How Much Does Spinning the Chamber Really Help?

Really? A math word problem in law school?

Ok, here’s a hypothetical problem for you: (Do not try this at home. Or anywhere.)

Let’s say it’s 1945 and you’re seventeen and  just sitting at the lunch counter with your 13 year-old friend. You’re bored. Television is just kicking off and the Simpsons are years away from being reality. So you start thinking. “Hmmm. What can two kids do to pass the time?”

“Well,” you say to yourself, “we can play Russian Poker,” which is more commonly known as Russian Roulette. Luckily you brought your uncle’s revolver and a bullet. So, you load the gun, putting the bullet in the last slot of the 5 chambers, point the gun at your friend’s right side and pull the trigger three times. It’s not going to fire until the fifth time. Right?. Oops. On the third pull, the gun fires, and two days later your friend dies.

Who’d be dumb enough to do that, you ask? Well, a kid named Malone. Commonwealth v. Malone, 47 A.2d 445 (PA 1946). This was a case we discussed in class tonight. Part of the discussion focused on Malone’s failure to spin the chamber in between trigger pulls, which is the proper way to play Russian Roulette, apparently. So, as I was walking to the trainer, tonight, I decided to figure out the difference in the odds.

Malone’s Odds

So, we need to calculate the odds of at least one of the three trigger pulls firing. That is calculated as:

=1 – odds of all three trigger pulls NOT firing

because if all three didn’t fire then at least one had to fire. So, in this case, that would be:

=1 – (0.8) (0.75) (0.67)

Why do the odds decrease each trigger pull? Because it’s sampling without replacement. So, first trigger pull, there’s an 80% chance the gun doesn’t fire. (It’s a five chamber gun, remember.) But, the second time you pull the trigger, there’s only 3 empty chambers out of the 4 remaining, so only a 75% chance the gun doesn’t fire this time.

So, the odds that Malone’s going to blow away a “vital part” of his friend’s body are:

59.8%

Would Spinning the Chamber Have Helped?

So, the question was whether it would have helped if Malone had spun the chamber in between trigger pulls.

Well… First off, it probably would have given his friend a little time to reconsider his decision to tell Malone to go ahead and play the game. But, that’s beside the point.

In this case, the odds would now be calculated as:

=1 – (0.8) (0.8) (0.8)

Since each trigger pull would be an independent event (now we’re sampling with replacement), the odds of the gun not firing each time remains 80%. So now Malone’s odds of blowing away part of his friend are only:

48.8%

Well, that’s on the right side of 50/50, at least. In fact, that’s a relative decrease of 18%. But, would that have been enough to change gross recklessness to recklessness and gotten Malone down to involuntary manslaughter?

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2 responses

  1. Great example of probability! I’ll use this in my class when I teach probability next. But you only answered the math question and not the legal question. Does that say something about the feasibility of concrete answers in law vs. mathematics?

    1. No one knows. In the original case, Malone was convicted of depraved heart murder. The court did talk about the probability (they rounded to 60%) that his friend would get hurt was high enough to prove his “wicked disposition.” But, they didn’t say at what probability it would be low enough to mitigate it down to involuntary manslaughter. My personal guess is that 48.8% is still pretty close to 50/50 and if someone was tang a 50/50 chance on killing someone, that’s still pretty “wicked.”

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