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johan
#1 Posted : Monday, April 04, 2011 1:24:46 PM
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Assuming both players take turns what is the probability the player who goes first will lose at Russian roulette using a gun with six chambers?
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Slinger17
#2 Posted : Wednesday, April 06, 2011 11:16:30 AM
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50%. The first person loses if the bullet is in chambers 1,3, or 5. 3/6 = 50%
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bobbym
#3 Posted : Thursday, April 07, 2011 2:29:25 AM
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Hi;

If you tree this process you will see that going first is a severe disadvantage.

 

\left(\frac{1}{6}\right) + \left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^2+\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^4 + \ldots +

Summing the series to infinity.

\sum _{k=0}^{\infty } \frac{1}{6} \left(\frac{5}{6}\right)^{2 k} = \frac{6}{11}

So the probability of the first player being killed is 6 / 11. Make sure you go second.

This assumes that the game is they spin the chamber before each shot.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Success is the ability to go from one failure to another with no loss of enthusiasm.
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Slinger17
#4 Posted : Thursday, April 07, 2011 10:04:17 AM
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I was assuming that they spun once and took turns from there. It's been a while since I've played Russian Roulette
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mohit
#5 Posted : Thursday, April 07, 2011 8:30:54 PM
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i think 6/11 is the correct answer...
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mohit
#6 Posted : Thursday, April 07, 2011 9:26:19 PM
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lemme xplain it in my way..m assuming tht there is only one bullet in the chamber and u spin the chamber after everytym u pull the trigger...let the probability of person losing the game be p which can be done only in one way nw if he doesnt lose then the probability will be 1-p caz nw the second person has probability "p" of losing the game nd this can b done when the first person shots an empty bullet means in 5 ways so total probability will be p = 1/6 + 5(1-p)/6...solving it for p will give u the answer to be p = 6/11...
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