MIND YouR Ps AND Qs ABSTRACT

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2 . ABSTRACT I the Qwerty problem [ 1] usig two fair dice, the Qwertias will give you 2 kg of super eutro fuel each time you throw a total of 2, 3, 4, 1 0, 11 or 12 ad you will give them 2 kg otherwise. You start with 6 kg of fuel. If you, gai a further 6 kg of fuel, your space ship ca just get you back to Earth. If you lose the 6 kg of fuel you started with, you will be a slave o Qwerty for 10,000 years. But you must gait ~kg before you lose 6 kg! What is your chace of freedom? From past experiece, we are pretty sure that we kow what your chace pis (ad it ai't good). We have discovered that Here f is the umber of ways of throwig a sequece of wis() ad losses ( )*of 2kgs of fuel, so that we ever lose 6 kgs before we gai 6 kg. We kow that ] 0 = 1, } 1 = 3 ad ] 2 = 9 ad we thik that J = 3". But is it? We also tried usig the followig tree diagram. It tured out to be of some use. 0 Now, as they say, read o. MIND YouR Ps AND Qs We've bee hell bet o tryig to fid the wretched f. Maybe we ca go roud it somehow. Maybe we do't have to fid a specific expressio for f (like 3"). Perhaps we ca creep up o p via the back door '" *We show how to get } 2 o p.77. :

3 So how could we do that? What do we kow about p that we could use? What do we kow about the whole Qwerty problem that we have't used yet? Whe you thik about it, we've oly cocetrated o p, the probability of escapig. We've give o thought to q, the probability of beig eslaved for a small period of 10 milleia. Maybe q is actually easier to fid tha p. If ihs the, as they say, we're laughig. Clearly p q = 1. You ca oly escape or be a llave. You may be able to stave off the exit hour by several millio throws of the dice but they'll eittfer get you evetually or you'll escape. (Of course, we're assumig you wo't die first. But who wahts to worry about that possibility?) \ Naturally we ca fid q usig a system of equatios as we did with p. That will clearly give us q =.! but that does't give us ay ew isight ito the problem (It might be a useful 9 exercise though to check that q =.! ad so practice the process show i the last article.) How 9 else could we get q? Hmmm. We kow this does't help much but q = fg(!) a( 2 ) b, we mea it's certai =O 3 3 that q has the same form as p whe expressed as a ifiite sum. Of course, i this case g is the umber of ways of gettig to a -6 situatio by "throwig" pluses ad miuses. Here, istead of there beig miuses, there are pluses. So here there must be three more miuses tha pluses. That meas a = ad b = 3, which gives So what is g? Now that's the wrog questio. If we could fid g directly, we'd be able to fid f directly ad we would't be i this pickle i the first place. OK, thik. How ca we fid g without fidigg? Is there some lik betwee g ad f (like the lik betwee p ad q, p q = 1), that we ca exploit? What is g? What is j? What? The tree diagram was useful before (refer [ 1 ]). Maybe we ca use it agai. Now f could be thought of as the umber of ways of gettig to a "good" ode, havig "lost" the toss times. I the same way, g is the umber of ways of gettig to a "bad" ode, havig "wo" the toss times. Hey. Surely ot. Could g be equal to f? Is that possible? If it is, the there is a good reaso for it. Symmetry!? Is't it all symmetric? Suppose we've got to a good ode usig miuses. What happes if we chage all the pluses to miuses ad all the miuses to pluses? Surely that shows us how to get to a bad ode. Ad vice-versa. For every way of gettig to a bad ode, there's a mirror-image way of gettig to a good oe. " " Fatastic! So f = g! So what? It's ice to kow, but how ca we use it? What is it that we've actually got?

4 Let's recap for a miute. We kow that ad ad f = g. So Ah but the, surely, q = 8p? Yes, that looks good to us. Ah but ow we're i with a show because we kow that ad q = 8p. Naturally the q = 8p = t 1 so p =.!.. Bigo! 9 But, but. It would be easy to give up at this stage. We've solved the Qwertia problem i two ways already, maybe three, or four. But we still do't kow for sure that f = 3". All that garbage at the ed of the last article [ t] about two ifiite sums is garbage. Clearly, well maybe ot, but it ai't "" "" true ayway, it is't the case that if La = Lh, the a= b. =O =O It's easy to see if you thik about it. Let a 2 = h 2 l ad a 2 l = h 20 That should do it. We apologise for eve brigig it up. But we still do't have a explicit value for fo. What to do?

5 Let's go back ad look at some examples. This is always a good way to start. With = 2 we have the followig ie cases Throw 1st 2d 3rd 4th 5th 6th 7th Ca we tell aythig from that? Probably ot. Why do't you go ad work out j 3? The rules are that there are three miuses, that the sum of all pluses ad miuses is 3, ad that ever before the ed do we get a partial sum of 3 or -3. So we ca't iclude ay of the followig i our tally for f 3 : Throw 1st 2d 3rd 4th 5th 6th 7th 8th 9th So what isf/ We've predicted thatj 3 = 3 3 = 27 adf 4 = 3 4 = 81. Is that what you've foud? If ot, sped a little time ad effort ad see what you get. (For the lazy oes amogst you we'll move right alog. For the others, we'll still be here whe you get back.) So how did you go? Tur out the way we expected? Did you lear aythig? (You probably leared that beig systematic was a great idea.) Have you doe eough to work out what j might be? " Oe thig that seems to be happeig is that we always ed with a. Does that have to be the case? If so, why? If ot, why ot? OK but suppose we stick to the cojecture thatf = 3. How could we prove somethig like this? How does 3 come up? It seems to us that there are at least two ways. You ca get 3 because there are three objects, ay of which ca go ito places. You ca also get 3 by showig that f = 3 f. 1 This is because Which oe of these is worth tryig? How ca we put three objects ito places? What would the three objects be? We oly seem to have worked with two thigs - pluses ad miuses. Fair eough. So how ca we show that f = 3 f. 1? That's of course, assumig that it is.

6 It's always worth tryig thigs out o a example first. All the ie arragemets above that make up f 2 ed i. That has to be the case. If they eded i aythig else the partial sum would have bee 3 earlier o i the sequece. Just before the last there seem to be three other paired possibilities. They are -,- ad. Why o--? We thik it's clear that if a sequece eded i --, the 3 would have happeed earlier. "" So let's pull out the - ad see what we have left. i ( -) ( -) ( -) the iterestig thig here is that removig - same thig happe for -? leads to the three possibilities for f 1 Does the (- ) (- ) (- ) It worked agai. It looks as if we may have a way to go from f 2 to f 1 or vice-versa. Ahwait though. Deletig is't goig to take us dow a step. To go fromf 2 to f 1 we eed to remove a mius sig. Let's do it ayway ad see what we get ( ) ( ) ( ) Well, there are certaily three reduced sequeces here but they are't f 1! They would be though, if we iterchaged pluses ad miuses i the first three terms. That all souds a bit loopy. Will it always work? First will isertig a ~ or - ito a proper arragemet with - 1 miuses always give us a proper arragemet? Let's start with the - ad see what happes. Suppose we have a proper sequece of - 1 miuses ad 2 pluses. We kow that at o stage do the partial sums add to 3 or -3, except at the ed whe they are 3. We also kow that the sequece fiishes with. So just before the, the sum is precisely 1. If we isert -before the ed 1 the the fial partial sums are 2, 1, 2, 3. So we do produce a proper arragemet with miuses. [ ] proper sequece with - 1 miuses isert ( -) [ ] ( -) proper sequece with miuses So for every (- 1 )-miuses sequece we ca get a -miuses sequece. Oh ad vice versa. The arrow i the above diagram ca go back the other way. Hece we have a equivalece betwee - 1 sequeces ad sequeces with - ext to the ed. This meas there must be f. _ 1 sequeces with '' miuses which ed. -.

7 But exactly the same argumet ca be applied to sequeces with miuses which ed. -. So there are f" _ 1 of them too. Now all we eed to be able to do is to master the dodgy argumet of the. sequeces. We're a bit wary about iterchagig the - ad sigs. Let's take a deep breath ad give it a go. ] sequece with - t miuses ad pluses i the square bracket Swap ad- [ ]* - t pluses ad ad miuses i i the square bracket ]* 3 pluses altogetheri miuses altogether The diagram above tells us what we have to do. It also shows that we ca go via this fiddle, from a (- t )-miuses sequece to a -miuses sequece, so that part of the book-keepig is OK. Does everythig else work out? Do we ever get 3 or -3 before the ed of the -miuses sequece? Look at the square bracket. No partial sums i that reach 3 or -3 ad it eds with a sum of 1 so that the fial give a total of 3. So i the square bracket with a asterisk, the partial sums owhere reach -3 or 3 ad the fial sum is - t. (By iterchagig ad - we just iterchage the sig of the sums.) Whe we fially add to the ed we have o 3 or -3 i the asterisked square bracket ad the fial five sums are -t, 0, t, 2, 3. So we do produce a good arragemet with miuses. We hope it's ow clear that for every (- t )-miuses sequece, iterchagig appropriate pluses ad miuses ad isertig gives us a -miuses sequece, ad vice versa. So there are as may -miuses sequeces edig as there are (- t )-miuses sequeces. Ad that'sj" _ 1 Puttig -,- ad together we've proved our cojecture. It's clear that f really does equal 3 f _. 1 So f = 3". At last we've justified it. But is there a easier way? Footote This whole problem is part of quite a large piece of literature i a area called Gambler's Rui. People have log bee iterested i wiig large amouts of moey ad by cotiually bettig agaif1st a adversary, perhaps doublig the stakes as they go, is oe way of tryig to achieve this. Cosequetly, mathematicias have had great pleasure i aalyzig such situatios. A readable itroductio ca be foud i The Theory of Stochastic Processes by D. R. Cox ad H.D. Miller ( t 965), while if you're feelig adveturous you could delve ito the delights of a detailed aalysis of this problem i Chapter XIV of William Feller's excellet A Itroductio to Probability Theory ad Its Applicatios (Third Editio, 1968, Wiley). Without lookig at a referece though see if you ca hadle the Qwertia problem if we remove the symmetry. How easy is it to escape if you oly eed a further 4kg of super eutro fuel istead of the 6 kg above? Referece [ t] D. Holto ad D. Fletcher, The Qwertias Te Thousad Years of Slavery, Mathematical Medley, Volume 26 No. t ( t 999), 9- t 3.