Russian roulette
March 21st, 2020 at 2:49:01 AM
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In the Bollywood film Sholay, three persons are forced to play a kind of Russian roulette. Gabbar Singh (the bad guy) made sure that there were only three bullets in consecutive chambers. The cylinder is only spun before targeting at the first person, and thereafter the cylinder will not be spun again. There was fear in the first mans eyes, the second man looked even more scared and the third man was completely terrified. What is the probability that the first person will survive? What is the conditional probability that the second person will survive given that the first person survived and what is the conditional probability that the third person will survive given that the first two persons survived?
March 21st, 2020 at 6:44:24 AM
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Klopp,
Ok... I'll give it a shot 😄
Fun problem!
Dog Hand
Ok... I'll give it a shot 😄
First of all, I assume the cylinder has six chambers.
Let's say that chambers 1, 2, and 3 are filled, so 4, 5, and 6 are empty.
Assuming a fair spin, [bold]the first man has a 3/6 = 1/2 chance of survival.[/bold]
If he survives, then his "chamber" was 4, 5, or 6, with equal probability. If it was 4 or 5, then the second man will survive; if it was 6, then the second man dies. Thus, [bold]the second man has a 2/3 chance of survival.[/bold]
If the second man survives, then his chamber was either 5 or 6, with equal probability. If it was 5, the third man survives, so [bold]the third man has a 1/2 chance of survival.[/bold]
Let's say that chambers 1, 2, and 3 are filled, so 4, 5, and 6 are empty.
Assuming a fair spin, [bold]the first man has a 3/6 = 1/2 chance of survival.[/bold]
If he survives, then his "chamber" was 4, 5, or 6, with equal probability. If it was 4 or 5, then the second man will survive; if it was 6, then the second man dies. Thus, [bold]the second man has a 2/3 chance of survival.[/bold]
If the second man survives, then his chamber was either 5 or 6, with equal probability. If it was 5, the third man survives, so [bold]the third man has a 1/2 chance of survival.[/bold]
Fun problem!
Dog Hand
March 21st, 2020 at 9:33:18 AM
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Let's say you're watching a game of Russian roulette. Two guys, player A and player B, are sitting at a table and there's a pistol with six chambers, one of which contains a bullet. They spin before each pull, alternating turns, game ends at first bang. Player A will start the game.
You want to bet on player B winning (surviving) and the local bookie is quoting odds of -130. What's the vig on that bet?
You want to bet on player B winning (surviving) and the local bookie is quoting odds of -130. What's the vig on that bet?
Its all about making that GTA
March 21st, 2020 at 9:35:45 AM
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Number the three filled chambers 1, 2, 3, and the other three 4, 5, 6.
There are six equally possible positions for the first chamber:
1, 2, 3 - the first person dies
4 - assuming that, if all three draw blanks, the first player pulls the trigger again, then the first person dies (after each person draws a blank)
5 - the third person dies
6 - the second person dies
Probabilities of survival: 1/3 for the first person, 5/6 for each of the other two
March 21st, 2020 at 11:04:28 AM
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Morbid as it is (particularly these days), a fun problem.
Edit: My response is based on the idea that the game ends if any player dies.
That assumption was based on this rule:
Chambers 1, 2, and 3 are filled, so 4, 5, and 6 are empty.
If it's spun to 1, 2 or 3, player 1 dies. 4, 5 or 6, he lives. Therefore, he has a 3/6, 50% chance to die.
Then it rotates one spot.
If it was originally spun to 1, 2 or 3, game over, player 2 (and 3) lives. If it originally spun to 4 or 5, player 2 lives. 6 he dies. Therefore 1/6 chance of dying, 16.6% chance.
Then it rotates one spot.
If it was original spun to 1, 2, 3, or 6, game over, player 3 lives. If it originally spun to 4 he lives. 5 he dies. Again 1/6, 16.6% chance of dying.
The question is, what if they all live? Does the game end? If it goes around one more time, player 1 dies, therefore 4/6 or 66.7%.
If the game stops after one pass, there is 1 chance in 6 for everyone to live.
Summary. Chance of dying:
1: 3/6 or 4/6 depending on rules.
2: 1/6
3: 1/6
If it's spun to 1, 2 or 3, player 1 dies. 4, 5 or 6, he lives. Therefore, he has a 3/6, 50% chance to die.
Then it rotates one spot.
If it was originally spun to 1, 2 or 3, game over, player 2 (and 3) lives. If it originally spun to 4 or 5, player 2 lives. 6 he dies. Therefore 1/6 chance of dying, 16.6% chance.
Then it rotates one spot.
If it was original spun to 1, 2, 3, or 6, game over, player 3 lives. If it originally spun to 4 he lives. 5 he dies. Again 1/6, 16.6% chance of dying.
The question is, what if they all live? Does the game end? If it goes around one more time, player 1 dies, therefore 4/6 or 66.7%.
If the game stops after one pass, there is 1 chance in 6 for everyone to live.
Summary. Chance of dying:
1: 3/6 or 4/6 depending on rules.
2: 1/6
3: 1/6
Edit: My response is based on the idea that the game ends if any player dies.
That assumption was based on this rule:
Quote:...what is the conditional probability that the third person will survive given that the first two persons survived?
Last edited by: DJTeddyBear on Mar 21, 2020
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
March 21st, 2020 at 11:23:23 AM
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Quote: Ace2Let's say you're watching a game of Russian roulette. Two guys, player A and player B, are sitting at a table and there's a pistol with six chambers, one of which contains a bullet. They spin before each pull, alternating turns, game ends at first bang. Player A will start the game.
You want to bet on player B winning (surviving) and the local bookie is quoting odds of -130. What's the vig on that bet?
For this I get:
The chance of player A dying (so B winning) is the series 1/2 + 1/8 + 1/32 + . . .
Thats a geometric series with the common ratio of 1/4 and an initial term of 1/2.
Therefore: 1/2 / (1 - 1/4) = 2/3.
So if you bet 130 on player B, 2/3 of the time you win 100 and 1/3 of the time you lose 130. Thats an EV of 23.3333.
So far from a vig, the player has an edge of 23.3333 / 130 = 17.949%.
The fair line on B should be -200.
Thats a geometric series with the common ratio of 1/4 and an initial term of 1/2.
Therefore: 1/2 / (1 - 1/4) = 2/3.
So if you bet 130 on player B, 2/3 of the time you win 100 and 1/3 of the time you lose 130. Thats an EV of 23.3333.
So far from a vig, the player has an edge of 23.3333 / 130 = 17.949%.
The fair line on B should be -200.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
March 21st, 2020 at 12:18:09 PM
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By the way, if the bullets were NOT loaded consecutively, say, they were in chambers 1, 2 and 4, then...
Edit: My response is based on the idea that the game ends if any player dies.
Player 1 still has a 3/6 chance to die.
Player 2 has a 2/6 chance to die (if it was originally rotated to 3 or 6).
Player 3 has a 1/6 chance to die (if it was originally rotated to 5).
Although I didn't say if before, the original post's comments about how scared the players were acting, was backwards. Player 1 should have been most frightened, not least.
Player 2 has a 2/6 chance to die (if it was originally rotated to 3 or 6).
Player 3 has a 1/6 chance to die (if it was originally rotated to 5).
Although I didn't say if before, the original post's comments about how scared the players were acting, was backwards. Player 1 should have been most frightened, not least.
Edit: My response is based on the idea that the game ends if any player dies.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
March 21st, 2020 at 7:46:44 PM
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On the first pull, Player A has a 1/6 chance of losing and a 5/6 chance of passing the gun to Player B. Then theres a 5/6 chance Player B will return the gun.Quote: unJonFor this I get:
The chance of player A dying (so B winning) is the series 1/2 + 1/8 + 1/32 + . . .
Thats a geometric series with the common ratio of 1/4 and an initial term of 1/2.
Therefore: 1/2 / (1 - 1/4) = 2/3.
So if you bet 130 on player B, 2/3 of the time you win 100 and 1/3 of the time you lose 130. Thats an EV of 23.3333.
So far from a vig, the player has an edge of 23.3333 / 130 = 17.949%.
The fair line on B should be -200.
So As overall chance of losing can be expressed as
A = 1/6 + 5/6B
B = 5/6A
Solve for A and its 6/11
The vig for -130 is 1 - 6/11 * 23/13 = 5/143 =~3.5%
The fair line would be -120.
Its all about making that GTA
March 21st, 2020 at 8:01:15 PM
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Quote: Ace2On the first pull, Player A has a 1/6 chance of losing and a 5/6 chance of passing the gun to Player B. Then theres a 5/6 chance Player B will return the gun.Quote: unJonFor this I get:
The chance of player A dying (so B winning) is the series 1/2 + 1/8 + 1/32 + . . .
Thats a geometric series with the common ratio of 1/4 and an initial term of 1/2.
Therefore: 1/2 / (1 - 1/4) = 2/3.
So if you bet 130 on player B, 2/3 of the time you win 100 and 1/3 of the time you lose 130. Thats an EV of 23.3333.
So far from a vig, the player has an edge of 23.3333 / 130 = 17.949%.
The fair line on B should be -200.
So As overall chance of losing can be expressed as
A = 1/6 + 5/6B
B = 5/6A
Solve for A and its 6/11
The vig for -130 is 1 - 6/11 * 23/13 = 5/143 =~3.5%
The fair line would be -120.
Oh duh. In my mind there were three bullets in the chamber.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
March 21st, 2020 at 8:53:49 PM
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Um...Quote: unJonOh duh. In my mind there were three bullets in the chamber.
Quote: Klopp... made sure that there were only three bullets in consecutive chambers...
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
March 21st, 2020 at 9:19:50 PM
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So disturbing.
I am a robot.
