9 of 100: Bond. James Bond.

The revolver carousel spins in the same direction each time. I have arbitrarily chosen a counter-clockwise convention. I have numbered the empty chambers from 1 to 4. Here are the scenarios:

Chamber 1: Next chamber is a kill
Chamber 2: Next chamber is safe
Chamber 3: Next chamber is safe
Chamber 4: Next chamber is safe

So the probability of Bond dying if Scaramanga simply fires again is $\frac{1}{4}$ . If Scaramanga rotates the chamber randomly before firing again, the chance of the carousel landing on a loaded chamber is $\frac{2}{6} = \frac{1}{3}$ . Therefore, Bond is better off having Scaramanga fire the second shot straight away.

First, we will calculate the probability of spinning and shooting. If Bond chooses to spin and shoot, there is a $\frac{4}{6}$ = $\frac{2}{3}$ chance he will survive, as 4 out of 6 chambers are empty.

But if Bond chooses to shoot without spinning, there are 5 chambers left, so one would think there would be only a $\frac{3}{5}$ chance of survival. But since the two bullets are next to each other, Bullet B cannot be shot out of the gun, since it can only be fired after Bullet A has been fired. Therefore, there is a $\frac{3}{4}$ chance of survival if Bond chooses to shoot without spinning, due to the fact that the two bullets are next to each other and that Bullet B cannot be the one that is fired.

Since a $\frac{3}{4}$ chance of survival is better than a $\frac{2}{3}$ one, Bond should choose to shoot without spinning.

Once a blank is fired, (knowing that angular movement has unchangeable direction) there are four possibilities for the next chamber:

• first possibility is that there is a bullet in it, and

• other three that there is a blank.

Hence, the probability of finding the bullet is $\frac{1}{4}$ , implying that Bond has $\frac{3}{4}$ chances to survive. If he chooses to spin, then he has the same chances of surviving like he had at the beginning: $\frac{2}{3}$ . It's clear that, in order to increase his chances of survival, he should choose Scaramanga to shoot right away!

If Scaramanga spins again then the probability that a bullet will be fired is 2 (bullets)/6 (chambers) = 1/3. If he fires straight away then IT SEEMS TO ME that the bullets could have been in chambers (1 & 2) - but this must be eliminated as a possibility because that would mean that Bond would already have been killed with a bullet from the 1st chamber. But the other possibilities are (2 & 3), (3 & 4), (4 & 5) or (5 & 6) each with equal probabilities which add up to 1, so each with 1/4. And with only the 1st one of these does Bond die with the bullets been in chambers 2 & 3. So the probability is less (1/4) if Scaramanga fires immediately! Regards, David Ps If the bullets were not next to each other then the prob of death if just firing would be the number of bullets (2) divided by the number of chambers OTHER THAN THE 1ST ONE which we know was blank, AND HAS ALREADY GONE. So that is 5. So the prob of death would be 2/5 which is more than 1/3.

Let's translate this into an analysis of the possible scenarios 0a 0b 0c 1x 1y 0d Imagine each term as an cylinder, when it has 0 theres not bullet inside, when it has 1 there is a bullet Possible scenarios:

[0a 0b], [0b 0c], [0c 1x], [1x 1y], [1y 0d], [0d 0a]

Remember that when he shoot first time there wasn't any bullet, let's take the cases where he shoots nothing happens and when he shoots again James die. [ 0c 1x] Unique case from all the 6 possibilities, so the chance is $\frac{1}{6}$

taking the possible scenarios already "drawn" again let's analise the chances of a first try bullet when he spins the cylinder. [1x 1y], [1y 0d], 2 out of 6, $\frac{1}{3}$ , with all those informations we can conclude that the chance of a first try bullet is greater than a sequence of "Nothing happens, boom".

We know that 2 out of 6 chambers are filled, meaning that the random, spun chance of Bond's death is $\frac{2}{6}$ , this carries out no matter when it is spun, as long as the chances are truly random, and not selected by Scaramanga. From the given information of the first shot containing no bullet, we can deduce, because a revolver only spins in one direction, and the bullets are next to each other; not spread out, that the chance of the next shot containing a bullet is only $\frac{1}{4}$ because only 1 of the 4 empty chambers allows the next shot to be fatal. When the two fractions are multiplied to have the same denominator, such as 12, the chance of death on spun = $\frac{4}{12}$ , and the chance of death on shoot = $\frac{3}{12}$ , therefore the chance of Bond's death is larger in case of a randomly spun cylinder.

A random spin gives the chance of $\frac{2}{6}$ , or a third. If the next shot is fired, there are only five possibilities left, but you know that the second bullet won't be fired, as they are next to each other, so it is one in five

• If Scaramanga spins the cylinder so that a random chamber is on top, then there's a probability of $2/5$ that this chamber contains a bullet.

• Else, if Scaramanga intends to fires again right away, then:

  • as we know for sure that the current chamber is an empty one
  • and since, among the 4 empty chambers, only one has a succeeding chamber containing a bullet

  the probability for the next chamber - from which Scaramanga will fire - to contain a bullet is $1/4$ ( $< 2/5$ ).

Therefore, Bond had better have Scaramanga fire right away.

Similar to Monty Hall problem

How did such a big percentage of people fail? If he is going to be killed because the bullet is above and Scaramanga spins he is going to be killed anyway (as both of the bullets are one nexto to the other. Nevertheless, if he isn't going to be killed because the bullet is not above, by spining the cylinder the buller might be in the next position so it's positioned above and kills Agent Bond. I hope u've understood, I'm not english, but spanish haha.

Why you guys make it so complicated?.. Probability of the next chamber having a bullet is 1/6. Probability of the chamber having a bullet if you spin again is 2/6.

If you spin the chamber, there is a higher chance for Agent Bond to be killed. If you just fire, there's a lower probability for Bond to be shot.

This is simple yet complex. For the first shot, we know that it could have been from one of the four chambers without a bullet. We know that with only one of the chambers the next shot would be a bullet. So if Bond chooses to just fire, his chances of surviving are $\frac{3}{4}$ . 2 out of the 6 chambers, or $\frac{1}{3}$ of the chambers, have a bullet in them. This means that there is a $\frac{1}{3}$ chance of spinning and getting a bullet. So, if Bond chooses to spin, then fire, his chances of surviving are $\frac{2}{3}$ . $\frac{3}{4}$ is greater than $\frac{2}{3}$ , so Bond has a greater chance of surviving if he chooses to just fire .

Before firing, there are 2/6 positions that are lethal. Survival chances are 2/3. Firing a blank limits the available orientations to 4 only, one of which can result in a lethal strike, with odds of survival at 3/4. So the odds are: a) spin again (2/3 odds of survival) and b) fire a consecutive time (3/4 odds of survival). Bond should instruct Scaramanga to pull the trigger a second time.

When there is a random shot chosen, $4$ out of $6$ of the choices will result in Bond surviving, a $\frac23$ chance. We are given that Scaramanga already fired an empty barrel. If the empty barrel is next to a barrel with a shot in it and Bond just wanted Scaramanga to fire the next shot, he would die. There is one empty barrel which is before a barrel with a shot in it, out of $4$ barrels. Therefore, the chance of surviving when firing the next shot is $\frac34$ . $\frac34 > \frac23$ , so Bond has a greater chance of surviving if he tells Scaramanga to fire the next shot.

Just think that in other way You have 6 pen ,2 are good and 4 defective U take a pen and it comes out to be defective, U throw it away now u have 2 good 3 bad pens so u have more probability to pick good pen than again repeating the procedure

If Bond rotates and then fires, the probability of shooting a bullet is 2/6 = 1/3. In this case Bond has a 2/3 chance of surviving.

When Scaramanga pulled the trigger on an empty chamber, there is a 1/4 chance that the next chamber has a bullet. So there is a 3/4 chance of surviving since most empty chambers are next to other empty chambers. Bond's best option is to just pull the trigger.

If we spin the slot the probability for James to be not shot is 2/3. But if we know that the chamber that will be selected is the next of one of the empty chambers, then the probability is 3/4 for James to survive. Is better for him the first option.