That's exactly my point. After 5 clicks if you don't do that the other player will. Hence the necessary draw rule.

To make the analogy even better, I should have been talking about deuces since those are cards we don't want.

The simplest explanation is that the cards have an equal chance of being in each spot in the deck (and Bruce made the equivalent explanation with chambers). But what's tripping you up is something else which this gun-to-head scenario emphasizes more effectively than the hold'em problem. There's another way of understanding why it's equal which is perhaps more satisfying/convincing for the russian roulette case because it illustrates exactly why one's intuition is wrong.

Hold'em
- If 1st player gets a deuce, 2nd player's chances of getting one decrease.
- If 1st player doesn't get one, 2nd player's chances of getting on increase.

Russian Roulette
- If 1st player gets the bullet, 2nd player's chances of getting it decrease (to zero).
- If 1st player gets the bullet, 2nd player's chances of getting it increase.

The only differences are that there's only one bullet as opposed to 4 deuces, and there are 52 "chambers" in hold'em instead of 6.

That's why I suggested, suppose 3 of the deuces are removed from the deck. That doesn't change the fact that both players have an equal chance of getting the deuce.

The increases always cancel out with the decreases. While the decreases have a bigger impact, they happen much less often. The greater impact is exactly proportional to the lower frequency and so they cancel out.

In holdem with a 49 card deck:
48/49 of the time the 2nd player will have a 1/48 chance of getting the deuce (and 1/48 is an increase of 1/2352 from 1/49).
1/49 of the time he'll have a 0% chance (a decrease of 1/49 from 1/49).

The decrease of 1/49 in the 2nd scenario is 48 times as big an impact as the increase of 1/2352 in the first scenario. But the 1st scenario happens 48 times as often. So we have essentially 48 * 1/48 = 1, the relative impact and relative frequency canceling to relative sameness.

Pr(2nd player deuce) = 48/49 * 1/48 = 1/49 (same as 1st player's chance)

Not really. The button's only advantage is information. He gets to make his decisions with more information than his opponent, namely he already saw his opponent's move. In Russian roulette, if played fairly there is no skill, no choices, and therefore no advantage is gained by having better information.

It's an understandable instinct to not want to be the first to partially commit suicide. Myself, I thought about my preference before and decided I'd rather go first to have the better fractions. But thinking about it again now, I'm leaning to be 2nd player. There's only a 3.33% difference between 5/6 and 4/5. I feel like #2's only real sweat is the 2/3 pull, whereas 1st player has to sweat a 3/4 and a 1/2, plus go first. Sure the 2nd player has to face a 0 at the end but it probably won't get that far. Ofc these are all irrational thoughts because mathematically it's the same.

This is an interesting point, I guess Monty Hall isn't the only math problem that can present psychological barriers to people. Mixing in psychological aspects forces one to really grasp the math concepts; I will remember this if I'm ever teaching courses. I can make it a problem on the test. They'll already be nervous during a test, and now they'll have a question where they have to imagine firing a gun at their own head. Cruel but effective =)

Again, we're not talking about the case where each guy spins the cylinder. In that case the second guy would have the advantage for the reason you say. But if only the first guy spins it, then the chances are the same. The fact that the second guy doesn't have to pull the trigger if the first dies is exactly canceled by the fact that if he does have to pull, he has a higher chance of dying than the first guy had when he pulled.

You want that to happen so both have an equal chance of dying before the game starts. No draws. A better way is to spin the cylinder every time, and then spin the gun to see who shoots next. So the players don't necessarily take turns, the same player can go more than once in a row. But it's fair, and each player has a 1/6 chance each time he pulls the trigger. Then you can quit playing any time you want, and it's still fair, and possibly no one will die. Any number of players can play that way.

So I think its fair to say next Vegas meetup should have a Russian roulette tournament? Single elimination one presumes.

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Thanks for breaking that down to me guys. You put a question like that on a test and you will be looking for a new job in no time/in jail lol. This thread is awesome.

Note the non-round shape of this cylinder:




"Other magicians who have played Russian roulette were not so lucky, however; proving that even for a skilled technician, things can go horribly wrong. In 1976, such was the case for Finnish magician Aimo Leikas. One of the highlights of his act had been pulling six bullets out of a box that included both live and dummy ammunition, then playing the game on himself. He claimed to use telepathy in order to select only the dummy bullets, which were unmarked. But a stunned crowd saw him kill himself when he pulled the trigger with live ammunition in the next chamber. Leikas had been performing the stunt for about a year with no incidents until that time; apparently his telepathic skills were off-kilter for one day in a “no-bad-day-allowed” line of work.

...

In 2000, a man from Houston, Texas died in a game of Russian roulette where he never had a chance. The man tried to play the game with a semi-automatic pistol – meaning that every time the gun was cocked, a cartridge was automatically inserted into the firing chamber. Given the 100 percent chance that he would have a live round fired when he pulled the trigger, the man was named as a runner-up for a “Darwin Award” in 2000, an award designed to “commemorate those who improve our gene pool by removing themselves from it.”

http://www.casino.org/blog/russian-r...e-game-chance/

Lol

What percent of people can't answer this question correctly? I would guess > 50% . Responses in this thread confirm said theory.

More interesting question is to see how much of a disadvantage the first player has when you add more than one bullet to the gun. Odds are no longer 50/50.

Nobody is down lifetime from this game.

Win - $$$ up lifetime
Lose - die - up lifetime

I've seen this 3 times now and lol'd every time. The picture of the dude makes it perfect

People in general, or people with some familiarity with probability? If the former we're talking >95%.

Yep let's see!

There's a few ways to do this but I'll just do the semi-brute way I did before.

Pr(player 2 wins) = 2/6 + (4/6)(3/5)(2/4)[1+1/3] = 3/5

This is confirmed on page 28 of Bruce' excellent book.


And now I see what the others were saying. RR is only analogous to hold'em when there's 1 bullet in the chamber. With 2 bullets in the chamber, the 2nd player still won't ever be "dealt in" if 1st player gets a bullet.

For that reason, the chance of Player 2 losing on his first turn is only 4/15 instead of 1/3 because if he's lucky he won't even have to take his turn.

Maybe we require 1 more turn to be taken after the 1st bullet enters a skull. Then there'd be a chance of tying (both dying).

Pr(2nd wins) = (1/3)(4/5) + (2/3)(3/5)(2/4)(2/3) = 40%
Pr(1st wins) = (2/3)(2/5)(3/4) + (2/3)(3/5)(2/4)(2/3)(1/2) = 4/15 = 26.67%
Pr(tie) = 1/3

Still not fair so maybe we require both bullets to be used (where the living person takes the dead guy's turns for him by firing at his corpse). Only way to win is fire 2 bullets into villain.

I feel like that would be fair but it's a longer calculation and maybe more suitable for one of the other methods.
(This would maybe be Part B on the test, or extra credit.)

No it's not! Simple combinatorics.

Pr(player 2 wins) = Pr(both bullets in chamber #'s 1, 3 and 5)
= C(3,2) / C(6,2) = 3/15 = 1/5

Pr(player 1 wins) = Pr(both bullets in chambers 2,4,6) = 1/5

And so yes, that last variant I mentioned is fair in that it sucks equally for both players.

Pr(both die) = 1 - 2/5 = 60%
Combinatorially: C(3,1)² = 9 combos out of 15