On the probability of the first player reaching m first if, starting at 0, at each round they go one step to the right with probability , 1/2 or , 1, steps to the left with probabilty, 1/2 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 1/2 or move, 1, units left with probability , 1/2 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 2 2 2 (2 m + 3 m) f(m) + (2 m + 5 m + 2) f(m + 2) + (-12 m - 24 m - 10) f(m + 1) 8 = - ---- Pi subject to the appropriate initial conditions and in Maple format (2*m^2+3*m)*f(m)+(2*m^2+5*m+2)*f(m+2)+(-12*m^2-24*m-10)*f(m+1) = -8/Pi subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 2/3 or , 1, steps to the left with probabilty, 1/3 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 2/3 or move, 1, units left with probability , 1/3 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 2 2 (16 m + 48 m) f(m) + (73 m + 292 m + 308) f(m + 2) 2 2 + (-18 m - 81 m - 72) f(m + 3) + (m + 5 m + 4) f(m + 4) 2 + (-72 m - 252 m - 144) f(m + 1) = 0 subject to the appropriate initial conditions and in Maple format (16*m^2+48*m)*f(m)+(73*m^2+292*m+308)*f(m+2)+(-18*m^2-81*m-72)*f(m+3)+(m^2+5*m+ 4)*f(m+4)+(-72*m^2-252*m-144)*f(m+1) = 0 subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 3/4 or , 1, steps to the left with probabilty, 1/4 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 3/4 or move, 1, units left with probability , 1/4 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 2 2 (81 m + 243 m) f(m) + (-288 m - 1008 m - 576) f(m + 1) 2 2 + (238 m + 952 m + 988) f(m + 2) + (-32 m - 144 m - 128) f(m + 3) 2 + (m + 5 m + 4) f(m + 4) = 0 subject to the appropriate initial conditions and in Maple format (81*m^2+243*m)*f(m)+(-288*m^2-1008*m-576)*f(m+1)+(238*m^2+952*m+988)*f(m+2)+(-\ 32*m^2-144*m-128)*f(m+3)+(m^2+5*m+4)*f(m+4) = 0 subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 2/3 or , 2, steps to the left with probabilty, 1/3 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 2/3 or move, 2, units left with probability , 1/3 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 3 2 3 2 (96 m + 336 m + 256 m) f(m) + (-90 m - 405 m - 447 m - 78) f(m + 2) 3 2 + (12 m + 60 m + 86 m + 42) f(m + 3) 3 2 + (6 m + 33 m + 37 m + 4) f(m + 4) 3 2 648 (m + 1) + (-240 m - 960 m - 1156 m - 424) f(m + 1) = - ----------- Pi subject to the appropriate initial conditions and in Maple format (96*m^3+336*m^2+256*m)*f(m)+(-90*m^3-405*m^2-447*m-78)*f(m+2)+(12*m^3+60*m^2+86 *m+42)*f(m+3)+(6*m^3+33*m^2+37*m+4)*f(m+4)+(-240*m^3-960*m^2-1156*m-424)*f(m+1) = -648*(m+1)/Pi subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 3/4 or , 2, steps to the left with probabilty, 1/4 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 3/4 or move, 2, units left with probability , 1/4 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 2 2 (81 m + 324 m) f(m) + (-288 m - 1296 m - 864) f(m + 1) 2 2 + (256 m + 1280 m + 1536) f(m + 2) + (-18 m - 99 m - 54) f(m + 3) 2 2 + (-32 m - 192 m - 192) f(m + 4) + (m + 7 m + 6) f(m + 6) = 0 subject to the appropriate initial conditions and in Maple format (81*m^2+324*m)*f(m)+(-288*m^2-1296*m-864)*f(m+1)+(256*m^2+1280*m+1536)*f(m+2)+( -18*m^2-99*m-54)*f(m+3)+(-32*m^2-192*m-192)*f(m+4)+(m^2+7*m+6)*f(m+6) = 0 subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 3/4 or , 3, steps to the left with probabilty, 1/4 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 3/4 or move, 3, units left with probability , 1/4 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 4 3 2 (486 m + 3159 m + 6399 m + 4050 m) f(m) 4 3 2 + (-756 m - 5292 m - 13302 m - 14112 m - 5202) f(m + 1) 4 3 2 + (-462 m - 3465 m - 8285 m - 6738 m - 688) f(m + 2) 4 3 2 + (-168 m - 1344 m - 3268 m - 2788 m - 696) f(m + 3) 4 3 2 + (18 m + 153 m + 453 m + 594 m + 312) f(m + 4) 4 3 2 + (12 m + 108 m + 290 m + 252 m + 10) f(m + 5) 4 3 2 5184 (m + 2) (m + 1) + (6 m + 57 m + 145 m + 118 m + 24) f(m + 6) = - -------------------- Pi subject to the appropriate initial conditions and in Maple format (486*m^4+3159*m^3+6399*m^2+4050*m)*f(m)+(-756*m^4-5292*m^3-13302*m^2-14112*m-\ 5202)*f(m+1)+(-462*m^4-3465*m^3-8285*m^2-6738*m-688)*f(m+2)+(-168*m^4-1344*m^3-\ 3268*m^2-2788*m-696)*f(m+3)+(18*m^4+153*m^3+453*m^2+594*m+312)*f(m+4)+(12*m^4+ 108*m^3+290*m^2+252*m+10)*f(m+5)+(6*m^4+57*m^3+145*m^2+118*m+24)*f(m+6) = -5184 *(m+2)*(m+1)/Pi subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 4/5 or , 3, steps to the left with probabilty, 1/5 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 4/5 or move, 3, units left with probability , 1/5 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 2 2 (256 m + 1280 m) f(m) + (-800 m - 4400 m - 3200) f(m + 1) 2 2 + (625 m + 3750 m + 5000) f(m + 2) + (-32 m - 224 m - 128) f(m + 4) 2 2 + (-50 m - 375 m - 400) f(m + 5) + (m + 9 m + 8) f(m + 8) = 0 subject to the appropriate initial conditions and in Maple format (256*m^2+1280*m)*f(m)+(-800*m^2-4400*m-3200)*f(m+1)+(625*m^2+3750*m+5000)*f(m+2 )+(-32*m^2-224*m-128)*f(m+4)+(-50*m^2-375*m-400)*f(m+5)+(m^2+9*m+8)*f(m+8) = 0 subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 4/5 or , 4, steps to the left with probabilty, 1/5 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 4/5 or move, 4, units left with probability , 1/5 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 4 3 2 (512 m + 3840 m + 8704 m + 6144 m) f(m) 4 3 2 + (-576 m - 4608 m - 13360 m - 16608 m - 7040) f(m + 1) 4 3 2 + (-414 m - 3519 m - 9431 m - 8250 m - 424) f(m + 2) 4 3 2 + (-252 m - 2268 m - 5898 m - 4848 m - 966) f(m + 3) 4 3 2 + (-90 m - 855 m - 2275 m - 2190 m - 680) f(m + 4) 4 3 2 + (8 m + 80 m + 276 m + 436 m + 280) f(m + 5) 4 3 2 + (6 m + 63 m + 195 m + 198 m) f(m + 6) 4 3 2 + (4 m + 44 m + 126 m + 100 m + 14) f(m + 7) 4 3 2 8000 (m + 2) (m + 1) + (2 m + 23 m + 63 m + 58 m + 16) f(m + 8) = - -------------------- Pi subject to the appropriate initial conditions and in Maple format (512*m^4+3840*m^3+8704*m^2+6144*m)*f(m)+(-576*m^4-4608*m^3-13360*m^2-16608*m-\ 7040)*f(m+1)+(-414*m^4-3519*m^3-9431*m^2-8250*m-424)*f(m+2)+(-252*m^4-2268*m^3-\ 5898*m^2-4848*m-966)*f(m+3)+(-90*m^4-855*m^3-2275*m^2-2190*m-680)*f(m+4)+(8*m^4 +80*m^3+276*m^2+436*m+280)*f(m+5)+(6*m^4+63*m^3+195*m^2+198*m)*f(m+6)+(4*m^4+44 *m^3+126*m^2+100*m+14)*f(m+7)+(2*m^4+23*m^3+63*m^2+58*m+16)*f(m+8) = -8000*(m+2 )*(m+1)/Pi subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 6/7 or , 4, steps to the left with probabilty, 1/7 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 6/7 or move, 4, units left with probability , 1/7 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 2 2 (1296 m + 7776 m) f(m) + (-3528 m - 22932 m - 17640) f(m + 1) 2 2 + (2401 m + 16807 m + 24010) f(m + 2) + (-72 m - 612 m - 360) f(m + 5) 2 2 + (-98 m - 882 m - 980) f(m + 6) + (m + 11 m + 10) f(m + 10) = 0 subject to the appropriate initial conditions and in Maple format (1296*m^2+7776*m)*f(m)+(-3528*m^2-22932*m-17640)*f(m+1)+(2401*m^2+16807*m+24010 )*f(m+2)+(-72*m^2-612*m-360)*f(m+5)+(-98*m^2-882*m-980)*f(m+6)+(m^2+11*m+10)*f( m+10) = 0 subject to the appropriate initial conditions On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 5/6 or , 5, steps to the left with probabilty, 1/6 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 5/6 or move, 5, units left with probability , 1/6 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 6 5 4 3 2 (3750 m + 58125 m + 347500 m + 999375 m + 1381250 m + 735000 m) f(m) + ( 6 5 4 3 2 -3300 m - 52800 m - 345250 m - 1180600 m - 2218850 m - 2142800 m - 794400) f(m + 1) + ( 6 5 4 3 2 -2574 m - 42471 m - 275316 m - 876615 m - 1371216 m - 845364 m - 11664 ) f(m + 2) + ( 6 5 4 3 2 -1848 m - 31416 m - 203864 m - 629400 m - 924532 m - 542364 m - 72936) f(3 + m) + ( 6 5 4 3 2 -1122 m - 19635 m - 128716 m - 399817 m - 606542 m - 413648 m - 96720) f(m + 4) + 6 5 4 3 2 (-396 m - 7128 m - 47694 m - 153084 m - 248478 m - 193236 m - 56880) f(m + 5) + 6 5 4 3 2 (30 m + 555 m + 4130 m + 16125 m + 35620 m + 43020 m + 22320) f(m + 6) + 6 5 4 3 2 (24 m + 456 m + 3316 m + 11740 m + 20516 m + 14204 m - 336) f(m + 7) 6 5 4 3 2 + (18 m + 351 m + 2514 m + 8355 m + 12822 m + 7224 m + 576) f(m + 8) 6 5 4 3 2 + (12 m + 240 m + 1706 m + 5532 m + 8422 m + 5448 m + 1080) f(m + 9) 6 5 4 3 2 + (6 m + 123 m + 874 m + 2869 m + 4628 m + 3476 m + 960) f(m + 10) = 2 2 2 - 4746093750 GAMMA(m/5 + 9/5) GAMMA(m/5 + 8/5) GAMMA(m/5 + 7/5) 2 2 m (-m) / 3 GAMMA(m/5 + 6/5) GAMMA(m/5) 25 9 / ((m + 4) Pi (3 + m) (m + 2) / 2 2 2 (m + 1) GAMMA(m/3 + 2/3) GAMMA(m/3 + 1/3) GAMMA(m/3) ) subject to the appropriate initial conditions and in Maple format (3750*m^6+58125*m^5+347500*m^4+999375*m^3+1381250*m^2+735000*m)*f(m)+(-3300*m^6 -52800*m^5-345250*m^4-1180600*m^3-2218850*m^2-2142800*m-794400)*f(m+1)+(-2574*m ^6-42471*m^5-275316*m^4-876615*m^3-1371216*m^2-845364*m-11664)*f(m+2)+(-1848*m^ 6-31416*m^5-203864*m^4-629400*m^3-924532*m^2-542364*m-72936)*f(3+m)+(-1122*m^6-\ 19635*m^5-128716*m^4-399817*m^3-606542*m^2-413648*m-96720)*f(m+4)+(-396*m^6-\ 7128*m^5-47694*m^4-153084*m^3-248478*m^2-193236*m-56880)*f(m+5)+(30*m^6+555*m^5 +4130*m^4+16125*m^3+35620*m^2+43020*m+22320)*f(m+6)+(24*m^6+456*m^5+3316*m^4+ 11740*m^3+20516*m^2+14204*m-336)*f(m+7)+(18*m^6+351*m^5+2514*m^4+8355*m^3+12822 *m^2+7224*m+576)*f(m+8)+(12*m^6+240*m^5+1706*m^4+5532*m^3+8422*m^2+5448*m+1080) *f(m+9)+(6*m^6+123*m^5+874*m^4+2869*m^3+4628*m^2+3476*m+960)*f(m+10) = -\ 4746093750/(m+4)/Pi^3/(3+m)/(m+2)/(m+1)*GAMMA(1/5*m+9/5)^2*GAMMA(1/5*m+8/5)^2* GAMMA(1/5*m+7/5)^2*GAMMA(1/5*m+6/5)^2/GAMMA(1/3*m+2/3)^2/GAMMA(1/3*m+1/3)^2* GAMMA(1/5*m)^2/GAMMA(1/3*m)^2*25^m*9^(-m) subject to the appropriate initial conditions ----------------------------------- On the probability of the first player reaching m first if, starting at 0, a\ t each round they go one step to the right with probability , 6/7 or , 5, steps to the left with probabilty, 1/7 By Shalosh B. Ekhad Consider a game where players take turns and at each step either move righ\ t one unit, with probability, 6/7 or move, 5, units left with probability , 1/7 and whoever reaches the location m first is declared the winner. What can you say about the probability of the first player of winning the ga\ me? Theorem: The probabability of the first player winning is 1/2 + 1/2 f(m) where f(m) satisfies the recurrence 2 2 (1296 m + 9072 m) f(m) + (-3528 m - 26460 m - 21168) f(m + 1) 2 2 + (2401 m + 19208 m + 28812) f(m + 2) + (-72 m - 720 m - 432) f(m + 6) 2 2 + (-98 m - 1029 m - 1176) f(m + 7) + (m + 13 m + 12) f(m + 12) = 0 subject to the appropriate initial conditions and in Maple format (1296*m^2+9072*m)*f(m)+(-3528*m^2-26460*m-21168)*f(m+1)+(2401*m^2+19208*m+28812 )*f(m+2)+(-72*m^2-720*m-432)*f(m+6)+(-98*m^2-1029*m-1176)*f(m+7)+(m^2+13*m+12)* f(m+12) = 0 subject to the appropriate initial conditions