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# Russian roulette on steroids

## Question

Russian roulette normally has a single bullet in one of 6 chambers.

The barrel is spun and players take turns pulling the trigger, passing the gun to the other player should they survive.

The game stops when one player dies.

Russian roulette on steroids has bullets in 3 consecutive chambers.

The barrel is spun only once.

You have the choice of being the first or second to shoot.

Which do you choose?

Second.

Player 1 dies on his first turn, chamber 1 has a bullet [3 of 6 cases] AND on his second turn, when chambers 1 and 2 are empty but chamber 3 has a bullet [1 of 6 cases]

The other two cases Player 2 dies, either on his first turn [chamber 1 is empty] or second turn [chambers 1, 2, 3 are empty].

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Second,

50/50 chance the other guy gets it and ends the game. 100% survival in that case.

If he hits a blank you have a better than 50/50 chance of hitting another. I dont have time before work to do the math on this one but it seems logical with three consecutive blank chambers and three consecutive full chambers. Once you hit a blank you can eliminate #2 and #3 bullets being next. You can also eliminate the chamber the other guy just used. So I guess you would by process of elimination have a 2/3 chance of another blank.

Just my 158gr worth.

Peace

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You want to be second, for two reasons:

player one faces 3 empty chambers and 3 armed chambers, making his/her odds 3 in 6 of living:

3:6 * 100% =

.5 or 50%

After player one pulls the trigger, he/she either dies (1:2 chance), thus ending the game and saving your life, or lives, leaving you facing 5 chambers of which only 2 are empty. Thus your odds of surviving, as the second player are

1:2 * 100% + 1:2 * 2:5 =

.5 * 1 + .5 * .4 =

.7 = 70%

A 70% chance of living sounds better than a 50% chance to me.

Standing there watching player one willingly hold a gun to his/her head should give a reality check on how dumb such a thing is, time to think of something to live for, and a good distraction while you slip out the back door and hightail it away from these whack-a-doos

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You want to be second, for two reasons:

player one faces 3 empty chambers and 3 armed chambers, making his/her odds 3 in 6 of living:

3:6 * 100% =

.5 or 50%

After player one pulls the trigger, he/she either dies (1:2 chance), thus ending the game and saving your life, or lives, leaving you facing 5 chambers of which only 2 are empty. Thus your odds of surviving, as the second player are

1:2 * 100% + 1:2 * 2:5 =

.5 * 1 + .5 * .4 =

.7 = 70%

A 70% chance of living sounds better than a 50% chance to me.Standing there watching player one willingly hold a gun to his/her head should give a reality check on how dumb such a thing is, time to think of something to live for, and a good distraction while you slip out the back door and hightail it away from these whack-a-doos

It's 2/3 not 70% cuz player 1 surviving the first shot skews the likelihood of the other outcomes a little bit.

But your approach shows clearly that the scales are tipped, by however much, to player 2's favor.

Nice!Totally agree.

Even to the point that I could not bring myself to describe the game including the phrase "points the gun at his head".

Did Russians have so little to do that this seemed like a sport?

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It's 2/3 not 70% cuz player 1 surviving the first shot skews the likelihood of the other outcomes a little bit.
I didn't pay attention to the stated rules, and reverted to regular roulette (steroid-free). I also did not continue on after both players survive their first turns. My mistake - I was doing this while at work (shhhh!)

Did Russians have so little to do that this seemed like a sport?
No, it was a cleverly disguised form of eugenics designed to weed out those prone to gambling and the extremely gullible.
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Well Bonanova,

I realized recently that I never finished out the math, simple as it is, after the second pull of the trigger. See previous post to see where I am at this point.

player one survived first round.

player two survived first round.

we now know that 2 out of 3 empty chambers are no longer an option. (outta the equation.)

we also know that the second and third full chambers can't precede the first. (outta here)

player one on his second turn once again has a 50/50 chance of survival as there are only two chambers that could be next, one full, one empty.

option #1. If he hits the full chamber we know the game is over and player two wins (might need some new clothes though.)

option #2. If he hits the empty chamber by process of elimination we know that full chamber #1 HAS to be next. The weapon is handed to player #2 and he has three full chambers with which to fight his way out of the room.

in any scenario it seems best to be player #2 as foolish as the game may seem.

Gun control = maintaining your grip

Peace

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