Batman’s Choice

A hero can be anyone, even a man doing something as simple and reassuring as putting a coat on a young boy’s shoulders to let him know the world hadn’t ended (Batman in The Dark Knight Rises)

Joker has captured Batman and keeps him into a dark and cold dungeon of Gotham City. Showing his sadistic character, Joker proposes the following game to Batman:

This is a six shooter revolver with two bullets in the cylinder. Bullets are inside two consecutive chambers. I will spin the cylinder and will fire the gun aiming to my head. If I survive you will have to do the same but you decide if you want to spin the cylinder before firing or not. If you still keep you head over your shoulders after firing, you will be free.

Joker fires and nothing happens. He survives and passes the revolver to Batman. What should Batman do? Spinning or not? What would you do?

From my point of view, answer is quite anti-intutive because the best option is not spinning the cylinder again. Spinning case is clear: probability of losing the head is 2/6=33% but what about not spinning? Doing the next shoot directly eliminates two possibilities: the previous shot of Joker and the second bullet according to direction of cylinder rotation (remember two bullets are consecutive and Joker is unfortunately  still alive). It means there is only 1 chance to dead between 4, so probability of losing the head in this scenario is 1/4=25% which is significantly lower than the first one.

Here you can find the resulting graph of simulating the game up to 500 times:

batmans_choice2

Will it be the end of Batman? Not sure.

This is the code of this experiment:

library(ggplot2)
library(extrafont)
niter=500
results=data.frame()
for (i in 1:niter)
{
bullet1=sample(1:6,1)
Joker=sample((1:6)[-c(bullet1, bullet1%%6+1)],1)
#Option 1: Shooting
Batman1=Joker%%6+1
dead1=(Batman1 %in% c(bullet1, bullet1%%6+1))*1
#Option 2: Rolling and Shooting
Batman2=sample(1:6,1)
dead2=(Batman2 %in% c(bullet1, bullet1%%6+1))*1
results=rbind(results, c(i, dead1, dead2))
}
colnames(results)=c("iter", "dead1", "dead2")
results$csum1=cumsum(results$dead1)/as.numeric(rownames(results))
results$csum2=cumsum(results$dead2)/as.numeric(rownames(results))
theme_xkcd=theme(
panel.background = element_rect(fill="darkolivegreen1"),
panel.border = element_rect(colour="black", fill=NA),
axis.line = element_line(size = 0.5, colour = "black"),
axis.ticks = element_line(colour="black"),
panel.grid = element_line(colour="white", linetype = 2),
axis.text.y = element_text(colour="black"),
axis.text.x = element_text(colour="black"),
text = element_text(size=18, family="Humor Sans"),
plot.title = element_text(size = 50)
)
p=ggplot(data=results, aes(x=iter, y=csum1))+
geom_abline(intercept = 1/4, slope = 0, size = 0.4, linetype=2, colour = "black", alpha=0.8)+
geom_abline(intercept = 1/3, slope = 0, size = 0.4, linetype=2, colour = "black", alpha=0.8)+
geom_line(aes(y=csum2), colour="green4", size=1.5, fill=NA)+
geom_line(colour="green4", size=1.5, fill=NA)+
coord_cartesian(ylim=c(.0, 1), xlim=c(1, niter))+
scale_y_continuous(breaks = c(0,round(1/4, digits = 2),round(1/3, digits = 2),1))+
geom_text(data=results[niter*.75, ], family="Humor Sans", colour="green4", y=0.38, label="Rotating Cylinder and Shooting ...", size=4, adjust=1)+
geom_text(data=results[niter*.75, ], family="Humor Sans", colour="green4", y=0.20, label="Shooting without Rotating Cylinder ...", size=4, adjust=1)+
labs(x="Number Of Trials", y="Prob. of Losing The Head", title="Batman's Choice")+
theme_xkcd
ggsave("batmans_choice.jpg", plot=p, width=8, height=5)
Advertisements

6 thoughts on “Batman’s Choice

  1. Assumptions could be wrong.
    Bullets are heavy, so if rotate, they probably would be at bottom part of drum, when rotation stops, and hole is at the top, so probability of shot is less than 1/3.

    Batman need to experiment with the drum before calculations and shooting.

    1. Agree. Anyway, I am sure these calculations are right to compare strategies since gravity works for both of them. Batman cannot experiment before shooting. Would you let him if you were Joker? Thanks for your comment!

  2. The force of gravity and the bullet’s weight on the spin are probably going to have a negligible impact when compared to force Batman will spin it with; more so if they come the classic movie cliche of suddenly flipping the cylinder into place while its still spinning fast.

    Probably not enough to overcome the 7% difference in probabilities.

    P.S. – I think the description should be “spin” the cylinder instead of rotate. I initially read it as “rotate the cylinder one tick”

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s