Register to reply 
Russian roulette 
Share this thread: 
#1
Apr309, 01:19 PM

P: 180

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  Russian roullete. 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 Code:
Thanks, MG 


#2
Apr309, 01:24 PM

Sci Advisor
HW Helper
P: 8,953

Because it is considering the probability that they are already dead from an earlier round?



#3
Apr309, 01:52 PM

PF Gold
P: 1,059

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 positionunless the gun was spun upside down. I think I have some vague memory of someone observing that.) 


#4
Apr309, 02:13 PM

P: 180

Russian roulette
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
Apr309, 02:31 PM

Sci Advisor
HW Helper
P: 8,953

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
Apr309, 02:36 PM

P: 180

mgb_phys,
Thanks. 


#7
Apr309, 02:54 PM

P: 52

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
Apr309, 03:01 PM

P: 180

Thanks. 


#9
Apr309, 03:12 PM

P: 52

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
Apr309, 03:14 PM

P: 52

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
Apr309, 03:36 PM

P: 180

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
Apr309, 04:13 PM

P: 52

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
Apr509, 07:22 AM

P: 180

rbeale98,
Thanks a lot. Appreciate your help. MG. 


#14
Apr509, 11:41 PM

PF Gold
P: 1,059

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)^(n1)*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: 11/65/3625/216 = 191/216 = 125/216 = (5/6)^3. Here the "Survival probability" can be thought of the amount of game that is still left to be playedshould we elect to continue. Adding up to infinity all these death possibilities gives 1/6[1+5/6+(5/6)^2++++5/6)^(n1)+++]=[tex]\frac{1}{6}*\frac{1}{15/6} =1[/tex] 1 being the total probability, 100% of the chances. 


#15
Apr1409, 08:54 AM

P: 180

Robert Ihnot,
Great summary ! Thanks a lot. MG. 


#16
Apr1409, 11:44 PM

PF Gold
P: 1,059

Glad to have been useful!



#17
Apr1609, 04:41 PM

P: 101

My interpretation is that the pertrial column is just the incremental probability of getting hit in that trial, reflecting that the left is the cumulative probability



Register to reply 
Related Discussions  
Roulette probability  Set Theory, Logic, Probability, Statistics  10  
Roulette F=mv^2/r  Classical Physics  0  
Answers to a Roulette problem  General Physics  1  
Man bets all on Las Vegas roulette spin  General Discussion  5 