BS: Mathematical Probability Query II

Subject: RE: BS: Mathematical Probability Query II
From: Nigel Parsons
Date: 14 Jun 02 - 12:29 PM

Refresh: no comments yet from Lindsay or Zebedee!

Subject: RE: BS: Mathematical Probability Query II
From: Nigel Parsons
Date: 13 Jun 02 - 05:36 AM

To give a more complete (but not precise) answer to how to approach the other question.
What is the possibility of picking two consecutive numbers in a choice of 5 different random numbers from a selection of 35 consecutive numbers.

In 35 attempts to choose one number, there will be 2 occasions when the number will be 1 or 35, these are the only 2 numbers with only one chance for a sequential number. The chance of picking a second number, and having it sequential with the first is 1 in 17 (if the first no is not 1 or 35) or 1 in 34 (if it is 1 or 35) so the chance of picking a second number to be sequential with the first is( (33/17)+(2/34))/35. i.e. 2/35
With one number you have no chance of having a pair.
The second number has a 2 in 35 chance of being consecutive with the first. i.e. 2 in 35
The third number has a 2 in 35 chance of being consecutive with the first. i.e. 2 in 35
The fourth number has a 2 in 35 chance of being consecutive with the first. i.e. 2 in 35
The fifth number has a 2 in 35 chance of being consecutive with the first. i.e. 2 in 35
Thus you have a 8 in 35 possibility that there is a number chosen sequential with the first choice.
By the same reasoning you have a 6 in 35 chance that the third, fourth or fifth number is sequential to the second.
By the same reasoning you have a 4 in35 chance that the fourth or fifth number is sequential to the third.
By the same reasoning you have a 2 in 35 chance that the fifth number is sequential to the fourth.
Your chances should thus be 20 in 35.

The odds are not, however quite this good, as I have ignored the fact that any sequence of 3, 4, or 5 numbers will have been counted as a pair more than once. Also the selection of a sequence with two pairs of sequential numbers will have been counted as 2 'hits' rather than one. The chances should still work out as better than evens!
The last two reasons for the inaccuracy of my result are the reasons that 'The House' almost invariably wins at 'Crown & Anchor'. the punter assumes that naming one symbol on three identical dice gives him 3 1in6 chances for an even money bet. And the chance for greater profit if two or three dice show his symbol (paying 2:1 or 3:1) But any multiple showings reduce the amount paid back by 'The House', as their are more losers!

Subject: RE: BS: Mathematical Probability Query II
From: Nigel Parsons
Date: 13 Jun 02 - 04:17 AM

Wolfgang: now we're completely together on this, except for a little extra twist after the second shot. My complete answer to the version as set in New Scientist was.

It would seem clear that each duellist needs to 'take out' their most dangerous opponent at any opportunity. By this reasoning, George will shoot first at Ernest and, if successful will then be shot at by Fred. If George is unsuccessful Fred will shoot at Ernest, and if successful will be shot at by George. On this reasoning, Ernest's chance of being killed in the first round is 60% (killed by George) and 30% (75% of the remaining 40% shot by Fred) Thus, it appears that Ernest has only a 10% chance of getting to shoot at anyone.
However, looking at the odds for the 6 possible two man duels we see the chance for tactical shooting by George.
Ernest v Fred (with Ernest shooting first) Fred dies 100% of the time
Ernest v Fred (with Fred shooting first) Ernest dies 75% of the time, otherwise Fred dies on the next shot
George v Ernest (With Ernest shooting first) George dies 100% of the time
George v Ernest (with George shooting first) Ernest dies 60% of the time, otherwise George dies on the next shot
Fred v George (with Fred shooting first) George dies 75 % of the time, otherwise Fred dies 15%(60% of the remaining 25%). with the next round, George dies 7.5% and Fred 1.5% etc. Cumulatively, George's chance of survival is 16.66%
Fred v George (with George shooting first) Fred dies 60% of the time, otherwise George dies 30%(75% * remaining40%). With the next round, Fred dies 6% and George 3%. Cumulatively George's chance of survival is 66.6%.
From the above it is clear that George can massively improve his chances by ensuring that in any two man duel he gets the first shot. He can ensure this improvement to his chances by firing into the ground whilst there are still 3 men standing, and shooting at the remaining duellist once they are down to two. The only possibilities then are that Fred's first shot kills Ernest (75%) and George has a 66.6% survival rate. or Fred fails to kill Ernest, Ernest kills Fred and George has a 60% survival chance. George's overall chance of survival is thus (75%*66.6%)=50% + (25%*60%)=15% =65%

Shooting to miss cannot improve the chances of either of the other duellists. (without collusion). Of course, if George can work this out, then so can Ernest & Fred! As George must wait to shoot again, the other two could discuss these dirty tactics, and realise their own chances of survival.
If they both take their chances to shoot each other, Fred will survive into round two 75% of the time, and of this 75% will survive completely (75% * 33.33%)= 25% of the time.
If Ernest survives Fred's shot (25% of the time,) he can immediately kill Fred and have a 40% chance of survival against George. making Ernest's overall survival chance (25% * 40%)= 10%.
Thus Ernest's chances can be improved from 10% to 25% by the removal of George, with chances of Fred increasing from 25 % to 75%. Thus Ernest & Fred decide that George should die for cheating, Fred shoots into the ground and Ernest shoots George.
This relies on the honour of Ernest, after Fred shoots the ground, Ernest could shoot Fred and improve his own chances to 40%
It takes only a small alteration in the percentage rates in the question to throw up a large number of preferred strategies. At some closer percentages we could find all the duellists firing into the ground in turn!
Despite all my comments, I think the answer required is that George survives 65% of the time.

Subject: RE: BS: Mathematical Probability Query II
From: Wolfgang
Date: 13 Jun 02 - 03:55 AM

O.K., Nigel, I'm confident now I know what you mean (and why I didn't get it first):

On the assumption that they see what George does Fred shoots to miss and Ernest kills George (for sure). Next to shoot is Fred and they both have the same probabilities to survive as before just a bit higher for the elimination of George.

I have two problems with that solution:
(1) Ernest realising what has happened might shoot to miss as well for that increases his probability to survive (someone of the others might get nervous next time) and after some rounds they all find out that never hitting each other is the best possibility.
(2) I always thought that George's best strategy is to shoot to miss but in a way nobody of the other two realises what he does. A 60% shooter should be able to feign intention to hit when he hasn't. In that case, the above solution doesn't work for the other two are not able to spot George's intention.

You could say, however, that merely by thinking it through they should realise that George must have the intention not to hit the first for his own benefit...

Wolfgang

Subject: RE: BS: Mathematical Probability Query II
From: Nigel Parsons
Date: 12 Jun 02 - 05:33 PM

Wolfgang: consider the reactions of the two better shots once they realise that the third is deliberately missing, and how they Might act in unison!
My full response in 1 or 2 days, depending on interest! Nigel

Subject: RE: BS: Mathematical Probability Query II
From: IvanB
Date: 12 Jun 02 - 01:24 PM

Alex, there's a fuller explanation in a link posted above, but, essentially, the strategy is that he is more likely to survive if he leaves the two better shots alive. Their best strategy would then have to be to shoot at one another and the great probability is that one of them would be killed, leaving the poor shooter with only one target for which he has a 50% chance. If he shoots to kill on the first shot and is lucky, then the chance of his death on the second shot is either 80% or 100%, depending on who he killed.

Subject: RE: BS: Mathematical Probability Query II
From: mousethief
Date: 12 Jun 02 - 12:18 PM

Zeb, can you explain for the probability challenged WHY it's better for the worst shot to miss his shot if he goes first?

Thanks,
Alex

Subject: RE: BS: Mathematical Probability Query II
From: Wolfgang
Date: 12 Jun 02 - 11:33 AM

Within the rules? I don't see it. Of course, if all three shoot in the air all of the time all three survive. But that is trivially true of all duells.

Wolfgang

Subject: RE: BS: Mathematical Probability Query II
From: Nigel Parsons
Date: 12 Jun 02 - 07:12 AM

It was never going to be a fair shoot out with their differing abilities, it's just that the 'handicapping' system of the shooting order has overcompensated. But can Fred & Ernest improve their chances ?

Subject: RE: BS: Mathematical Probability Query II
From: GUEST
Date: 12 Jun 02 - 07:10 AM

Well, call it sour grapes, but I think this question is fundamentally flawed. If ALL of the cowboys can work out the odds, then Mr. 100% & Mr. 80% would realise that this is NOT in fact a "fair" shootout, since it allows the one with least talent to have the best chance of survival.

Subject: RE: BS: Mathematical Probability Query II
From: Nigel Parsons
Date: 12 Jun 02 - 06:34 AM

This "Cowboy" question has reappeared in The New Scientist, with Duellists George, Fred & Ernest, with respective accuracies of 60%, 75% & 100%. Shooting at any chosen target, in that order, until only 1 survives.
I know it's a catch question, but has anybody Really thought it through. I got to the answer of George surviving 65% of the time by shooting to miss, but we are assuming that the cowboys can work out the strategies. Can Fred & Ernest improve their chances?
I'll save my answer for now!

Nigel

Subject: RE: BS: Mathematical Probability Query II
From: DMcG
Date: 30 May 02 - 07:17 AM

The is a full analysis - about a chapters worth if I recall! - about exactly this problem in one of Martin Gardner's books. I believe it is called "Mathematical Puzzles and Diversions". Amongst other things, he discusses some of the mistakes people often make trying to solve this puzzle.

Subject: RE: BS: Mathematical Probability Query II
From: Nigel Parsons
Date: 30 May 02 - 07:12 AM

Zebedee asked for a little patience, 18 months?

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 17 Oct 00 - 06:23 PM

Wolfgang and Lindsay,

I've not forgotten you and will both post the answer here and send you personal messages explaining it in a few days time. Work this week has suddenly got horrendous, so I'd ask for a bit of patience.

Thank you,

zeb

Subject: RE: BS: Mathematical Probability Query II
From: GUEST,Murray MacLeod
Date: 17 Oct 00 - 05:45 PM

Another good one, Marion. I found three on first reading, four the next time, five after that and finally six ! (And to think I once thought of taking up proof-reading as a career)

Murray

Subject: RE: BS: Mathematical Probability Query II
From: Lindsay
Date: 17 Oct 00 - 08:58 AM

Wolfgang, I also figured that the odds added rather than multiplied. I ended up doing a horrendus LCD and 66%. Like you, I'm interested in where the smoke and mirrors are.

Subject: RE: BS: Mathematical Probability Query II
From: Wolfgang
Date: 17 Oct 00 - 06:44 AM

I tried, Zeb, but I saw no really easy way to do it (I'm lazy). I only came so far:
I assume you mean drawing without replacement (usual, but not totally necessary in lotteries) and I assume you mean 'at least two consecutive numbers' (for that makes it much more easy).

Then you only have to multiply the five probabilities to get no consecutive number in the first draw (1, obviously), in the second, in the third, in the fourth, in the fifth. And subtract what you get from 1.
My problem were the too many possibilities (cases) to consider. E.g., the second of the above probabilities is 29/31 (for both the number one larger than the first number and the number one smaller than the first are taboo if you do not want to have consecutive numbers). But that is only true if the first draw was not the 1 or the 32. And it gets more complicated later with more different cases (like, e.g., two numbers already drawn that differ by two) to carry on. I did a rough calculation (leaving out all the more difficult cases) to come to a probability for at least two consecutive numbers of more than 50%. But my rough calculation (1 - 29/31 x 26/30 x 23/29 x 20/28 = .54) necessarily leads to a too large probability. And I don't know how far off I was, though I thought not too much and am therefore not completely surprised by your figure above.
Most times however, there is an easy way to do it and I'm looking forward reading about it.

Wolfgang

Subject: RE: BS: Mathematical Probability Query II
From: Marion
Date: 16 Oct 00 - 09:31 PM

Here's one that's not really a math puzzle, but since we delved into psychological phenomena in the first thread, I figured it was fair game.

Count the number of times the letter F appears in the following sentence:

FINISHED FILES ARE THE RESULT OF YEARS

OF SCIENTIFIC STUDY COMBINED WITH THE

EXPERIENCE OF YEARS.

(just giving you some space here so you don't see the answer before solving the problem)

There are 6, but many people only find 3 at first. If you only found 3, try it again. If you still can't find all 6, keep trying until you do. There isn't a trick in the instructions; there really are 6 "F"s in the capitalized sentence.

If anyone really can't find them, I'll post an explanation pointing them out. I find this puzzle intriguing because I found 3 at first, then when I was told there were 6 I went over the sentence very carefully (I thought) and still didn't find the others. I'm curious to know how many you find initially, and if it wasn't all 6, how difficult it was for you to find the rest of them.

Marion

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 16 Oct 00 - 05:42 PM

Eddie,

I don't think you quite understand the problem. As has been shown, in certain circumstances, the best strategy for the 50% shooter is to miss. (He could shoot up in the air if he was worried about his aim being bad enough to accidently hit)

This does not however mean that the 100% or 80% shooters should change their own strategy. The whole point is that each will do the best for their own survival. Shooting the 50% guy because he's 'cheated' and then dying is hardly a great outcome.

Given the assumption that each participant does the best for his own survival, the explanations offered earlier are totally valid.

Zeb

Does no-one want to takle my 'lottery' query?

Subject: RE: BS: Mathematical Probability Query II
From: Wolfgang
Date: 16 Oct 00 - 07:08 AM

Crazy Eddie,
C doesn't violate the rules by aiming wide. In your reasoning, however, the other two are assumed to violate the rules if C goes for his advantage. But if violation of the rules is allowed you can get every kind of response. Note that for neither A nor B it is of an advantage to miss.

Wolfgang

Subject: RE: BS: Mathematical Probability Query II
From: Crazy Eddie
Date: 16 Oct 00 - 06:49 AM

Zebedee & Wolfgang, The idea that the 50% shooter should aim to miss, works on a purely logical basis, but not necessarily on a reasonable one. The 100% and 80% shooters obviously know a bit about shooting. The 50% guy would have to aim really wide to be CERTAIN of missing. Remember, he is a poor shot; If he aims to miss by twelve inches, he could hit by accident! If he aims very wide, the others may figure out what he is doing. Then they shoot him for cheating & have a near-even shoot-out between themselves

Subject: RE: BS: Mathematical Probability Query II
From: Wolfgang
Date: 16 Oct 00 - 05:11 AM

I still think Andy's notation is clearer and easier but with me making two typos in my solution it can become difficult too. Sorry.

C has best chances, B has least

Wolfgang

Subject: RE: BS: Mathematical Probability Query II
From: Wolfgang
Date: 16 Oct 00 - 04:27 AM

When people (like me) ask probability problems you can be sure of two things: (1) the least likely solution at the first glance is correct and (2) there is a slightly unfair trick hidden in the wording of the task.
Let's start with Andy's notation (for clarity and brevity). I'm convinced Andy's solution is correct starting with his assumptions (at least it looks very good to me without doing all the calculations). But there is a way, in which C (Joe) can push his chances of survival well above 50%. To do this, C (Joe) has not to follow Andy's assumption/advice three (If C shoots at B first and hits he faces certain death (from A) so he too must shoot A first), but whenever elected to shoot, as long as both opponents are still alive, he shoots into the air (a shooter with a 50/50 chance of hitting should be able to do that without fail). C only starts shooting with the aim to hit when the first of the two opponets is dead. He'll always be the first to shoot (in the remaining duell) this way.
With A remaining, C has a 50% chance to survive if he hits with the first shot, otherwise it's B surviving. With B remaining opponent, C has an above 50% chance to survive, for even if he doesn't kill with the first aimed shot, he has a 20% chance for a second, and so on.
With this reasoning, B has the worst chance of surviving (other than in Andy's scenario), for the possibility that C shoots first and hits A is taken away by C's reasoning. So without the exact probabilities (I didn't ask for them), C has the highest chance of surviving (above 50%), A has the second highest, and C the lowest.
Yes, it's a bit unfair, I know, but nowhere I had said explicitely that they may not shoot into the air if it is to their advantage.

By the way, with this thinking the truell is reduced to two successing duells, the first is A against B (A's chances higher) and then the winner shoots against C with the 50% shooter C starting first (C's chances higher)

Wolfgang

Subject: RE: BS: Mathematical Probability Query II
From: Lindsay
Date: 15 Oct 00 - 03:00 PM

Okay, are we talking about 66%? If so, I think I know where I went wrong.

BTW, feel free to email directly if you'd like. I've always enjoyed math, but the older I get the harder it is to remember the byways. This exercise is very refreshing.

Lindsay (as in Pvt Michael Lindsay, 42nd Regiment of Foote, circa 1758)

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 15 Oct 00 - 02:46 PM

I guess we misunderstood. Regarding the 5/32 consecutive numbers, the answer is actually over 50%.

I'll post the reasoning in a day or so.

Zeb

Subject: RE: BS: Mathematical Probability Query II
From: Lindsay
Date: 15 Oct 00 - 02:42 PM

My turn to apologize. I was not offended by your previous missive (though I did read the entire thread), I simply don't like to look at the solution until I am done playing with the puzzle.

My last message was my answer to the question of "5 out of 32, with two consecutive numbers". I figured the first number was a gimme, but each successive number would require figuring two options of each previous number, i.e. one higher and one lower.

Hou did a div?

Lindsay (If I walk the walk, I should talk the talk) :o)

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 15 Oct 00 - 02:29 PM

Sorry Lindsay for being rude to you. I apologise.

Zeb

Subject: RE: BS: Mathematical Probability Query II
From: Lindsay
Date: 15 Oct 00 - 02:26 PM

16 chances out of 31,315? I did not list my computations. If I'm right I don't want to spoil it for anyone else. If I'm wrong, I can blame it on a computational error.

Lindsay

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 15 Oct 00 - 01:00 PM

Lindsay, click on the link that I provided earlier with the solution and all should be clear.

What is the probability of someone posting here without reading the entire thread or thinking about it a bit?

Very high

Subject: RE: BS: Mathematical Probability Query II
From: Lindsay
Date: 15 Oct 00 - 12:54 PM

While, the probabilites listed by AndyG agree with mine, I cannot figure how they are modified by the odds of who gets the first shot. Obviously, if A gets the first shot, then his odds of surviving increase since he will shoot B first. If he is not killed by C, then C will die in the next round.

So how do we modify A's survival odds of 24% in light of the fact that he has a 33.3% chance of shooting first? or a 66.6% of not shooting first?

The same question applies to B and C. I believe the percentages listed by AndyG assume (no slight intended, pls correct me if I am wrong) B and C shoot before A.

Terry W.

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 15 Oct 00 - 12:50 PM

In the hope of getting this thread back to mathematical puzzles (which lots of us enjoy) and was the reason for the thread existing at all, I'll restate my question:

In a lottery where 5 numbers are drawn out of a pool of 32, what is the probability that two consecutive numbers will appear in the winning selection?

Zeb

Subject: RE: BS: Mathematical Probability Query II
From: Sourdough
Date: 15 Oct 00 - 12:32 PM

Amos and Mary in KY

I think the Greeks noticed this problem when they gavre the concept the name, hubris.

++++++++++

On a more down to beach level, I watched the summary propgram about "Survivor" and saw that this was basically the strategy used in that "game". Whenever possible, players ganged up on players that they saw as the greatest threats and removed them from competition. I found it an illuminating process broadening my understanding of why dog eat dog competition does not always lead to the "survival of the fittest". I think that if I were going to be on a desert island and my survival was at state, I would not have chosen Richard as the person I would want to entrust my own survival to.

For those outside the US who are not familiar with it, "Survivor" was a television entertainment/reality series in whoihc a group of people were "stranded" on a desert island for a matter of weeks. Each day they got to choose who they wanted to bounce off the island until there was just one remaining Survivor and he got the 1,000,000 dollar prize.

Sourdough

Subject: RE: BS: Mathematical Probability Query II
From: Mary in Kentucky
Date: 15 Oct 00 - 11:09 AM

One of my favorite quotes from a marine, "If your're out front, you're gonna get shot at."

Subject: RE: BS: Mathematical Probability Query II
From: Amos
Date: 15 Oct 00 - 09:50 AM

The psychological insight that you become a greater threat, and therefore a more likely target, the better shot you are is interesting. But I would hate to see it misused and applied in general to competitive behaviours (I think) -- "Don't be too good because the better you get, the more of a threat you are to others, and therefore the more likely they are to target you...".... makes pretty piss-poor general advice.

A

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 15 Oct 00 - 06:38 AM

Eddie,

If you click on the explanation link that I posted a few messages back, you'll see that under certain circumstances, it's better for (in our case) Joe to deliberatly miss his shot.

Zeb

Subject: RE: BS: Mathematical Probability Query II
From: Crazy Eddie
Date: 15 Oct 00 - 06:13 AM

AAARGH! I used to think I was pretty good at this kind of thing!
Over the weekend I've had time to think about these, and I realise that I got the cards and the cups totally WRONG!!!!
(Sound of head repeatedly hitting table)!
I like the sound of the cowboy shoot-out above (I mean the puzzle aspect not the violence) but no time to study it now.
However, I agree that everyone should try for the best shot (greatest threat to self) first.

There's a dart game called "Killer". Everyone is given a number. Every time your number is hit you lose a life (usually everyone starts with 3 lives). Where players know each other, the poorer players will "gang up" on the good player. Get rid of him, & the poor players then fight it out among themselves.
Question is, when do you change sides?????

I have seen people get really upset in this game either because (A) << Everyone is picking on ME>> OR << I helped you, now you try to kill me>>

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 15 Oct 00 - 04:37 AM

OK, here's another initially suprising one:

In a lottery where 5 numbers are drawn out of a pool of 32, what is the probability that two consecutive numbers will appear in the winning selection?

Zeb

Subject: RE: BS: Mathematical Probability Query II
From: GUEST,Dave (the ancient mariner at work)
Date: 14 Oct 00 - 07:28 PM

Imagine the world reduced to a population of 100, with all the existing ratio's remaining the same it would look like this.
57 Asians
21 Europeans
14 from N.& S. America
8 Africans.

70 Non White, 30 White, 70 Non Christian, 30 Christian, 50% of the wealth would belong to 6 people all US citizens. 70 would be illiterate; 50 would suffer malnutrition; 80 would live in sub standard housing and only one would have a college education. When one talks about survival, how many would know how to use the gun? how many would be strong enough to stand and pull the trigger? yours, Aye. Dave

Subject: RE: BS: Mathematical Probability Query II
From: Zebedee
Date: 14 Oct 00 - 01:18 PM

Joe has the best chance of survival, and Jim has the worst.

An explanation of the answer (using different names) can be found here

Zeb

Subject: RE: BS: Mathematical Probability Query II
From: Jon Freeman
Date: 14 Oct 00 - 09:26 AM

Wolfgang, I am sitting silent because I have no idea where to begin tackling this problem. I'll keep looking - please supply the answer at some point.

Jon

Subject: RE: BS: Mathematical Probability Query II
From: GUEST,Stavros
Date: 13 Oct 00 - 03:39 PM

if jim goes first:

jack is dead, and jim has a 50/50 chance of survival in the first round. Joe dies on the first shot of the second round (if he doesn't kill jim first)

Subject: RE: BS: Mathematical Probability Query II
From: Mary in Kentucky
Date: 13 Oct 00 - 03:39 PM

I agree with Andy. Provided they can choose who to knock off, and the person allowed to shoot each time is determined by chance, Joe has the best chances of survival. Intuitively, Jim and Jack will knock each other off if given the chance. Jack is a better shot than Joe, but not that much better.

Subject: RE: BS: Mathematical Probability Query II
From: AndyG
Date: 13 Oct 00 - 03:11 PM

I've renamed my cowboys (for clarity) and made the assumption (not clear) that each man is free to choose his own targets,
A has 100% accuracy
B has 80% accuracy
C has 50% accuracy

If that is so then:
A must target B before C to maximise his chance of survival
B must target A first or face certain death (as a consequence of A's targeting criteria)
If C shoots at B first and hits he faces certain death (from A) so he too must shoot A first.

Given that I estimate that after 2 rounds of fire,
A has ~24% chance of survival which is fixed
B has ~29% chance of survival which is slowly increasing
C has 44% chance of survival which is slowly decreasing
and, at this point there's a 2.5% chance that both B & C are still alive.

AndyG

Subject: RE: BS: Mathematical Probability Query II
From: Mary in Kentucky
Date: 13 Oct 00 - 02:50 PM

Obviously, nobody wants Jim shooting at them! Question: you determine the first shooter by chance, then do the others go in turn clockwise or counterclockwise, or is the second turn determined by chance again? With my luck, I'd never get a chance to shoot!

Subject: RE: BS: Mathematical Probability Query II
From: MMario
Date: 13 Oct 00 - 12:03 PM

too many variables! does not compute! page fault error! Your program has caused an illegal action in kernel!

Subject: RE: BS: Mathematical Probability Query II
From: Mary in Kentucky
Date: 13 Oct 00 - 12:00 PM

lunchtime!...later....

Subject: Mathematical Probability Query II
From: Wolfgang
Date: 13 Oct 00 - 05:14 AM

a foolow (a typo I leave in for it has a kind of significance for the first problem) up to this thread.

Jim, Jack, and Joe are three cowboys who have come to the conclusion that the earth is not big enough for all of them. Two of them have to die. But they want to do it in a fair shootout with all three of them placed in a kind of triangle (so one shot can only kill one of the opponent for the other one is located at a different position), you might want to call what follows a triell. They let chance decide who gets the first shot (no interference, no simultaneous shooting; the second in line just has to wait patiently), who gets the second and the third. If after the first three shots, there is more than one surviving, they go on shooting in the same order as in the first round. Jim has a 100% probability of killing an opponent with a single shot, Jack has an 80% probability of killing, Joe has only a 50% probability of killing.

Who of the has the highest probability of surviving and who has the lowest (assuming that hey are good in doing probabilities, know about their respective hitting probabilities and do the best they can for their own survival)?

Wolfgang