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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. If you are Mr. Black, where should you shoot first for the highest chance of survival?

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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Note:

Duel = Two People (a duo)

Truel = Three People (a trio)

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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

I would shoot at Mr. White because if you shot Mr.Gray and killed him Mr. White would go next and always hits the person so you would be dead, but if you shoot Mr.White and hopefully kill him, you go up against Mr. gray who only hits it 2/3 of the time so he might miss giving Mr. Black another shot against Mr. White.

*Last edited by im really bored (2005-07-19 18:18:18)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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There is also the possibility of shooting in the air.

Mr Grey would then most likely try to shoot Mr White. If he misses, Mr White would probably get him, and then you would go for Mr White.

Assuming Mr Grey and Mr White do as suggested, the "shooting in air" probabilities are:

Possible outcomes:

Grey Kills White (2/3)

then ... Black Kills Grey (1/3)

or ... Black Misses Grey (2/3)

... then ... Grey Kills Black (2/3)

... or ... Grey misses Black (1/3)

... ... then ... Black Kills Grey (1/3)

... ... or ... Black Misses Grey (2/3)

etc !!!

Grey Misses White (1/3)

then ... White Kills Grey (fer sure!)

then ... Black Kills White (1/3)

or ... Black Misses White (2/3)

... then ... White Kills Black (fer sure!)

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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

In my idea it would put Mr.black and Mr.Gray against each other in the end, that is best possible senario for mine, instead of Mr. black up against Mr. White as the last two to where if you miss Mr. White then you die next shot, but if its upagainst Mr. Gray as in mine, if you miss then there is a chance he might miss too giving you another shot.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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I am still stuck working out the probabilites!

But I am curious to see if the strange idea of shooting in the air has any merit ... or not

I will try to add up the probabilites just to get an estimate:

Grey Kills White (2/3)

then ... Black Kills Grey (1/3) OVER with 2/3 * 1/3 Probability

or ... Black Misses Grey (2/3)

... then ... Grey Kills Black (2/3) OVER with 2/3 * 1/3 * 2/3 Probability

... or ... Grey Misses Black (1/3)

... ... then ... Black Kills Grey (1/3) OVER with 2/3 * 1/3 * 2/3 * 1/3 Probability

... ... or ... Black Misses Grey (2/3)

etc !!!

So the chances of surviving after Grey kills White are (2/3 * 1/3 + 2/3 * 1/3 * 2/3 * 1/3 + ...)

Which equals 2/9 + (2/9)² + (2/9)³ + ..., ∼ 2/7 (or = 2/7?)

Grey Misses White (1/3)

then ... White Kills Grey (fer sure!)

then ... Black Kills White (1/3) OVER with 1/3 * 1/3 Probability

or ... Black Misses White (2/3)

... then ... White Kills Black (fer sure!) OVER with 1/3 * 2/3 Probability

So the chances of surviving after Grey misses White are 1/3 * 1/3 = 1/9

So I figure the chances of surviving by shooting in the air are 2/7 + 1/9 pr about 39.7%

(But I just hammered this answer out on the keyboard, and may have made a mistook)

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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

wow well thats alot to take in at once, I would try and verify mine like you did yours but im just too tired right now.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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Cool, look forward to it !

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

MathsIsFun wrote:

So the chances of surviving after Grey kills White are (2/3 * 1/3 + 2/3 * 1/3 * 2/3 * 1/3 + ...)

Which equals 2/9 + (2/9)² + (2/9)³ + ..., ∼ 2/7 (or = 2/7?)

After Grey kills White, Black would shoot at Grey, having a 1/3 chance of success.

If Grey survives, he would shoot at Black, having a 2/3 chance of success, but as he also has a 2/3 chance of staying alive, he actually has a 4/9 chance of success. If he also misses, then they start again as if none of that happened (assuming they have infinite bullets!) This means that the chances of them succeeding this time are a smaller ratio of the existing chances and if they both miss again, then the chances of them succeeding in the third round will be a smaller ratio still and so on.

What this means is that we can simply multiply up the original successes (1/3 and 4/9) by the same thing so that they add to give 1. Adding them now gives 7/9, so we must multiply them by 9/7 to make them add to 1. We are only interested in Black's chance of success, which we can now work out as 1/3*7/9=3/7. However, there was only a 2/3 chance of Grey killing White originally, so the chance is actually 2/7, as MathsIsFun guessed. That was just to confirm to everyone that he was right.

To im really bored, you only need to work out the chance of success if Black kills White, because if he misses then it will be as if he has shot into the air. So, if the chance of Black winning assuming that White dies is higher than 2/7+1/9=25/63, then shooting at White is the better strategy.

Incidentally, I just realised that aiming at Grey is just as bad as aiming at himself!

Why did the vector cross the road?

It wanted to be normal.

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