Russian Roulette: Survival & Hit Probability

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In summary, the game of Russian roulette involves spinning a revolver cylinder with a single round and placing it against one's head before pulling the trigger. The chances of getting hit in each trial are 1/6, and the survival probability decreases with each trial. The "Per Trial Hit" probability is lower than 1/6 because it takes into account the fact that the participant must survive the previous trials in order to reach that specific trial.
  • #1
musicgold
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Hi,

My question is about Russian roulette, a lethal game of chance. In brief, the participant place a single round in a revolver cylinder (which has 6 chambers), spins the cylinder, place the muzzle against his head and pull the trigger. For detailed information please follow this link - http://en.wikipedia.org/wiki/Russian_roulette" [Broken].

The following table describes the chances of getting hit if the game is conducted for 6 trials. I understand the survival probability column, which is calculated based on (1/6) ^ number of trial.

PHP:
Trial 	Survival probability	           Per trial hit probability
1	                      83.3%	           16.7%
2	                      69.4%	           13.9%
3	                      57.9%	           11.6%
4	                      48.2%	            9.6%
5	                      40.2%	            8.0%
6	                      33.5%	            6.7%

What I don’t understand is the Per Trial Hit probability. For example, the number is 13.9% for the second trial. It is the chance that the participant will get hit after the first trial and before the end of the second trial, i.e. 83.3% - 69.4% = 13.9%. I am trying to understand why the chance of getting hit between two trials is less than the chance of getting hit in every trial, i.e. 1/6 = 16.7%.

Thanks,

MG
 
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  • #2
Because it is considering the probability that they are already dead from an earlier round?
 
  • #3
Consider that the first person has 1/6 of a chance of getting hit. Then we could look at 5/6 chance for survival and the second person has the position: (5/6)(1/5) since there are only 5 more chanbers giving 5/6*(1/5) = 1/6. The next case would be 5/6*4/5*1/4 = 1/6. This whole pattern could go on in exactly the same way giving 1/6 of a chance for each player. This being some kind of fatalistic situation.

HOWEVER, it does not work that way in this problem, and the key phrase is "SPINS THE CYLINDER." Thus in the second case, there is still 1/6* of a chance that he might get shot, and the math is: 5/6*1/6 =5/36.

* If each participant spins the cylinder, I doubt is exactly a random event. It would depend on how hard or easy the cylinder swings, or whether it has sticking points, etc. (And since the one chamber with the bullet is heavier than the others,it would likely wind up near the bottom of the gun and rarely be in the topmost position--unless the gun was spun upside down. I think I have some vague memory of someone observing that.)
 
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  • #4
Thanks folks.

Sorry about the confusion. I did not make it clear that I am assuming that only one person is doing that experiment (?) . Let us say, this is a hostage situation, and the hostage is given 6 trials of this game. In that case, I don't understand how the 'Per trial hit' probability is lower than 1/6.
 
  • #5
Assuming you don't spin it between trials shouldn't the probability go up?
First go you have a 1/6 chance of 'winning'
Next go you know it wasn't in the first chamber so it is in one of the next 5 so chance=1/5.
Then 1/4,1/3,1/2,1!

(on the other hand the apriori Bayesian probability of me getting a probability question correct is low!)
 
  • #6
mgb_phys,

Thanks.

mgb_phys said:
Next go you know it wasn't in the first chamber so it is in one of the next 5 so chance=1/5.

I am not sure if this assumption is correct. If you see the survival probability numbers in the above table, the chance of getting hit in each trial is the same, 1/6. The person has to spin the cylinder after every trial, provided he did not get hit in the last trial.
 
  • #7
the assumption above is obviously that the person spins the chamber every time. the first column is the probability of surviving that many rounds, the second column is the probability of getting shot after surviving all of the previous rounds.

say for example you were forced to play the game three times. the probability of surviving is the first column, or 57.9%, or (5/6)(5/6)(5/6). the probability that you will get shot on the third try is 11.6%, or (5/6)(5/6)(1/6)
 
  • #8
rbeale98 said:
the probability that you will get shot on the third try is 11.6%, or (5/6)(5/6)(1/6)
Exactly. I am trying to understand how this chance has become 11.6%, if each time I am facing a 1/6 (16.7%) chance of getting hit.
Thanks.
 
  • #9
it is the probability you will get shot on the third try before the first game even started. so getting shot on the third try is dependent upon the fact that you survived the first and second shot. same is true for survival, surviving the third shot is dependent upon the fact that you survive the first and second, hence (5/6)(5/6)(5/6).
 
  • #10
think about if 100 people stood in a line and played the game. the probability of each person getting shot diminishes as you go down the line because it is assumed that the game ends once someone gets shot
 
  • #11
rbeale98,

Thanks lot. Let me see if I have got this right.

1.The probability figures in the second column are sort of long term average numbers. If the same experiment is carried out on, let us say, 1000 statues, with 6 trials per statue, then the chance of getting a statue hit in a 3rd trial is 11.6%.

2. The reason there is only 6.7% chance of getting hit during the 6th trial (as per the second column) is because the player has very small chance of reaching the 6th trial.


Is my understanding correct?

Thanks again.

MG.
 
  • #12
you are correct.

all probabilities are long term average numbers. the second column basically says if there were 100 people lining up to play this game, you want to be as far away from the first person as possible. the first person has the highest probability of getting shot.
 
  • #13
rbeale98,

Thanks a lot. Appreciate your help.

MG.
 
  • #14
rebeal98: it is the probability you will get shot on the third try before the first game even started. so getting shot on the third try is dependent upon the fact that you survived the first and second shot. same is true for survival, surviving the third shot is
dependent upon the fact that you survive the first and second, hence (5/6)(5/6)(5/6).


To recapitulate, you have to ask what is happening in the problem to get the math results shown. In case 1, we have 5/6 chance to live and 1/6 chance to die. This checks with the chart. By the second case, we are to consider the 5/6 chance for life as being carried over, since, if the 1/6 case occurs THE GAME IS OVER.

So for case two, the figures are 5/6*1/6 to die and (5/6)^2 to live, since cylinder is rotated and there is 1/6 of a chance for death. By the third case it is (5/6)^2(5/6) = (5/6)^3 to live and (5/6)^2*(1/6) to die. The pattern repeats endlessly. For the nth case, the chance to live is (5/6)^n and to die at that point is (5/6)^(n-1)*1/6. The math done here checks to the limits of the chart.

To look at this from another way, 1/6 of game is over in death in case 1, then in case 2, we carry over the 5/6 and in 1/6 of the cases that is the end, which equals (5/6)(1/6) = 5/36. SO HOW MUCH OF THE GAME IS LEFT? It is 1 (for the full game) -1/6 for the first death possibility -(5/6)(1/6) = 5/36 for second death possibility. Thus to put this in condensed form it is: [tex] 1-\frac{1}{6}-\frac{5}{36} = 1-\frac{11}{36} = \frac{25}{36} [/tex]

This result (5/6)^2 is same result that is more easily obtained using the method of reberal98. But once you see that the death part is dropped because the game ends, so at each new step we are dealing with smaller numbers. So, by the third case we have (5/6)^3 and (5/6)^2*(1/6). If you like you can go at this the long way and get:

1-1/6-5/36-25/216 = 1-91/216 = 125/216 = (5/6)^3. Here the "Survival probability" can be thought of the amount of game that is still left to be played--should we elect to continue.

Adding up to infinity all these death possibilities gives 1/6[1+5/6+(5/6)^2++++5/6)^(n-1)+++]=[tex]\frac{1}{6}*\frac{1}{1-5/6} =1[/tex] 1 being the total probability, 100% of the chances.
 
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  • #15
Robert Ihnot,

Great summary ! Thanks a lot.

MG.
 
  • #16
Glad to have been useful!
 
  • #17
My interpretation is that the per-trial column is just the incremental probability of getting hit in that trial, reflecting that the left is the cumulative probability
 

1. What is Russian Roulette: Survival & Hit Probability?

Russian Roulette: Survival & Hit Probability is a game of chance where players take turns spinning a revolver cylinder with one bullet in it. The players then take turns pulling the trigger and shooting the gun at their own head. The goal is to avoid getting shot and be the last player standing.

2. Is Russian Roulette: Survival & Hit Probability a real game?

No, Russian Roulette: Survival & Hit Probability is not a real game. It is a fictional game often depicted in movies, books, and TV shows. It is extremely dangerous and should never be attempted in real life.

3. What are the odds of surviving Russian Roulette?

The odds of surviving Russian Roulette depend on the number of players and the number of bullets in the revolver. For example, if there are 6 players and 1 bullet in the revolver, the odds of surviving are 5 out of 6 or about 83%. However, these odds are not exact as they also depend on the players' willingness to take risks.

4. Why is Russian Roulette: Survival & Hit Probability considered a dangerous game?

Russian Roulette: Survival & Hit Probability is considered a dangerous game because it involves putting a loaded gun to one's head and pulling the trigger. This game has no winners, only survivors, and it can result in serious injury or death. It should never be played in real life.

5. Are there any variations of Russian Roulette: Survival & Hit Probability?

Yes, there are variations of Russian Roulette that have been created for entertainment purposes only, using fake or modified guns. However, these variations should still be approached with caution as they can still result in injury. It is always best to avoid playing any version of Russian Roulette in real life.

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