BitPistol.com (OP)
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August 14, 2016, 10:24:23 AM |
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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 ...
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BitPistol.com (OP)
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August 14, 2016, 02:31:05 PM |
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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
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BitPistol.com (OP)
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August 14, 2016, 05:32:36 PM |
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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?
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actmyname
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August 15, 2016, 12:10:56 AM |
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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?
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HCP
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August 15, 2016, 11:24:59 AM |
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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??
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onleines
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Don't buy Bitcoin - Earn it
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August 16, 2016, 01:36:33 PM |
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BitPistol.com (OP)
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August 20, 2016, 04:30:03 PM |
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Great job! Promo of 1 mBTC was credited to your in game account!
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actmyname
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August 20, 2016, 06:15:33 PM |
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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.
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BitPistol.com (OP)
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August 20, 2016, 10:44:50 PM |
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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
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BitPistol.com (OP)
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August 21, 2016, 09:26:42 AM |
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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
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BitPistol.com (OP)
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August 21, 2016, 02:33:41 PM |
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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?
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The End Is Neigh
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August 21, 2016, 02:40:59 PM |
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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? https://i.imgur.com/AJagSye.jpgIt 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
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JasonXG
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August 21, 2016, 05:58:25 PM |
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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.
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BitPistol.com (OP)
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August 21, 2016, 10:15:04 PM |
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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
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HCP
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August 22, 2016, 10:19:10 AM |
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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
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BitPistol.com (OP)
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August 22, 2016, 05:22:32 PM |
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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!
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JasonXG
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August 22, 2016, 05:52:21 PM |
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How do you know bullet 2 hit first ? I'm not convinced. If it was a stub nose or hallow tip you wouldn't know, the whole window would shatter. I think there will be more cracks then that. Most windows are shatter proof so they shatter like small spider webs all around the window. Either that or it shatters completely. That is why when someone breaks into a car you can't use a bat you need something with a small tip. Small tips with a lot of force behind it will shatter a shatter proof window. Without shatter proof the whole thing would just shatter from anything no matter wide or narrow. So a bat would shatter the whole window.
Also recoil climbs and goes side ways. Bullet number 1 looks like it would hot first then the recoils starts to climb. Bullets never ever go top to bottom .
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BitPistol.com (OP)
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August 23, 2016, 09:06:53 PM |
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How do you know bullet 2 hit first ? I'm not convinced. If it was a stub nose or hallow tip you wouldn't know, the whole window would shatter. I think there will be more cracks then that. Most windows are shatter proof so they shatter like small spider webs all around the window. Either that or it shatters completely. That is why when someone breaks into a car you can't use a bat you need something with a small tip. Small tips with a lot of force behind it will shatter a shatter proof window. Without shatter proof the whole thing would just shatter from anything no matter wide or narrow. So a bat would shatter the whole window.
Also recoil climbs and goes side ways. Bullet number 1 looks like it would hot first then the recoils starts to climb. Bullets never ever go top to bottom .
This is getting interesting, basically where i found it, answer wasn't given, but the most comments were for bullet #2. However your explanation will give this puzzle a fresh breath
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HCP
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August 25, 2016, 10:19:08 AM |
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How do you know bullet 2 hit first ?
Did you read my explanation? It is all to do with the way the cracks in the glass have spread. If you follow the way the cracks spread from each hole, you can see that some of the cracks are "stopped" by other cracks... Logically, this means that the crack that STOPS the 2nd crack was already there. The best example is the one going vertically up from #3. It looks like it runs into the crack from #2 (that is running, more or less, from left to right) and stops. Logically, this would mean the crack from #2 was already there, so hole #2 was made before #3. The same happens with the little short crack from #1, travelling to the left... it is "stopped" by the crack coming down vertically from #2. Therefore, #2 was before #1. This means it must have been the first one, because it was before both #1 and #3. You're reading WAY too much into the puzzle with your comments about bullet types and recoil etc.
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