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Lexiatel


August 14, 2016, 03:35:47 AM 

Ladies (if there are some) and Gentlemen!
I would like to introduce to you brand new deadly Bitcoin game  BitPistol!
Load your gun >>> Shoot! >>> Win Bitcoins!
As game is about shooting from 6 bullet pistol we have a great puzzle for you to solve!Imagine a 6 bullet revolver as it is in BitPistol, which is loaded with 2 bullets in a row:Now imagine you are playing russian roulette with your opponent! He just made an empty shot and passes revolver to you!QUESTION: Would you spin the cylinder to randomize your turn or trigger the pistol right away without spinning? We are waiting for correct answers! I'm not sure it'll matter either way? The odds of hitting red are the same either way, am I right? Well, I certainly wouldn't play this 'game' irl, I mean, who invented such an awful event? I have known of people who played this game and died from it... I'd also like to know how one reacts after seeing their friend blow their brains out, because someone is going to die (or end up messed up in the head at the very least).




BitPistol.com


August 14, 2016, 09:52:12 AM 

Answer i sent pm
I haven't got any messages with answer, only with bonus attenting, where i couldn't find 3 in a row as well




BitPistol.com


August 14, 2016, 10:24:23 AM 

Hint: The problem is in how you calculate odds
Ah. My bad. You want to go second. I hadn't miscalculated necessarily, but it was in isolated cases. Here's the revised edition: Round 1: P1 has a 1/6 chance of dying Round 2: has a 5/6 chance of occurring, P2 has a 1/5 chance of dying, which equates to a 1/6 chance of dying. Round 3: has a 4/6 chance of occurring, P1 has a 1/4 chance of dying, which equates to a 1/6 chance of dying. Round 4, 5, and 6 follow the same format. Looks like I screwed up! And you want to be second, since though the odds are the same, P1 COULD shoot themselves on the first round. Though if you average it out it doesn't really matter. Either works.
Alternate solution: go first and shoot your opponent. And yet i can't agree with your solution. Probably there is some simplier way to solve it and maybe the correct answer was already mentioned in this thread ...




BitPistol.com


August 14, 2016, 02:31:05 PM 

Ladies (if there are some) and Gentlemen!
I would like to introduce to you brand new deadly Bitcoin game  BitPistol!
Load your gun >>> Shoot! >>> Win Bitcoins!
As game is about shooting from 6 bullet pistol we have a great puzzle for you to solve!Imagine a 6 bullet revolver as it is in BitPistol, which is loaded with 2 bullets in a row:Now imagine you are playing russian roulette with your opponent! He just made an empty shot and passes revolver to you!QUESTION: Would you spin the cylinder to randomize your turn or trigger the pistol right away without spinning? We are waiting for correct answers! I'm not sure it'll matter either way? The odds of hitting red are the same either way, am I right? Well, I certainly wouldn't play this 'game' irl, I mean, who invented such an awful event? I have known of people who played this game and died from it... I'd also like to know how one reacts after seeing their friend blow their brains out, because someone is going to die (or end up messed up in the head at the very least). The correct answer to this puzzle you can read a few replies above, however odds are not the same if we are speaking about 2 bullet puzzle, it is worth shooting without spinning as you have 25% of probability to be shot. I believe also the equal chances is the solution to the second puzzle, where people are shooting each after other 1 bullet loaded revolver without spinning the chamber. I also must agree with you that this game itself is very thrilling when played in real live, i am sure the one will be shocked when he sees what you suggested




BitPistol.com


August 14, 2016, 05:32:36 PM 

Would it not be a pure 50/50 chance (ie. "fair") regardless of where you started?
There are an even number of chambers and 2 players, so each player will have 3 chambers that they could potentially use. Regardless of who goes first, each player has an equal number of chambers assigned to them, so after the spin, both players have an equal chance of having a bullet in one of their chambers.
And because you can't change which chambers you have (no spinning) the odds of getting a bullet overall don't change from 50/50.
That is correct answer! But we got a greater challenge for you: This one is die hard. Imagine that you and your opponent start with 1 bullet loaded into the 6 slot. After each try you load one more bullet to the chamber, spin again and pass the pistol to your opponent. QUESTION: Do you prefer to start as first, or shoot in second turn?




actmyname
Legendary
Offline
Activity: 952
Large scale, green crypto mining ICO


August 15, 2016, 12:10:56 AM 

Would it not be a pure 50/50 chance (ie. "fair") regardless of where you started?
There are an even number of chambers and 2 players, so each player will have 3 chambers that they could potentially use. Regardless of who goes first, each player has an equal number of chambers assigned to them, so after the spin, both players have an equal chance of having a bullet in one of their chambers.
And because you can't change which chambers you have (no spinning) the odds of getting a bullet overall don't change from 50/50.
That is correct answer! But we got a greater challenge for you: This one is die hard. Imagine that you and your opponent start with 1 bullet loaded into the 6 slot. After each try you load one more bullet to the chamber, spin again and pass the pistol to your opponent. QUESTION: Do you prefer to start as first, or shoot in second turn? Going to try the thing I did last time. Tired so I'm probably wrong, rounded to first decimal. R1: P1 has a 1/6 chance to die. (16.6%) R2: 5/6 chance to happen, 2/6 chance to shoot, 10/36 > P2 has a 5/18 chance to die. (27.7%) R3: 5/9 chance to happen, 3/6 chance to shoot, 15/54 > P1 has a 5/18 chance to die. (27.7%) R4: 5/18 chance to happen, 4/6 chance to shoot, 20/108 > P2 has a 5/27 chance to die. (18.5%) R5: 5/54 chance to happen, 5/6 chance to shoot, 25/324 > P1 has a 25/324 chance to die (7.7%) R6: 5/324 chance to happen, 6/6 chance to shoot, 5/324 > P2 has a 5/324 chance to die (1.5%) Tallying up gives: P1 has a ~52.2% chance of death P2 has a ~47.8% chance of death
Probably wrong, maybe. Someone else can do this better than me. Go second?




HCP


August 15, 2016, 11:24:59 AM 

Seems to me like you'd want to go first... because at most, you get 3 attempts... and by going first there is always a chance you don't die on each one of your attempts, whereas by starting 2nd, you end up at 100% chance of death on your 3rd try... plus your odds are technically better for each one of your attempts if you are 1st... 1st attempt: Player A 1/6 die, Player B 2/6 die 2nd attempt: Player A 3/6 die, Player B 4/6 die 3rd attempt: Player A 5/6 die, Player B 6/6 die Although, I'm not exactly sure that is a mathematically sound explanation... perhaps it is more of a "gut instinct" answer??






BitPistol.com


August 20, 2016, 04:30:03 PM 

Great job! Promo of 1 mBTC was credited to your in game account!




actmyname
Legendary
Offline
Activity: 952
Large scale, green crypto mining ICO


August 20, 2016, 06:15:33 PM 

Seems to me like you'd want to go first... because at most, you get 3 attempts... and by going first there is always a chance you don't die on each one of your attempts, whereas by starting 2nd, you end up at 100% chance of death on your 3rd try... plus your odds are technically better for each one of your attempts if you are 1st... 1st attempt: Player A 1/6 die, Player B 2/6 die 2nd attempt: Player A 3/6 die, Player B 4/6 die 3rd attempt: Player A 5/6 die, Player B 6/6 die Although, I'm not exactly sure that is a mathematically sound explanation... perhaps it is more of a "gut instinct" answer?? However, the reality is that for player B to shoot, it requires player A not to die. That would change the probability. For example, for the first attempt: There are Player A dies: 1/6 chance Player A lives, Player B lives: 5/9 chance Player A lives, Player B dies: 5/18 chance Following this, what I'd previously posted begins to flip the odds in B's favor (provided my math is correct) quickly. Right now it seems as if Player A is the better option: and it would be, if it was only one round.




BitPistol.com


August 20, 2016, 10:44:50 PM 

Seems to me like you'd want to go first... because at most, you get 3 attempts... and by going first there is always a chance you don't die on each one of your attempts, whereas by starting 2nd, you end up at 100% chance of death on your 3rd try... plus your odds are technically better for each one of your attempts if you are 1st... 1st attempt: Player A 1/6 die, Player B 2/6 die 2nd attempt: Player A 3/6 die, Player B 4/6 die 3rd attempt: Player A 5/6 die, Player B 6/6 die Although, I'm not exactly sure that is a mathematically sound explanation... perhaps it is more of a "gut instinct" answer?? If i would to choose without calculating odds, then emotionally i picked second turn, as shooting with 2 and 4 bullets sounds for me less fatal then with 1 3 and 5 and i would be pretty sure that i wont have to shoot last 6 bullet turn




BitPistol.com


August 21, 2016, 09:26:42 AM 

Would it not be a pure 50/50 chance (ie. "fair") regardless of where you started?
There are an even number of chambers and 2 players, so each player will have 3 chambers that they could potentially use. Regardless of who goes first, each player has an equal number of chambers assigned to them, so after the spin, both players have an equal chance of having a bullet in one of their chambers.
And because you can't change which chambers you have (no spinning) the odds of getting a bullet overall don't change from 50/50.
That is correct answer! But we got a greater challenge for you: This one is die hard. Imagine that you and your opponent start with 1 bullet loaded into the 6 slot. After each try you load one more bullet to the chamber, spin again and pass the pistol to your opponent. QUESTION: Do you prefer to start as first, or shoot in second turn? Going to try the thing I did last time. Tired so I'm probably wrong, rounded to first decimal. R1: P1 has a 1/6 chance to die. (16.6%) R2: 5/6 chance to happen, 2/6 chance to shoot, 10/36 > P2 has a 5/18 chance to die. (27.7%) R3: 5/9 chance to happen, 3/6 chance to shoot, 15/54 > P1 has a 5/18 chance to die. (27.7%) R4: 5/18 chance to happen, 4/6 chance to shoot, 20/108 > P2 has a 5/27 chance to die. (18.5%) R5: 5/54 chance to happen, 5/6 chance to shoot, 25/324 > P1 has a 25/324 chance to die (7.7%) R6: 5/324 chance to happen, 6/6 chance to shoot, 5/324 > P2 has a 5/324 chance to die (1.5%) Tallying up gives: P1 has a ~52.2% chance of death P2 has a ~47.8% chance of death
Probably wrong, maybe. Someone else can do this better than me. Go second? Hmm... Your calculations seem correct, and sum of percentages to die each round is 100%, looks like you solved it! Now if i stuck in such a challenge, i would go shooting second without hesitations, and for you, nutcracker, i will come up with some new puzzle shortly




BitPistol.com


August 21, 2016, 02:33:41 PM 

Seems to me like you'd want to go first... because at most, you get 3 attempts... and by going first there is always a chance you don't die on each one of your attempts, whereas by starting 2nd, you end up at 100% chance of death on your 3rd try... plus your odds are technically better for each one of your attempts if you are 1st... 1st attempt: Player A 1/6 die, Player B 2/6 die 2nd attempt: Player A 3/6 die, Player B 4/6 die 3rd attempt: Player A 5/6 die, Player B 6/6 die Although, I'm not exactly sure that is a mathematically sound explanation... perhaps it is more of a "gut instinct" answer?? However, the reality is that for player B to shoot, it requires player A not to die. That would change the probability. For example, for the first attempt: There are Player A dies: 1/6 chance Player A lives, Player B lives: 5/9 chance Player A lives, Player B dies: 5/18 chance Following this, what I'd previously posted begins to flip the odds in B's favor (provided my math is correct) quickly. Right now it seems as if Player A is the better option: and it would be, if it was only one round. Ok, here is one more puzzle for those who solved the probability cases, but now it is more physics: QUESTION: What bullet hits glass first on the following image?




The End Is Neigh
Jr. Member
Offline
Activity: 47


August 21, 2016, 02:40:59 PM 

Seems to me like you'd want to go first... because at most, you get 3 attempts... and by going first there is always a chance you don't die on each one of your attempts, whereas by starting 2nd, you end up at 100% chance of death on your 3rd try... plus your odds are technically better for each one of your attempts if you are 1st... 1st attempt: Player A 1/6 die, Player B 2/6 die 2nd attempt: Player A 3/6 die, Player B 4/6 die 3rd attempt: Player A 5/6 die, Player B 6/6 die Although, I'm not exactly sure that is a mathematically sound explanation... perhaps it is more of a "gut instinct" answer?? However, the reality is that for player B to shoot, it requires player A not to die. That would change the probability. For example, for the first attempt: There are Player A dies: 1/6 chance Player A lives, Player B lives: 5/9 chance Player A lives, Player B dies: 5/18 chance Following this, what I'd previously posted begins to flip the odds in B's favor (provided my math is correct) quickly. Right now it seems as if Player A is the better option: and it would be, if it was only one round. Ok, here is one more puzzle for those who solved the probability cases, but now it is more physics: QUESTION: What bullet hits glass first on the following image? It would be #2 based on the cracks of the glass. Right? I don't see any other way it would make sense to me so it must be true




JasonXG


August 21, 2016, 05:58:25 PM 

Why are people sending screenshots ? I thought that promo was over !? Or can I still participate ? Do I get rewarded fpr answering any of these questions ? I've read the op over and over and looked through the messages buy I still nothing of this screenshot promo. I only see people unloading but I'm not sure what for.




BitPistol.com


August 21, 2016, 10:15:04 PM 

Why are people sending screenshots ? I thought that promo was over !? Or can I still participate ? Do I get rewarded fpr answering any of these questions ? I've read the op over and over and looked through the messages buy I still nothing of this screenshot promo. I only see people unloading but I'm not sure what for. This promo is over for you, you was already rewarded, check 4 page of this post. You can participate in this promo only once. Questions are just mind games, nothing to do with promo action




HCP


August 22, 2016, 10:19:10 AM 

It would definitely be bullet #2 that hit the glass first... followed by bullet #3 and last was bullet #1. The reasoning being that #2 doesn't have any cracks that are "stopped" by existing cracks, #3 only has a crack stopped by cracks from #2, and #1 has cracks that are stopped by both #2 and #3. That was a pretty easy one compared to the maths involved with the probabilities of dying in games of Russian roulette




BitPistol.com


August 22, 2016, 05:22:32 PM 

Seems to me like you'd want to go first... because at most, you get 3 attempts... and by going first there is always a chance you don't die on each one of your attempts, whereas by starting 2nd, you end up at 100% chance of death on your 3rd try... plus your odds are technically better for each one of your attempts if you are 1st... 1st attempt: Player A 1/6 die, Player B 2/6 die 2nd attempt: Player A 3/6 die, Player B 4/6 die 3rd attempt: Player A 5/6 die, Player B 6/6 die Although, I'm not exactly sure that is a mathematically sound explanation... perhaps it is more of a "gut instinct" answer?? However, the reality is that for player B to shoot, it requires player A not to die. That would change the probability. For example, for the first attempt: There are Player A dies: 1/6 chance Player A lives, Player B lives: 5/9 chance Player A lives, Player B dies: 5/18 chance Following this, what I'd previously posted begins to flip the odds in B's favor (provided my math is correct) quickly. Right now it seems as if Player A is the better option: and it would be, if it was only one round. Ok, here is one more puzzle for those who solved the probability cases, but now it is more physics: QUESTION: What bullet hits glass first on the following image? It would be #2 based on the cracks of the glass. Right? I don't see any other way it would make sense to me so it must be true Your answer is correct!




