Simon Posted April 28, 2007 Report Posted April 28, 2007 The following probability riddle is adapted from a debate about individual risk vs frequency of risk. An example was given where a trigger happy soldier played Russian Roulette indefinitely. Before every round, the cylinder of his gun contained 4 blank bullets and one live. If he survived, he reloaded and spun the cylcinder - resetting the odds. Before every round, he claimed: "I only have a 1/5 chance of getting killed, no matter how many rounds went before. The past has gone and can't affect the present. I am alive. The only risk I take is now." His mates try to reason with him: "What about the rounds in the future. Can't you see? The more rounds you play, the greater the danger?" He replied: "No. If I play a round in the future, it means I definitely survive this one. So where is the greater danger? It's still only a 1/5 chance, both then and now!" Common sense tells us that the more mortal risk someone has been exposed to, the more likely it is he will be dead. But is this really true? Or is the soldier making some kind of sense? Here is a fuller version of the puzzle that explores this paradox: A soldier is about to enter a private room to play Russian Roulette by pointing a gun at his head. The gun's cylinder contains five bullets. Four are blank. One is live. The soldier has spare blank ammunition to reload if required. Before the game commences, the soldier is randomly given one of five cards. Each card has a different number - 1, 2, 3, 4 and 5. The card he receives tells him how many rounds he must attempt to play. There will be no cheating. He will follow the card's instruction unless death intervenes. No-one will know what is on the card except him. For every round he plays, there must always be four blank and one live bullet in the cylinder. If he survives a round and has to play again, he will replace the blank with a spare and spin the cylinder - always resetting the odds. The game is over when the quota on the card has been completed or the soldier is dead. Either or both these conditons may finish the game. The soldier enters the room and shuts the door. We don't know what card he was given. The only information we definitely recieve is how many shots are fired. Depending on the number of shots that were heard from inside the room, what is the percentage probability that the soldier is alive? Consider your answer: a) If one shot is heard.:) If two shots are heard.c) If three shots are heard.d) If four shots are heard.e) If five shots are heard. Simon Quote
ronthepon Posted April 28, 2007 Report Posted April 28, 2007 It's good to simplify it all into expressions. Probability of getting card numbered 'x' where x can vary from 1 to 5, 'P(x) = 0.2 Probability that the soilder shoots himself in the head while he fires the nth bullet (n begins from one, and ends at 5) is 1/(5-n+1) = 1/6-n Obviously the events are as independent as can be. Now the events. Oh, but now it gets tough. Quote
ronthepon Posted April 28, 2007 Report Posted April 28, 2007 One shot was fired. CasesHe got a 'one' card.He got a higher card and died on the first shot.He got a one card: 1/5. He got a higher card: 4/5. If he's alive, then the case will be a 4/5 thing; once he's got a one. So his chances of being alive = (chances of getting a 1)*(chances of shooting a blank). = (1/5)(4/5) = 4/25 Obviously, if he's shot only one shot, the chances of him being alive is small. Quote
ronthepon Posted April 28, 2007 Report Posted April 28, 2007 Not that what I've got is something of a conditional probility. We're assuming that one shot is fired only before we sit down to find his chances of remaining alive. Quote
ronthepon Posted April 28, 2007 Report Posted April 28, 2007 Now if he's got two shots fired, eitherHe's got a 2 cardOr he's got a higer one, and died.Assuming he got a 2 (probability of 1/5, but that's not important) he's alive by a factor of 4/5. But he might have got 3,4,5, and these cases kinda compete with the probability of 2 actually coming. Hence, the probability of him having got a two is not 1/4, and not 1/5 as it was earlier. The the probability that he's still alive is (4/5)(1/4) = 4/20 Going on a similar path, I'd say that the probabilites of him remaining alive when we had heard 3, 4 or 5 shots are (1/3)(4/5) = 4/15 for three shots. (1/2)(4/5) = 4/10 for four shots. (1/1)(4/5) = 4/5 for five shots. So the more number of shots we hear fired, the higher hopes we can place about seeing him alive.:) However, there remains the question of wether he will actually be able to make those shots. That analysis is another matter, and the results derived above can be used to quickly answer that. Quote
Simon Posted April 28, 2007 Author Report Posted April 28, 2007 The more shots we hear fired, the higher hopes we can place about seeing him alive. Fascinating! Taken literally: you're saying that the more risk the soldier is exposed to, the greater his chance of survival. At this point I won't confirm or dispute this remark. All I will say is that some people may have a problem with it. As to your actual answers, expressed as percentages they read as follows: a) If one shot is heard, 16% chance the soldier is alive.:) If two shots are heard, 20% chance the soldier is alive.c) If three shots are heard. 26.66% chance the soldier is alive. d) If four shots are heard, 40% chance the soldier is alive.e) If five shots are heard, 80% chance the soldier is alive. By my calculations, only one of these answers is correct. Also your initial formulas interest me. Probability of getting card numbered 'x' where x can vary from 1 to 5, P(x) = 0.2 Probability that the soldier shoots himself in the head while he fires the nth bullet (n begins from one, and ends at 5) is 1/(5-n+1) = 1/6-n The first one is self-evidently correct. The second one puzzles me. It seems potentially ambiguous what "n" refers to. My interpretation was that n = the number of shots fired. Thus, you say the probabilty that the soldier is dead after the nth shot is "1/6-n". So for example: If we know that only 1 shot is fired, you assess the probabilty the soldier is dead as 1/6-1 which is 1/5 or 20%. The soldier has an 80% chance of being found alive. Whether you're correct or not, this sounds plausible. On the other hand, if we know that 5 shots are fired, you assess the probabilty the soldier is dead as 1/6-5 which is 1/1 or 100%. The soldier has a 0% chance of being found alive. Does this sound plausible? This is simply based on what you first came up with. Correct me if I misunderstood you. Simon Quote
ronthepon Posted April 28, 2007 Report Posted April 28, 2007 Say, you're right! I completely ignored the fact that the bullets were limited! It appears that I've not been subtracting the bullets from the gun.:doh: We'll have to edit out the mistake from the solution then. Now if he's got two shots fired, either He's got a 2 cardOr he's got a higer one, and died.Assuming he got a 2 (probability of 1/5, but that's not important) he's alive by a factor of 3/4. (four shots are left now, and one is lethal. Note that it's [math]1 - \frac{3}{4}[/math]) But he might have got 3,4,5, and these cases kinda compete with the probability of 2 actually coming. Hence, the probability of him having got a two is not 1/4, and not 1/5 as it was earlier. The the probability that he's still alive is (3/4)(1/4) = 3/16 Going on a similar path, I'd say that the probabilites of him remaining alive when we had heard 3, 4 or 5 shots are (1/3)(2/3) = 2/9 for three shots. (1/2)(1/2) = 1/4 for four shots. (1/1)(0/1) = 0 for five shots. Aaah. That changes everything. Thanks simon, this'll keep me from repeating this mistake. But let's see the outcome. a) If one shot is heard, 16% chance the soldier is alive.b) If two shots are heard, 18.75% chance the soldier is alive.c) If three shots are heard. 22.22% chance the soldier is alive. d) If four shots are heard, 25% chance the soldier is alive.e) If five shots are heard, 0% chance the soldier is alive. He's dead. I'm pretty confident that this is what you got. About the 1/6-n thing, it came from the fact that after the first shot, there would be five bullets with only one good one.In each shot only one good bullet would remain, thus the numerator would remain one. The denominator would reduce by one after each step, and it had to start with 5. So... er... you make a shortened function that just happens to be agreeing with the function you look for. Quote
ughaibu Posted April 28, 2007 Report Posted April 28, 2007 You two might be interested by this one: consequently.org/news • On the Cable Guy Paradox Quote
Simon Posted April 28, 2007 Author Report Posted April 28, 2007 Sorry Ron, but I'm afraid you have wandered further off the path! :) Having had a brainwave, you said: I completely ignored the fact that the bullets were limited! I've not been subtracting the bullets from the gun. Unfortunately that was the wrong 'eureka'! Your original assumption was correct first time. The gun always does have the same number of bullets when fired! The puzzle made this clear: For every round he plays, there must always be four blank and one live bullet in the cylinder. If he survives a round and has to play again, he will replace the blank with a spare and spin the cylinder - always resetting the odds. Having mistakenly thrown out a premise that was in fact true, none of your last answers are correct. I refer back to your previous set of answers. By my calculation, one of them was correct, the other four were wrong Simon Quote
CraigD Posted April 28, 2007 Report Posted April 28, 2007 Without delving into any logical thought of the Conundrum, I wrote a simple program to play the “Russian roulette in a room” game according to the rules given, and tally the number of winning (live) and losing (dead) soldiers, and the number of shots heard. Here are the results:Shots Live Dead Chance of a Live guy heard guys guys 1 4 5 .4444444444444444444 2 16 16 .5 3 64 48 .5714285714285714286 4 256 128 .6666666666666666667 5 1024 256 .8Here are the results for a slight variation of the game, where each soldiers is given one of 10 cards numbered 1 throught 10Shots Live Dead Chance of a Live guy 1 4 10 .2857142857142857143 2 16 36 .3076923076923076923 3 64 128 .3333333333333333333 4 256 448 .3636363636363636364 5 1024 1536 .4 6 4096 5120 .4444444444444444444 7 16384 16384 .5 8 65536 49152 .5714285714285714286 9 262144 131072 .6666666666666666667 10 1048576 262144 .8In either case, the more shots are heard, the greater the chance that the soldier lived. Interestingly, the chance of the soldier living when the maximum number of shots is heard is the same (.8). Other interesting coincidences are left to the reader to notice. Further interesting coincidences may be seen when considering the simple probability of winning the Russian roulette game for a given number of shots, which is simple the compound probability of winning a single round ([math]p_1=\frac45[/math]), to the power of the number of shots ([math]p=p_1^n[/math]):Shots Chance of Chance of Losing Winning 1 1/5 4/5 .8 2 9/25 16/25 .64 3 61/125 64/125 .512 4 369/625 256/625 .4096 5 2101/3125 1024/3125 .32768 6 11529/15625 4096/15625 .262144 7 61741/78125 16384/78125 .2097152 8 325089/390625 65536/390625 .16777216 9 1690981/1953125 262144/1953125 .134217728 10 8717049/9765625 1048576/9765625 .1073741824The probability of a win when the maximum number of shots are heard is the same as the probability of winning on round of the game. (which follows as a special case of the “Russian roulette in a room” game, when the soldier is give 1 of 1 cards numbered 1 through 1) Here’s the little MUMPS program that produced the data in the first 2 listings:r X2,!,X3,! f C=1:1:A i '$p(B,",",C) s B=$p(B,",",1,C-1) q x X2 s:$p(B,",",C-1)=1 $p(B,",",C-1)=2 f s E=$p(B,",",C),E=$s(E<M:E+1,1:""),$p(B,",",C)=E,C=C+$s(E:1,1:-1) q:C<1!(C>A)!(E=1) k C s M=5,N=5 f A=1:1:N s B="" f x X3 q:'B x X2 s:'$p(B,",",C) C=C-1 s C(C,B[1)=$g(C(C,B[1))+1 f C=1:1:N,0 w $j(C,4),$j(C(C,0),10),$j(C(C,1),10)," ",C(C,0)/(C(C,0)+C(C,1)),!PS: If you check my work, you’ll notice it's WRONG – the probability of the soldier drawing each of the cards is not equal, as specified, but proportional to the number of possible winning and losing plays. I’ll post correct numbers soon. PPS: Approaching the calculation with equal probabilities for each card, the results are unchanged. So, though sheer luck or profound unconscious intuition ;) , the results above are correct Here’s the problem expressed as subdivided probabilities:The probabilities of a card being drawn:1=.2 2=.2 3=.2 4=.2 5=.2 The probability of losing in the first round, or winning if the card drawn was “1”:1=.2 1-1=.04 1-2:5=.16 2=.2 2-1=.04 3=.2 3-1=.04 4=.2 4-1=.04 5=.2 5-1=.04Continuing until reaching the probability of losing or winning in the 5th round:1=.2 1-1=.04 1-2:5=.16 2=.2 2-1=.04 2-2:5,1=.032 2-2:5,2:5=.128 3=.2 3-1=.04 3-2:5,1=.032 3-2:5,2:5,1=.0256 3-2:5,2:5,2:5=.1024 4=.2 4-1=.04 4-2:5,1=.032 4-2:5,2:5,1=.0256 4-2:5,2:5,2:5,1=.02048 4-2:5,2:5,2:5,2:5=.08192 5=.2 5-1=.04 5-2:5,1=.032 5-2:5,2:5,1=.0256 5-2:5,2:5,2:5,1=.02048 5-2:5,2:5,2:5,2:5,1=.016384 5-2:5,2:5,2:5,2:5,2:5=.065536Collecting the probabilities of winning or loosing in a given round, summing and dividing to give the probability of winning for a given number of shots heard:1 .16/(.04*5+.16) = 4/9 = .4444444444444444444 2 .128/(.032*4+.128) = 1/2 = .5 3 .1024/(.0256*3)+.1024) = 4/7 = .5714285714285714286 4 .08192/(.02048*2+.08192) = 2/3 = .6666666666666666667 5 .065536/(.016384*1+.065536) = 4/5 = .8 Quote
Simon Posted April 28, 2007 Author Report Posted April 28, 2007 Thank you Craig. Your results tally precisely with mine. Curiously, I thought of exploring a 10 round scenario - but with 10 bullets in the gun. In any event the principle is the same in all such variations. This game appears to offer an impossible conclusion in terms of risk frequency. Less risk = more danger. More risk = less danger. And yet if only one shot is heard, we know for certain that only one round was played. The individual risk of someone getting killed by one round is only 20%. Nevertheless, if no other shots are fired, the chance of him really being dead from that single round is about 55%. The odds have been changed by the soldier's possible commitment to future rounds - though we know those rounds did not take place. The additional probabilty of his death has nothing to do with the amount of risk he was exposed to. Instead it reflects the risks not taken and the soldier's unknown intentions. In other words, the increase from 20% to 55% probability of death by one round is based on the non-occurance of other rounds. What if only one shot was fired, but the soldier had the capacity to play up to 10 rounds, as in Craig's scenario? The probability of his death would become 71%. What if only one shot was fired, but the soldier might have played anything up to a million rounds? The unplayed future rounds would raise the odds of his death to 99%. As far as counter-intuition goes, this is 'Alice in Wonderland' ! Simon Quote
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