Elementary Theory of Russian Roulette

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1 Elemetary Theory of Russia Roulette -iterestig patters of fractios- Satoshi Hashiba Daisuke Miematsu Ryohei Miyadera Itroductio. Today we are goig to study mathematical theory of Russia roulette. If some people may feel bad about the Russia roulette game, we wat to say sorry for them, but as a mathematical theory Russia roulette has a very iterestig structure. We are sure that may of the reader ca appreciate it. High school studets made this theory with a little help by their teacher, so this article shows a woderful possibility of a research by high school studets. I a Russia roulette game persos play the game. They take turs ad take up a gu ad pull a trigger to themselves. The game eds whe oe of the players gets killed. Note that i this versio of the game, oe does ot rotate the cylider before he pulls the trigger. I this article we ofte state the mathematica fact without proofs. Proofs are give at the appedix. Problem. Suppose that we use a revolver with chambers ad bullet. Calculate the probability of death of the first ma mathematically.

We are sure that may of the reader ca appreciate it.

2 Elemetary Russia Roulette.b Aswer. Suppose that players A ad B play the game, ad A is the first player. I the first roud A takes up the gu ad pulls the trigger to himself. This time the probability of his death is. If A survives, the i the secod roud B takes up the gu ad does the same ad if B survives, the i the third roud A takes up the gu ad does the same thig. Let s calculate the probability of A's death of third roud. If A is to die i the third roud, A has to survive the first roud. The probability of survival for A i the first roud is, ad after that B has to survive i the secod roud. Sice there are oly chambers ad bullet, so the probability of survival is. The there are chambers ad bullet, ad resultig probabily of death is. Therefore the probability of death of A i the third roud is ä ä. As to the probability of A's death i the fifth roud we ca do the almost the same calculatio ad we get ä ä ä ä. Fially the probability of death of the first player A is + ä ä + ä ä ä ä = =. Problem. Suppose that we use a revolver with chambers ad bullets. Calculate the probability of death of the first ma mathematically. Aswer. Sice there are bullets, the probability of his death i the first roud is, ad probability of his survival is. We ca use almost the same method we used i problem i the rest of the solutio, ad the aswer is ÅÅ + ä ä + ä ä ä ä = 9 =. So far we studied the case of chambers, but i mathematics we ca thik of a gu with ay umber of chambers ad bullets. For example we ca study a gu with chambers ad bullets. This is ot a absurd idea eve i a real life, because this may be a machie gu. We deote by F[,m] the probability of the first ma's death whe we use a revolver with -chambers ad m-bullets. For example by Problem we have F[,] =, ad by Problem F[,] =. Similarly we have F[,m] = m + Å -m ä ÅÅ -m- - ä Å - m + Å -m ä ÅÅ -m- - ä ÅÅ -m- - ä ÅÅ -m- - ä Å - m + By this formula you ca calculate F[,m] for ay atural umbers ad m. If you fid this formula too difficult to uderstad, do ot worry about it. If you ca uderstad Problem ad, you ca

Let s calculate the probability of A's death of third roud. If A is to die i the third roud, A has to survive the first roud.

3 Elemetary Russia Roulette.b Similarly we have F[,m] = m + Å -m ä ÅÅ -m- - ä Å - m + Å -m ä ÅÅ -m- - ä ÅÅ -m- - ä ÅÅ -m- - ä Å - m + By this formula you ca calculate F[,m] for ay atural umbers ad m. If you fid this formula too difficult to uderstad, do ot worry about it. If you ca uderstad Problem ad, you ca uderstad the rest of our article. You just have to uderstad that there is a way to calculate F[,m] for ay ad m. Next is the best part of our article! With F[,m] for may ad m we make a triagle. {F[,]} Figure() {F[,],F[,]} {F[,],F[,],F[,]} {F[,],F[,],F[,],F[,]} {F[,],F[,],F[,],F[,],F[,]} {F[,],F[,],F[,],F[,],F[,],F[,]} {F[7,],F[7,],F[7,],F[7,],F[7,],F[7,],F[7,7]} By calculatig F[,m] we get the followig triagle from the above triagle. Let's compare these triagles. F[,] is the third i the th row of the above triagle. I the same positio of the triagle below we have ÅÅ. Therefore F[,]= ÅÅ. 8< Figure HL 9 Å, Å = 9 Å, Å, Å = 9 Å, Å, Å, Å = 9 Å, Å, Å 7, Å, Å = 9 Å, Å, Å, Å, Å, Å = 9 Å 7, Å 7, Å, Å, Å, Å 7, = Problem. Ca you fid ay patter i figure ()? Aswer. Let's compare Figure () to the followig Figure (). If you reduce the fractios i Figure (), the fractios geerated will form Figure (). The patter is quite obvious i Figure (). For example look at th row. F[,] = 9 ad F[,] = ÅÅ are the secod ad third oes i the row. F[7,] = ÅÅ Å, which is the third = + 9+

You just have to uderstad that there is a way to calculate F[,m] for ay ad m. Next is the best part of our article! With F[,m] for may ad m we make a triagle.

4 Elemetary Russia Roulette.b Aswer. Let's compare Figure () to the followig Figure (). If you reduce the fractios i Figure (), the fractios geerated will form Figure (). The patter is quite obvious i Figure (). For example look at th row. F[,] = 9 ad F[,] = ÅÅ are the secod ad third oes i the row. F[7,] = ÅÅ = Å + 9+, which is the third oe i the 7th row. This remids us of Pascal's triagle. I geeral there exists the same kid of relatio amog F[,m],F[,m+], F[+,m+] for ay atural umber ad m with m. For proof of the relatio see Appedix. {} Figure(), <,, <,,, <, ÅÅ, ÅÅ 7,, <, ÅÅ 9, ÅÅ, ÅÅ,, < ÅÅ, ÅÅ, ÅÅ, ÅÅ, 7, < 8 7, Problem. Ca you fid ay other patter i figure ()? Aswer. I fact there are several patters. Please look at F@,D + F@,D the followig Figure (). Å = H + ÅÅ L = ÅÅ 7 = F@, D. F@,D+F@ +,D I geeral we ca prove that =F[+,]. As to the proof for this relatio wee Appedix. There are also other patters. Look at the Figure () ad (). Ca you fid ay patter i Figure () ad ()? {} Figure(), <,, <,,, <, ÅÅ, ÅÅ 7,, <, ÅÅ 9, ÅÅ, ÅÅ,, < ÅÅ, ÅÅ, ÅÅ, ÅÅ, 7, < 8 7, {} Figure(), <,, <,,, <, ÅÅ, ÅÅ 7,, <, ÅÅ 9, ÅÅ, ÅÅ,, < ÅÅ, ÅÅ, ÅÅ, ÅÅ, 7, < 8 7,

This remids us of Pascal's triagle. I geeral there exists the same kid of relatio amog F[,m],F[,m+], F[+,m+] for ay atural umber ad m with m. For proof of the relatio see Appedix.

5 Elemetary Russia Roulette.b {} Figure(), <,, <,,, <, ÅÅ, ÅÅ 7,, <, ÅÅ 9, ÅÅ, ÅÅ,, < ÅÅ, ÅÅ, ÅÅ, ÅÅ, 7, < 8 7, {} Figure(), <,, <,,, <, ÅÅ, ÅÅ 7,, <, ÅÅ 9, ÅÅ, ÅÅ,, < ÅÅ, ÅÅ, ÅÅ, ÅÅ, 7, < 8 7, Remark. We ca also study the Russia Roulette game with more tha persos. For example if persos play the game, the the probability of death of the third player form the followig triagle. Ca you fid ay patter i this triagle? Perhaps you will fid this very similar to the Figure (). <, <,, <,,, <, ÅÅ, ÅÅ,, <, ÅÅ, ÅÅ, ÅÅ,, < 7, ÅÅ, ÅÅ, ÅÅ, ÅÅ, 7, < Figure(7) Appedix. If you kow combiatorics ad how to calculate C m, the you ca read the proofs of mathematical facts preseted i this article. A proof for the fact preseted at Problem. To prove the existece of the relatio of F[,m] we eed a differet way to calculate F[,m] from the way we used i Problem ad. Let me illustrate it by usig problem.

Ca you fid ay patter i this triagle? Perhaps you will fid this very similar to the Figure (). <, <,, <,,, <, ÅÅ, ÅÅ,, <, ÅÅ, ÅÅ, ÅÅ,, < 7, ÅÅ, ÅÅ, ÅÅ, ÅÅ, 7, < Figure(7) Appedix.

6 Elemetary Russia Roulette.b A proof for the fact preseted at Problem. To prove the existece of the relatio of F[,m] we eed a differet way to calculate F[,m] from the way we used i Problem ad. Let me illustrate it by usig problem. Sice we have chambers, the chambers ca be represeted as {,,,,,} where we put bullets, ad there are C ways to do that. The bullet which is i the chamber with a small umber comes out first. If oe bullet is i the chamber ad the other is i a chamber whose umber is bigger tha, the the first oe will die. We have C cases of this kid. If oe bullet is i the chamber ad the other is i a chamber whose umber is bigger tha, the the first ma will die. We have C cases of this kid. If oe bullet is i the chamber ad the other is i chamber, the the first ma will die. We have C case of this kid. Therefore F[,] = C + C + C. Similarly we ca prove that C F[,]= C + Å C ad F[7,] = C C + C + C. By the famous equatio 7 C p C q = p- C q + p- C q- 7 C = C + C ad C + C + C =( C + C )+( C + C )+ C, where we used a trivial fact that C = C. Therefore this is the reaso of the existece of relatio amog F[,], F[,] ad F[7,]. Similarly we ca prove that F[,m]= C - m- + - C m- + - C m- +, C m F[,m+]= C - m + - C m + - C m + ÅÅ ad F[+,m+]= C C m + - C m + - C m + ÅÅ Å. By m+ + C m+ the equatio p C q = p- C q + p- C q- we have + C m+ = C m+ + C m ad C m + - C m + - C m + =( - C m + - C m- )+( - C m + - C m- )+( - C m + - C m- )+. Appedix. If You kow how to calculate k= k ad k= the you ca prove the fact preseted at Problem, amely F[,]+- F[+,]=F[+,]. A proof of a formula F[,]+F[+,]=F[+,]. F[,] = - H - k+l H - kl H -L H -L ) - C + - C + - C + =( ÅÅ C k= )/( Å = - k= ( -(k-)+k(k-))/(h - L H - L) = - k= HH + L - H + L k + k L/(H - L H - L) = (H + L H - L - H + L µ Å H-L H-L H -L + µ ÅÅ ) / ( H - L H - L) = Here if we put + ito, we have F[+,] = k,

The bullet which is i the chamber with a small umber comes out first. If oe bullet is i the chamber ad the other is i a chamber whose umber is bigger tha, the the first oe will die.

7 Elemetary Russia Roulette.b 7 = - k= HH + L - H + L k + k L/(H - L H - L) = (H + L H - L - H + L µ Å H-L H-L H -L + µ ÅÅ ) / ( H - L H - L) = Here if we put + ito, we have F[+,] = F[+,]= C + - C + - C + ÅÅ Å + C = k= H - k + L H - k + L ê H + L H - L = k= HH + + L - H8 + L k + k L ê HH + L H - L L = HH + + L - H8 + L µ Å H+L H+L H +L + µ ÅÅ L ê HH + L H - L L = Therefore we have F[,]+F[+,]= F[+,].

L H - L) = 8 + -. Here if we put + ito, we have F[+,] = + 8 +.

In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008

In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008 I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces