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Combinations of bullet placements:
1,2
2,3
3,4
4,5
5,6
6,1
If you are player 1, u die under 4 of the above scenarios out of 6.
I don't understand. How can player 1's probability of dying be 2/3 when there are only 2 bullets in the 6-chambered gun? Checkerboard, can you please explain.
Loh K L
Combinations of bullet placements:
1,2
2,3
3,4
4,5
5,6
6,1
If you are player 1, u die under 4 of the above scenarios out of 6.
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I'm afraid I still don't understand. In your 6 scenario's, the player will die in all of them. If someone else understands what Checkerboard is saying, please explain it to me. I'm a bit thick between the ears, sorryOriginally Posted by checkerboard
Loh K L
My guess he is referring to Player 1 will die in 4 out of the 6 scenarios, assuming no re-spin.
Considering the 6 scenarios :
L1 L2 E3 E4 E5 E6 - Player 1 will die at first shoot
E1 L2 L3 E4 E5 E6 - Player 2 will die at 2nd shoot
E1 E2 L3 E4 E5 E6 - Player 1 will die at 3rd shoot
E1 E2 E3 L4 L5 E6 - Player 2 will die at 4th shoot
E1 E2 E3 E4 L5 L6 - Player 1 will die at 5th shoot
L1 E2 E3 E4 E5 L6 - Player 1 will die at first shoot
Jacian is right in the explanation. Thanks.
It goes like this:
1) You put 2 successive bullets in.
2) Now look at the barrel and label them, 1-6, with 1 being the position where the first bullet that will be fired at.
Now, if you are player 1. You will die on the first shot if the bullet is 1,2 or 6,1. You will die on the second try if the bullets are placed 3,4 and you will on the last attempt if the bullet is placed 5,6.
If you are player 2, you will only die if the bullets are placed 2,3 or 4,5.
I hope you get it now and sorry if I wasn't clear.
Last edited by checkerboard; 10th Mar 2008 at 11:30.
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do not do to others what you will not want done to you!
be kind! =)
checkerboard, I understand what you are saying but I think that's not what timebomb is saying. I think he refers only to the first trigger by the first player.
why I don't do garden hybrids and aquarium strains: natural species is a history of Nature, while hybrids are just the whims of Man.
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Okay, I think I see what you mean but the maths is seriously flawed. I think you are getting confused. The scenario is very simple. It looks like this:
E1, E2, E3, E4, L5, L6
Don't confuse players 1 and 2. Don't try to work their odds at the same time.
Just work the probability for player 1 first. In the above scenario, there are 6 chambers, 2 of which are loaded. It can't be simpler - Player 1 has a 2/6 or 1/3 (33.3%chance of dying. After Player 1 has taken his turn at the gun and survives, it would mean he has clicked on E1, E2, E3 or E4.
You are Player 2. It's your turn with the gun. If you spin the barrel again, you will be resetting the gun. In other words, you will wipe the slate clean; the probability for you dying will be exactly the same as when Player 1 picked up the gun, at 33.3%.
The question is - If you choose not to spin, what are your chances?
Loh K L
Just a quick guess, only 25% of dying if not choose to spin?
Good guess
Now, do you see why you were all confused the by scenario's that Checkerboard put up. The mistake he made was to try and work the probability for both the players at the same time. You know, this is funny. Because Checkerboard is the one who kept insisting that sequence affects probabilities. In the case of this puzzle, he's right, you know. So how can we work the probability for player 2 when player 1 hasn't taken his turn at the gun yet?
Loh K L
Question, in this case, if you are given a choice of choosing to be player 1 or player 2, will you choose to be player 2 as it has a higher chance of surviving or does it not matter? Thanks.
I think I know what is the issue. Timebomb is considering the game at each shoot, whereas, I am considering the game in totality.
When you say that the first guy has 1/3 chance of surviving the first shoot, I do not disagree with that, what I am saying is, over the course of the entire game, he has a 2/3 chance of dying while player 2 has a 1/3 chance of dying without a re-spin.
So when you say player 2 picks up the gun if he re-spin, his probability is 1/3. I do not disagree that for that 1 shoot, it is 1/3. But for the whole game, he is now effectively Player 1 in the previous game taking the whole game into context.
I think it is a question of interpretation, and timebomb, my analysis earlier did factor in that Player 1 has shot. Under your form of analysis, yes Player 2 has 25% chance of surviving the next shoot as Jacian pointed out.
So the difference is, you are going a shoot by shoot analysis, I am going by the full remaining game analysis. I hope you can see it. An interesting question, which I will think about would be, if we increase the number of bullets, will the full remaining game type of analysis lead to a different decision than the shoot-by-shoot analysis? If it does, which analysis, would you then prefer? I think I will prefer the full game analysis since that's the way I view things.
To Jacian question: yes, I will choose to be player 2, and I will not re-spin if given the chance.
Last edited by checkerboard; 10th Mar 2008 at 15:17.
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do not do to others what you will not want done to you!
be kind! =)
Timebomb: "So how can we work the probability for player 2 when player 1 hasn't taken his turn at the gun yet?"
Yes you can. Just think of the simple case of 1 bullet.
I am player 2. If player 1 survives his shoot, I have 3/5 chance of dying in the remainder of the game. If he is dead, I know I have survived. I do not need player 1 to shoot at his head to know that.
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do not do to others what you will not want done to you!
be kind! =)
Yes, I can see the point of your argument. We can work the probability for Players 1 and 2 at the same time but the probability for Player 2 will change as new information is revealed. In the case of this puzzle, the new information is the fact that "Player 1 didn't die". You ignored this piece of information in your calculation for Player 2's probability of dying. You have to remember this - as new information is revealed, the probabilities for the players change.
This is the part you couldn't understand about the 3 prisoners' puzzle. New information was revealed to Tom and his probability should change if not for the fact that he learned nothing that concerns him. He did though learned something about Harry. You thought that if it's either Tom or Harry, the probabilities have to be equal between the 2, 50-50. But the probability does not have to be equally shared. It could be, as the logic indicates, that Tom has a 1/3 chance while Harry has 2/3.
Now back to the Russian Roulette puzzle - I have worked the probabilities for any number of bullets in the gun, inserted in successive chambers. It could be 2, 3 or 4 bullets, it's always better for the second player to not spin the barrel. If there are 5 bullets in the gun and Player 1 survives his turn, just pick up the gun and shoot him
Your argument about full-analysis and shoot-by-shoot analysis is also flawed. Let's imagine a very special gun that has 100 chambers. We insert 98 bullets in the gun, leaving only 2 successive chambers empty. Your opponent goes first. What are his chances? It's not very good. He has a 98% chance of blowing his brains off. But he's one lucky dude. Despite the odds, he survives; he clicked on one of the 2 empty chambers.
It's your turn. What do you want to do? If you spin the barrel again, you reset the probability back to 98% chance of dying. If you do not spin, your chances are 1/2 (50%). Going by your full-analysis, you should spin the gun because after Player 1 has taken his turn at the gun and survived, your probability of dying has got even got worse, from 98% to 99%.
Checkerboard, this is not a language puzzle. There can only be one interpretation and one correct answer.
"Going by your full-analysis, you should spin the gun because after Player 1 has taken his turn at the gun and survived, your probability of dying has got even got worse, from 98% to 99%."...
The above statement is not correct:
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Ok, the full game analysis for the 100 chamber game.
Player 1 survives on 2 scenarios, where the empty chamber is the last one and the first one, and when it is the first and second one.
So, given that Player 1 survives, it has to be one of the 2 scenarios. In one scenario, I survive, in another, I am dead. So my prob is 50%.
Prob of player 1 surviving the next shot if I survive is zero.
If I respin, my prob of surviving is only 2/100 and even if I survive this one, the chance of me not surviving thereafter is 50%, because player 2 chance of survival will be 50%. So my prob of surviving another round is essentially 50%, but chance of me reaching there is only 1 in 50.
So I will not re-spin. Not re-spinning strategy dominates re-spinning in this case. You are wrong to say the full analysis strategy will be to re-spin. I do not think you quite understand the full game strategy, but maybe because I am really bad at explaining.
Last edited by checkerboard; 10th Mar 2008 at 18:49.
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do not do to others what you will not want done to you!
be kind! =)
Anyway, I think I am confident of my answers just as you are, which is fine. I agree it is not a language puzzle.
I do not necessarily think that the shoot-by-shoot analysis is optimal. It may though I am not sure, give the same answers as the full game analysis on a case by case basis.
Let me see if I can think of a prove by contradiction.
For my perspective of this ongoing debate: I do hope you can go back to the previous scenario/story I posed on the 3 prisoner problem and prove by contradiction why it is not half. You have yet to do that and continue to insist I am wrong. Like I say, I will be more than happy to acknowledge that you are right if you can disprove that scenario.
Last edited by checkerboard; 10th Mar 2008 at 18:50.
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do not do to others what you will not want done to you!
be kind! =)
"In the case of this puzzle, the new information is the fact that "Player 1 didn't die". You ignored this piece of information in your calculation for Player 2's probability of dying. You have to remember this - as new information is revealed, the probabilities for the players change. "
I have included this new information already...
"If player 1 survives his shoot, I have 3/5 chance of dying in the remainder of the game"
So, if player 1 survives, my chance of dying is 3/5. Like I say, I do not need player 1 to really survive, to know that.
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do not do to others what you will not want done to you!
be kind! =)
Interesting thread. Frankly speaking, I did not read through all the examples and understand the 'real arguements' here but my guess is about whether the sequence is important.
If I take KL example on the 5 matchsticks, counting the sequence or not will end with the same results.
For first player, his chance is obvious 1/5
For second player, his chance is (4/5) for first player not getting the short matchstick and him getting it (1/4) = (4/5)(1/4) = 1/5
For third player, using same logic, = (4/5)(3/4)(1/3) = 1/5
Now, we take a look at checkerboard example
For the first ball, it is clearly 4/24. For the second ball, if we are to consider the sequence, it will be (20/23) for B not to get red, (19/22) for C not to get red, (18/21) for D not to get red and (3/20) for A to get red in 2nd try and so on.
The final equation will be
(4/24)(20/23)(19/22)(18/21)(3/20)(17/19)(16/1(15/17)(2/16)(14/15)(13/14)(12/13)(1/12)
which works out to be (4/24)(3/23)(2/22)(1/21).
You are right on the ball example. Let me see if it is a universal solution for different scenarios because i think we may be lucky that it is one ball drawn, 4 red ball, 4 players, such that the fractions cancel off nicely. Will be interesting if it is universally applicable.
Thank you for pointing it out
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do not do to others what you will not want done to you!
be kind! =)
Thanks Jacian, So far i did some trials and it seems to really cancel off nicely.Wow!
This is interesting, cause it what appears a conditional case yields the same results as though the first person can draw it 4 times striaght. I will need to think what are the conditions that made it work in such a way as conditional probabilities are certainly different from independent ones.
Many thanks. I learn something new today.
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do not do to others what you will not want done to you!
be kind! =)
What about this? Tom is waiting to board a ferry. The ferry has 100 seats and Tom is last in the line. The first person randomly takes one of the 100 seats which by chance could be his assisned seat.
For each subsequent passenger, if his assigned seat is vacant, he will take it. If his assigned seat is already taken, he will take one of the remaining seats randomly.
So, for Tom, who is the last passenger to board the ferry, what is the chance of him finding his assigned seat vacant?
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