later again to cover the situation when someone else shoots first and misses and Tarquin is to fire next)

(Aside: Also because we'll need it several times later we consider what happens when Tarquin is dead, and F. is to shoot next at Tsunami. As neither of them is 100% accurate it could take many shots before one of them is successful. We calculate the probability of F. surviving each 'round' (sic). At each stage there are different possibilities

Stage 1: F. shoots and has a prob of 1/2 of killing Tsunami, and 1/2 of the truel continuing. Then Tsunami shoots and there is a (1/5 x 1/2) = 1/10 chance of the truel continuing.

Stage 2: now we are back where we began, both alive and F.'s turn to shoot, but now he has a (1/2 x 1/10) = 1/20 chance of finishing the truel. and a 1/20 of it continuing.

This pattern then repeats for ever.

So to find F.'s probability of survival in a straight duel with Tsunami we need to add up

1/2 + 1/20 + 1/200 + 1/2000 + ........ = 5/9

The easiest way to see this is to write it as
0.5*(1+0.1+0.001+0.0001+ ... ) = 0.5*(1.11111111)=0.5*10/9 =5/9.

So if F.'s chances of surviving are 5/9, then Tsunami's must be the remaining 4/9

End of aside)


2 Tsunami to shoot first

He aims at Tarquin.

( 4/5 x 1/3) = 4/15 of the time he hits and then it is F.'s turn in what is now a straight duel with Tsunami. The outcome we just worked out above , so the survival probabilities from this brach are (4/15 x 5/9) = 4/27 for F. and (4/15 x 4/9) =16/135 for Tsunami.

However he misses (1/5 x 1/3) = 1/15 of the time and now all three truellists are still standing. If it is F to fire next he will fire in the ground (see Paul's logic above) then Tarquin will kill Tsunami for sure leaving a straight duel between F and Tarquin, which we considered above. So the survival probabilities here are (1/15 x 1/2) = 1/30 for both Tarquin and F.

3 F. to fire first.

He will fire into the ground. leaving all three men standing.
There is now a 50/50 chance who is to fire next. This means each branch has a total probability of (1/3x 1/2) = 1/6.

If Tarquin fires next he will kill Tsunami and leave again a straight duel with F. so their chances of surviving this scenario are ( 1/6 x 1/2) = 1/12 each.

If Tsunami shoots next he has 1/5 chance of missing - in which case it will be Tarquin to shoot next and the scenario in the line above will play out. but this contributes only (1/5 x 1/12) = 1/60 to the survival probabilities of F. and Tarquin.

If on the other hand he succeeds in hitting Tarquin (1/6 x 4/5) = 2/15, we are back in the case treated in the aside above and the outcome is (2/15 x 4/9) = 8/135 for Tsunami and (2/15 x 5/9) = 2/27 for F.


This exhausts all the possibilities ( and us) .

So adding up all the routes to survive we have for F.

1/6 + 4/27 + 1/30 + 1/12 + 1/60 + 2/27 = 0.522222

For Tarquin,

1/6 +1/30 + 1/12 + 1/60 = 0.3

For Tsunami,

16/135 + 8/135 = 0.1777777

Fortunately these add up to 100%

There must be a better way........

Ian

quote:

There must be a better way........

I don't think there is (well, Paul's is alittle more concise).....so sleep well tonight.

I will post my two solutions later.

Cheers

Don

quote:

Originally posted by Don Atkinson:

I will post my two solutions later.

Cheers

Don

Please, don't - or I will shoot myself - and miss - or not..

quote:

Originally posted by Don Atkinson:

quote:

There must be a better way........

I don't think there is (well, Paul's is alittle more concise).....so sleep well tonight.

I will post my two solutions later.

Cheers

Don

Yep Paul's is a bit better in using the fact right from the start that F will always be in the final duel.

I'm on holiday this week so I'm allowed to stay up late

Ian

quote:

Originally posted by IanGtoo:



Yep Paul's is a bit better in using the fact right from the start that F will always be in the final duel.


Ian

Not a fact, surely; we are talking probablilty?

quote:

Not a fact, surely; we are talking probablilty?

Definitely. It has a probability of 1. ie A "Dead Cert"

Cheers

Don

quote:

Definitely. It has a probability of 1. ie A "Dead Cert"

ie to avoid any possibility/probability of any doubt whatsoever, it really is an absolute certainty that F will always be in the final duel, and what's more, he will, with absolute certainty, have the first shot in that duel.

No doubt about it whatsoever

(in all probability)

Cheers

Don

Given all shoot according to their optimal strategy...

Paul

Given all shoot according to their optimal strategy...

Good point Paul.

That original text did include that concept (phew! - you had me wondering for a moment)

Assume all three adopt their best strategy for survival, and that nobody is killed by a stray shot.....

Cheers

Don

quote:

Nigel Cavendish
quote:
Originally posted by Don Atkinson:

I will post my two solutions later.

Cheers

Don


Please, don't - or I will shoot myself - and miss - or not..

Nigel, I'm pretty sure that you are as good a shot as Tarquin......

Shall I?......... or shall I not?.......

Cheers

Don

Let us have it Don.

http://www.youtube.com/watch?v=4PFWtZIqcKc

Aargh...

Paul

quote:

Originally posted by IanGtoo:
3am in the morning .... thanks Don!

I get F surviving 52.22% of the time, Tarquin next with 30% and finally poor old Tsunami on 17.77%.

Pure bollocks. I would not have entered a dual with Tarquin and fred without A/ loading Tarquin's pistol with blanks B/ filing the firing pin down on F's pistol C/ Wearing a bullet proof vest D/ Having a marksman on a grassy knoll to help if needed and E/ Life policies for T and F naming myself as sole beneficiary.

Long-Division - with a slight difference

A few years ago, Miss Sally - my youngest, came home from school fully excited at having started long-division and eager to show me the problem her teacher had asked them to copy out that morning as an example.

Unfortunately, a mixture rain and wind on the way home had obscured most of the figures young Sally had written down and she was distraught.

Fortunately, we could detect the general outline of the long-division such that we were confidently able to place an X in place of each digit, and, even more delightfully, Sally was able to decipher the 8 in the quotient. In other words we have an eight-digit number, divided by a three digit number, giving a five-digit number with no remainder. The middle digit in the five digit quotient is 8.

A bit of thought soon had the unique, original problem and its solution re-constructed, perhaps you would like to try?

Cheers

Don

quote:

Let us have it Don.

Apologies, though I had better do it on the computer rather than by hand - will try to find time tomorrow night.

Cheers

Don

quote:

Originally posted by Don Atkinson:
Long-Division - with a slight difference

.....
Don

Hmmmmmmm.......

quote:

Hmmmmmmm.......

is that followed by a smile...........or a grimace?

Cheers

Don

10020316/124=80809 seems to match.

Seems to , want to explain your working for us ?

cheers

ian

quote:

10020316/124=80809 seems to match.

It matches perfectly. Superb - you haven't forgotten the basic princples of how to do long division!

Even with the answer, it should still allow Ian (or others) to get the old grey matter swirling around as to HOW to decode the the various elements......so perhaps we could wait before posting explanations?

Cheers

Don

I did forget the principles of long division. That was what I spent the first 15 minutes remembering. Then maybe half an hour of messing about.