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Author Topic: BitPistol - Are you smart enough to solve these puzzles? >>>TEST your brains <<<  (Read 5567 times)
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August 14, 2016, 12:20:55 AM
 #121

Answer i sent pm

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August 14, 2016, 03:35:47 AM
 #122

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).


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August 14, 2016, 09:52:12 AM
 #123

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
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August 14, 2016, 10:24:23 AM
 #124

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 ... Roll Eyes
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August 14, 2016, 02:31:05 PM
 #125

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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August 14, 2016, 05:32:36 PM
 #126

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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August 15, 2016, 12:10:56 AM
 #127

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?

Known alts of actmyname (unofficial list by members with -2 trust or more): DarkStar_, Lauda, Lutpin, The Pharmacist, satoshi, theymos, thermos, the Monopoly man, Charlie Sheen, Shaquille O'Neal/s
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August 15, 2016, 11:24:59 AM
 #128

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?? Huh

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August 16, 2016, 01:36:33 PM
 #129

Good looking site.
Funny idea.

I did add it to Gaming4coin
http://gaming4coin.com/bitpistol-russian-roulette-652.html

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August 20, 2016, 03:11:21 PM
 #130

Good looking site.
Funny idea.

I did add it to Gaming4coin
http://gaming4coin.com/bitpistol-russian-roulette-652.html

Thanks!  Wink
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August 20, 2016, 04:30:03 PM
 #131

I done it

http://prnt.sc/c6qa00

Thanks Wink

Great job! Promo of 1 mBTC was credited to your in game account!
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August 20, 2016, 06:15:33 PM
 #132

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?? Huh
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.

Known alts of actmyname (unofficial list by members with -2 trust or more): DarkStar_, Lauda, Lutpin, The Pharmacist, satoshi, theymos, thermos, the Monopoly man, Charlie Sheen, Shaquille O'Neal/s
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August 20, 2016, 10:44:50 PM
 #133

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?? Huh

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  Tongue
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August 21, 2016, 09:26:42 AM
 #134

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  Cheesy
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August 21, 2016, 02:33:41 PM
 #135

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?? Huh
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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August 21, 2016, 02:40:59 PM
 #136

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?? Huh
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  Tongue
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August 21, 2016, 05:58:25 PM
 #137

Why are people sending screenshots ? I thought that promo was over !? Huh
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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August 21, 2016, 10:15:04 PM
 #138

Why are people sending screenshots ? I thought that promo was over !? Huh
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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August 22, 2016, 10:19:10 AM
 #139

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 Tongue

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August 22, 2016, 05:22:32 PM
 #140

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?? Huh
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  Tongue

Your answer is correct!
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