Client consulted me on Roulette strategy. Am I correct?

[canis582]

[ QUOTE ]
You always go last in this game no matter what!

I want to know what idiot would pull the trigger if they were sixth in a six shot game??!?

[/ QUOTE ]

Last time I went sixth, the gun mis-fired.

[WhiteWolf]

[ QUOTE ]
[ QUOTE ]
You always go last in this game no matter what!

I want to know what idiot would pull the trigger if they were sixth in a six shot game??!?

[/ QUOTE ]

Last time I went sixth, the gun mis-fired.

[/ QUOTE ]

Last time I played, I told my captors to load up three bullets, then I turned the gun on them and used it to escape the prison camp.

[JellyFishy]

Hi,

Don't you agree that any Professional in the field of risk analysis and
accounting should always make the best informed choice to help client
choose his options base on both the value and risk of client's options.

You only considered the value of my client's option.

Assuming that my client is rational risk averse, he will opt to compensate
risk/variance of lost for more EV.

Probability of survival is 50/50 for 6 shot gun whether going 1st or 2nd.
Assuming both prefers survival to death. EV for going 1st or 2nd must be the
same for both players if they both wants to live.

How much they value their life is irrelevant as long as both prefers to
survive, their goal is the same and they cannot trade their positions. Since
EV is the same, it is irrelevant too. So everything depends on variance or
risk.

But risk or variance of choice is not the same. Given 2 package of "3 shots
of fun" the risk averse person will prefer the lower risk of lost package.

The highest risk of death possible is 100% lost of life which is in the
package that includes the last shot. Any rational risk averse person will
prefer any other package holding EV constant.

Fishy

I assumed my Russian client was risk averse.
If he is extremely risk loving then he must be making a grave mistake right
now using my strategy.
Just because he didn't pay for my "risk preference analysis service" to save
a few bucks.

[hedgeyerbets]

No risk averse person would play the game to begin with. If forced to play, there is no preferred "rational" strategy if you use the standard economic definition of rationality. Going 1,2,3 is no better or worse than going 1,2,6 or 1,5,6 or 2,3,4 or .....

[JellyFishy]

My adviser presented me with a solution to counter my logic.
I don't have time to confirm it yet but just in case he is correct, please stop using my strategy until everything is clarified.

Anyway here is my advisers solution:

If you are rational risk averse, you do not play this game if you value
your life more than the other guy. PERIOD. I know this for a fact.

Let's make this simple. Say you value your life at 100 and your
opponent's life at 80.

EV playing game = 50% * -100 + 50% * 80 = -10

If your opponent also believes your life is worth 100 and his is only
worth 80.

Opponent EV playing game = 50% * 100 + 50% * -80 = 10


If anything, you want to go 2nd because if you go first and don't die,
your opponent may choose to quit the game, a source of counterparty
risk. Therefore your equity is (assuming you value your life and your
opponent's both at 100):

EV = [-100 * (1/6)] + [100 * (1/5) * (prob of opponent not defaulting on
claims)] + [rest of EV series]

This number will always be <=0, therefore at best breakeven and most
likely negative EV to go first assuming even the smallest bit of
counterparty default risk.

You say that "The highest risk of death possible is 100% lost of life
which is in the package that includes the last shot. Any rational risk
averse person will prefer any other package holding EV constant."

This thinking is seriosly flawed as you need to discount this 100% loss
of life by the probability that it will reach the 6th shot which is
about 1/6.

100% chance of dying on last shot * prob of reaching last shot =
100%*(1/6)= -16.7%
(1/6) change of dying on first shot * prob of reaching first shot =
(1/6) * 100% = -16.7%

Therefore you are indifferent to either. In fact mathematically, the EV
of the 1,2,3,4,5, or 6th shot are equal at -16.7% for the player taking
the shot and +16.7% for the player not taking the shot.

Assuming the real world and how much counterparty risk there is, I would
definitely take the 2nd shot. If it gets down to the last shot, I can
always default. In fact, I can default on the second shot gaining me
16.7% EV. Last time I checked, games of Russian Roulette aren't
enforceable.

[hedgeyerbets]

Dude you just quoted yourself and said that your argument was seriously flawed. Count me confused.

[JellyFishy]

Hi,

The response from adviser 1, he quoted me. I didn't quote myself. Here is another response from adviser 2.

Adviser 2 solution:

I might as well throw my hat in here (even though the example is pretty macabre).

Fishy, as I was reading your last message, I thought the same thing as Tony did: that you are making an implicit assumption that the bullet was in the sixth chamber of the gun. This is clearly incorrect. If we knew that with certainty, then this would be a moot point because we would shoot first and take the 1, 3, 5 package instead of the 2, 4, 6 package and live with certainty (assuming the game is always played to completion).

The EV of this game is:

EV= .5(-infinity-->assuming we die) + .5(0-->assuming we get zero utility from living) = .5(-infinity)= -infinity

An interesting aside is to think about though (completely unrelated) is what we should do if we must play this game but then get the option of switching positions with our opponent. For example, shoot first then elect to take the 2, 4, 6 package or continue with the 1, 3, 5 package. Then this becomes similar to a Monte Hall problem and it is always better to shoot first and then switch to the 2, 4, 6 package since if we fire the first shot and then do not die, then given the fact that there is no bullet in chamber one the conditional probability of it being in chamber 3 or 5 is greater than the conditional probability of it being in chamber 2, 4, or 6.

This kind of gets back to what adviser 1 was saying in that as the game goes on, it becomes more and more likely that somebody will "default" and not play since the sample size of barrels is getting smaller and smaller and the conditional probability of the bullet being in the next chamber increases after each time somebody pulls the trigger and doesn't die.

Adviser 1, don't mean to be a nit (yes I do hehe), but if we assume people are risk averse then we must multiply all your probabilities by the natural log of the value of one's life.

If we absolutely must play this game (i.e. somebody is LITERALLY holding a gun to our head) then perhaps the best option would be to elect to shoot first, then grab the gun and continually fire at the other guy until we kill him.

(A better use of this guy's money would have been to go to a good psychiatrist).

[Kaeser]

Why does everyone always set death as -infinity? Even Pascal gave us a 3-1 shot at +infinity. [img]/images/graemlins/grin.gif[/img]

[WhiteWolf]

[ QUOTE ]


An interesting aside is to think about though (completely unrelated) is what we should do if we must play this game but then get the option of switching positions with our opponent. For example, shoot first then elect to take the 2, 4, 6 package or continue with the 1, 3, 5 package. Then this becomes similar to a Monte Hall problem and it is always better to shoot first and then switch to the 2, 4, 6 package since if we fire the first shot and then do not die, then given the fact that there is no bullet in chamber one the conditional probability of it being in chamber 3 or 5 is greater than the conditional probability of it being in chamber 2, 4, or 6.




[/ QUOTE ]

Not true. If we know the bullet is not in chamber 1, then there is a 20% chance it is in chamber 2, the same 20% chance it is in chamber 3, and so on. Switching to the chamber 2, 4, 6 cycle gives us a 60% chance of finding the bullet, whereas staying with 3,5 gives us only a 40% chance of doing so...