Exploration of the three-person duel

Transcription

1 Exploation of the thee-peson duel Andy Paish 15 August The duel Pictue a duel: two shootes facing one anothe, taking tuns fiing at one anothe, each with a fixed pobability of hitting his opponent. The complete analysis of this two-vaiable poblem is simple. Let s take a look. Let the fist shoote hit the second with pobability p, and miss with pobability p. Of couse we have p + p = 1. Likewise, define q and q. So the fist shoote wins exactly when he lands the fist shot. Thus, the pobability that the fist shoote wins is given by the following geometic sum: p + pqp + p 2 q 2 p + p q p +... = p And the pobability that the second shoote wins is given by: pq + p 2 qq + p q 2 q + p 4 q q +... = pq It should eassue the eade that, since exactly one of the shootes must win the duel, the fomulas do indeed add to 1. Using these fomulas, analysis is staightfowad. Eithe playe inceases his chance of winning by inceasing his chance of hitting. By satisfying p = q 1 + q we equalize each playes pobability of winning that is, they each will win with pobability The tuel poblem Now that we can handle two shootes, what happens if we add a thid? The pictue gets moe fuzzy. Is it even clea who to shoot? To answe this and othe questions, fist we must specify some ules. Each membe of this thee-peson duel o tuel goes in a pedefined ode playe 1, then 2, then, and epeat as necessay fiing a shot at one of the othes. Each is equied to do his best to hit the opponent of his choice, athe than 1 2 1

2 stategically fie into the ai. The tuel continues until one combatant emains standing. Say the pobability that playe 1 hits his taget is p, with his chance of missing q 1 again satisfying p + p = 1. Likewise we have q, q,, and, with all of these quantities univesally known. Each paticipant is standing sufficiently fa apat so that they cannot accidentally hit the wong taget. Now we can ask, and answe, who each playe s taget should be. Conside the playe cuently attempting his shot. If he takes misses, no matte his taget, the tuel continues with the next playe. If he shoots and hits his taget, then he finds himself in a duel with the emaining playe, with the latte going fist. Since it is always pefeable to duel someone with wost possible maksmanship, any playe s natual taget is whicheve peson has the best maksmanship. Thus, independent of who is aiming at a paticula playe, we see that his taget is fixed. With this woked out, the following fomulas and moe can be geneated fo each playes chances at winning. These apply wheneve p > q > : p p P 1wins = 4 1 p q pq + pq P 2wins = 5 P wins = All fomulas may be found in the appendix Equilibium solutions p 1 p + pq + pq2 Fom hee, we can seach fo tiples p, q, such that each playe will win a thid of the time, similaly to. Fo instance, conside 7 1, 1, 5 7 9, which is oughly.55, 1,.26. Hee, 1 will aim fo 2, and hit 55% of the time, and will poceed to win the duel with about 60% of the time, meaning 1 will win oughly 4% of the time. Meanwhile, if 1 misses 45% of the time, then 2 will suely hit him and poceed to beat 74% of the time, making 2 the winne % of the time. This leaves % of the victoies to playe. So oughly they wok out close, and in fact the exact numbes wok out pefectly. It is supising that you can have an objectively fai tuel in which one of the playes neve misses! Thee exist these so-called equilibium solutions in each egion of shooting pattens that is, fo p > q >, p > > q, etc. While it is not obvious that each egion should have these solutions, the following points show that they do some solutions ae appoximate. Note that 4, 5, and 6 apply only when p > q >, so these cannot be used to veify all tiples given. p > q > : , , p > > q: 0.250, 0.115, q > p > : 7 1, 1,

3 q > > q:, 1, > p > : 0.418, 0.250, 1 > q > p: 0.26, 0.86, 1 Now, looking to calculus, the Implicit Function Theoem eveals that, in fact, each of these points exists on a continuum of such solutions. This is intuitive when viewing the fomulas as 2 equations in vaiables the thid equation is foced by the necessity that all thee must add to 1. 4 Gaphical analysis Although we have been able to find many equilibium solutions, and have shown the existence of infinitely many, this still does not give a stong undestanding of the dynamics of this poblem. If I m in a tuel, with my fixed maksmanship, how good do I want my opponents to be? This is not at all clea. Of couse if I m much bette than both opponents then I should be okay, but beyond that it s had to say anything with cetainty. While fomulas help to bing pecision to a poblem, nothing beats pictues at showing whats eally going on. I have ceated sets of gaphs, one set fo each shoote in a tuel. In each gaph, we fix one maksmanship, and allow the othe two to vay. Each colo depicts which shoote has the highest chance of suvival. Befoe looking at what these gaphs show us, its woth making some notes on intepeting them. The gaphs ae available at Each gaph is divided into egions of shooting pattens. Fo the gaph of Pi hits = ρ, the bounday lines ae given by y = x, y = ρ, and x = ρ. In the bottom left of the gaph, we get the case when i is the best shoote, with the line y = x detemining which of the othe two shootes is next best. Although this egion is navely though of as the best fo i, it is impotant not to ule out the consequences of two slightly wose shootes teaming up against a common taget. In the top ight of the gaph, we find the egion in which i is the wost shot of the, with y = x seving the same pupose. Again we find that the nave expectation that this is a bad position fo i is not always the case hee. Instead, being the wost shoote allows i to guaantee his will not be the fist to lose, and in fact will often have the fist shot advantage in the esulting duel. Finally, In the top left egion of this gaph, we see the esults of when the fist shoote is the second best shot. He theefoe shoots at the best, but is shot at by the best as well. The thid shoote also aims fo the best. The bottom ight of the gaph is the same stoy, with the othe two shootes oles evesed. These egions ae not typically the best place fo i, because he finds himself to be the taget of the best shoote of the tuel. Howeve, the ight conditions can make this set-up vey comfotable. So now that we know how to intepet them, what is thee to lean fom these gaphs? Fist, lets examine them with espect to equilibium solutions when each shoote has the same chance of suvival. These solutions ae actually vey clealy

4 pesented in the gaphs. They occu any time that seas of ed, geen, and blue in the same egion all meet at a common point. Thus the fist gaph, P1 hits = 0.2, shows 6 equilibium solutions one in each egion. In paticula, it tells us thee is an equilibium nea 0.1, 0.2, Sue enough, the pobabilities of suvival at that point ae oughly 0.4, 0.2, 0.4. This is a vey simple way of ball-paking these solutions! Futhemoe, looking at the pogession of gaphs fo a given shoote, we can watch the movements of a cetain family of equilibium points as a given shoote inceases/deceases his maksmanship. Although analytic methods show that such families of solutions must exist, it is ewading to see them in these gaphs. Now, lets look the diffeent egions of a paticula shootes victoies. Fo a given playe, i, his gaph shows potentially 6 egions of victoy one fo each egion of shooting pattens. The two in the bottom left of the gaph when i is the best shot ae always connected to one anothe, and lives in the bottom ight of the egion when i is much bette than the othe two. This matches intuition, as two shootes slightly wose than i should be able to team up and beat i. Howeve, if we look at this egion fo shoote 1, then 2, then, all at a fixed maksmanship level, well see the coesponding egions shink fom 1s domination down to s moe wose showing. This is a good pictue of the advantage of going ealie in the ound. Shoote 1 wins most of the time when he is the stongest, but loses most of the time, while 2 falls somewhee in the middle. Similaly, we can estict ou inteest only to the egions whee i is the wost shoote. Hee, we see shoote 1s suvival-of-the-weakest egion gow with his maksmanship, until it finally tops off and dies away into nothing by the time hes a.6 shot. Meanwhile, shootes 2 and find thei coesponding egions gow to fill the entie egion! This may seem stange, but as I said befoe it actually makes good sense. As the wost shoote, they definitely make it to the duel when the fist man dies. Howeve, since theyll typically go fist hee, thei 0.5-o-bette shot makes them the favoite to win. So in this case, paallel situations give the fist shoote an almost sue loss, and the second and thid a pobable win. Finally, vey diffeent things occu in the othe egions depending on position in the tuel. Shoote 1 finds himself winning most tuels in which he is the middle shot, though sufficiently stong opponents can always bing him down. Meanwhile, shoote 2 will almost neve win when hes second to 1, but almost always when second to. Shoote has the same touble when a stong 1 fies at him, but can often beat a stonge 2. 5 Impovements What ae some impovements which could make these gaphs moe useful? Instead of a winne-take-all coloing scheme, a moe sophisticated gaphics pogam could mix colos 50% blue, 0% ed, and 20% geen, fo example to give a stonge sense of how good each playes chances ae. This would tun all 4

5 equilibium points gay, and help give a cleane pictue of the continuity within a given gaph. Additionally, it would be vey inteesting to have enough of these gaphs in a flipbook say 100 fo each shoote to watch a clea, continuous animation of victoy pattens. While it isn t too difficult to intepolate these gaphs mentally, watching the ates of change could be enlightening. Coupling this with a continuum of colos would make fo a vey detailed, compehensive, and attactive, pesentation of this tuckload of infomation. 6 Appendix: Fomulas We get a set of thee fomulas fo each odeing of p, q, and. p > q > : p p P 1wins = 1 p q pq + pq P 2wins = p P wins = 1 p + pq + pq2 p > > q: p P 1wins = q p P 2wins = P wins = q > p > : q > > p: P wins = P 1wins = qp + pq + pq pq + pq 2 p p + pq P 1wins = 1 p q pq P 2wins = p 1 p + pq + p2 q 1 p p p 1 p + pq + pq 1 p

6 q p 2 q P 2wins = p + p 2 q P wins = 1 p > p > q: p P 1wins = q p P 2wins = P wins = pq + pq + p2 q + pq pq 2 > q > p: p pq P 1wins = + pq + pq 1 p q p + p 2 q P 2wins = p 2 q P wins = 1 p