{
  "index": "1960-A-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "6. A player throwing a die scores as many points as on the top face of the die and is to play until his score reaches or passes a total \\( n \\). Denote by \\( p(n) \\) the probability of making exactly the total \\( n \\), and find the value of \\( \\lim _{n \\rightarrow \\infty} p(n) \\).",
  "solution": "First Solution. If by definition \\( p(0)=1 \\) and \\( p(n)=0, n<0 \\) then the following relations are easy to verify.\n\\[\n\\begin{array}{l}\np(0)=1 \\\\\np(1)=\\frac{1}{6} p(0)\n\\end{array}\n\\]\n(1)\n\\[\n\\begin{array}{l}\np(2)=\\frac{1}{6}[p(1)+p(0)] \\\\\np(3)=\\frac{1}{6}[p(2)+p(1)+p(0)] \\\\\n\\vdots \\\\\np(n)=\\frac{1}{6}[p(n-1)+p(n-2)+\\cdots+p(n-6)], n>0 .\n\\end{array}\n\\]\n\nAdding these equations and then canceling, we obtain\n\\[\np(n)+\\frac{5}{6} p(n-1)+\\frac{4}{6} p(n-2)+\\cdots+\\frac{1}{6} p(n-5)=1 .\n\\]\n\nNow if it is assumed that \\( p(n) \\rightarrow p \\) as \\( n \\rightarrow \\infty \\), then it follows that \\( (21 / 6) p=1 \\) and hence that \\( p=2 / 7 \\).\nWe shall now show that, regardless of the initial conditions, a sequence \\( \\{p(n)\\} \\) that satisfies the recursion (1) is convergent.\n\nLet\n\\[\n\\begin{aligned}\np(n) & =\\min \\{p(n-i): i=1,2, \\ldots, 6\\} \\\\\n\\bar{p}(n) & =\\max \\{p(n-i): i=1,2, \\ldots, 6\\}\n\\end{aligned}\n\\]\n\nFrom (1) we obtain\n\\[\n\\frac{1}{6}[5 \\underline{p}(n)+\\bar{p}(n)] \\leq p(n) \\leq \\frac{1}{6}[5 \\bar{p}(n)+\\underline{p}(n)] .\n\\]\n\nHence \\( \\underline{p}(n) \\leq \\underline{p}(n+1) \\leq \\bar{p}(n+1) \\leq \\bar{p}(n) \\), so the sequence \\( \\{\\underline{p}(n)\\} \\) is non-decreasing and \\( \\{\\bar{p}(n)\\} \\) is non-increasing. Let\n\\[\n\\begin{array}{l}\n\\bar{q}=\\lim _{n-\\infty} \\bar{p}(n) \\\\\n\\underline{q}=\\lim _{n-\\infty} \\underline{p}(n)\n\\end{array}\n\\]\n\nThen \\( \\underline{q} \\leq \\bar{q} \\).\nNow suppose \\( \\bar{q}>\\underline{q} \\) and set \\( \\epsilon=\\frac{1}{5}(\\bar{q}-\\underline{q}) \\). Since \\( \\epsilon>0 \\), for some \\( N \\)\nnd all \\( k \\geq N \\)\n\\[\n\\underline{q}-\\epsilon<\\underline{p}(k) \\leq \\underline{q}\n\\]\nso\n\\[\n\\frac{1}{6}[5(\\underline{q}-\\epsilon)+\\bar{q}]<\\frac{1}{6}[5 \\underline{p}(k)+\\bar{p}(k)] \\leq p(k),\n\\]\ni.e., \\( \\underline{q}<p(k) \\). From this follows \\( \\underline{q}<\\underline{p}(k) \\) for \\( k>N+5 \\), a contradiction since \\( \\underline{p}(k) \\) increases to \\( q \\).\nWe conclude \\( \\bar{q}=\\underline{q} \\). Then it follows immediately that \\( \\lim _{n-\\infty} p(n)= \\) \\( \\bar{q}=\\underline{q} \\).\n\nSecond Solution. Define \\( p(0)=1, p(n)=0 \\) for \\( n<0 \\), and for each \\( n=0,1,2, \\ldots \\) let\n\\[\n\\alpha_{n}=(p(n-5), p(n-4), p(n-3), p(n-2), p(n-1), p(n))^{T} .\n\\]\n\nThen for all \\( n \\) we have\n\\begin{tabular}{|l|l|}\n\\hline & \\( \\alpha_{n+1}=A \\alpha_{n} \\) \\\\\n\\hline where & \\\\\n\\hline & \\( A=\\left(\\begin{array}{llllll}0 & 1 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 1 \\\\ \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6}\\end{array}\\right) \\). \\\\\n\\hline\n\\end{tabular}\n\nHence \\( \\alpha_{n}=A^{\\prime \\prime} \\alpha_{10} \\).\nNow \\( A \\) is row-stochastic (i.e., it has non-negative entries and each row sums to one), and it is easy to see that \\( A^{0} \\) has all positive entries. It is shown in the theory of Markov processes that, if some power of a rowstochastic matrix has all positive entries. its powers converge to a rowhas all rows the same and\n\\[\n\\lim \\alpha_{n}=\\left(\\lim A^{n}\\right) \\alpha_{0}=B \\alpha_{0},\n\\]\na vector with all components the same, say \\( p \\). We have\n\\[\n\\lim _{n} p(n)=p\n\\]\n\nIf \\( \\beta=(1,2,3,4,5,6) \\), then \\( \\beta A=\\beta \\). Therefore \\( \\beta A^{\\prime \\prime}=\\beta \\), and taking limits \\( \\beta B=\\beta \\). Then\n\\[\n21 p=\\beta\\left(B \\alpha_{0}\\right)=\\beta \\alpha_{0}=6\n\\]\nso \\( p=2 / 7 \\).\nFor the theorem on row-stochastic matrices see, for example, P. A. P. Moran, Introduction to Probability Theory, Clarendon Press, Oxford, 1968, page 112.\n\nThird Solution. We now roll up the really heavy artillery and argue as follows.\nThe probability of obtaining a total of \\( n \\) in \\( k \\) throws (exactly) is the coefficient of \\( x^{\\prime \\prime} \\) in\n\\[\n\\left[\\frac{1}{6}\\left(x+x^{2}+\\cdots+x^{0}\\right]^{4} .\\right.\n\\]\n\nSince obtaining \\( n \\) in \\( k \\) throws and obtaining \\( n \\) in \\( l \\) throws for \\( k \\neq l \\) are mutually exclusive events, the probability of ever obtaining a total of \\( n \\) is the coefficient of \\( x^{\\prime \\prime} \\) in\n\\[\n\\sum_{k=0}^{\\infty}\\left[\\frac{1}{6}\\left(x+x^{2}+\\cdots+x^{x}\\right)\\right]^{x} .\n\\]\n\nHence\n\\[\n\\begin{aligned}\n\\sum_{n=0}^{\\infty} p(n) x^{n} & =\\frac{6}{6-\\left(x+x^{2}+\\cdots+x^{6}\\right)} \\\\\n& =\\frac{6}{(1-x)\\left(6+5 x+4 x^{2}+3 x^{3}+2 x^{4}+x^{5}\\right)} \\\\\n& =\\frac{2}{7(1-x)}+\\frac{2}{7} \\frac{15+10 x+6 x^{2}+3 x^{3}+x^{4}}{6+5 x+4 x^{2}+3 x^{3}+2 x^{4}+x^{5}} .\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\sum_{n-0}^{\\infty}\\left(p(n)-\\frac{2}{7}\\right) x^{\\prime \\prime}=\\frac{2}{7} \\frac{15+10 x+6 x^{2}+3 x^{3}+x^{4}}{6+5 x+4 x^{2}+3 x^{3}+2 x^{4}+x^{5}} .\n\\]\n\nThus far the argument has been formally combinatoric in character. If we now regard \\( x \\) as a complex variable and show that the denominator of the last fraction does not vanish inside or on the unit circle in the complex plane, then it follows that for some \\( x>1, \\sum_{n-0}^{\\infty}(p(n)-(2 / 7)) x^{\\prime \\prime} \\) converges and hence \\( \\lim _{n-\\infty} p(n)=2 / 7 \\).\nLet\n\\[\n\\begin{array}{c}\nD=6+5 x+4 x^{2}+3 x^{3}+2 x^{4}+x^{5} \\\\\n\\text { If } x=1, D \\neq 0 \\text {. Assume now that } x \\neq 1 \\text { but }|x| \\leq 1 \\text {. Then } \\\\\nD(1-x)=6-x-x^{2}-x^{3}-x^{4}-x^{5}-x^{6}\n\\end{array}\n\\]\n\\[\n|D||1-x|>6-6|x| \\geq 0 .\n\\]\n\nThus the zeros of \\( D \\) are outside the unit circle.\nRemark. This is one of the problems criticized by L. J. Mordell in his article \"The Putnam Competition,\" American Muthemutical Monthly, vol. \\( 70(1963) \\), pages \\( 481-490 \\). The published solution in the Monthly is a ticated for an undergraduate competition. The examiners no doubt envisioned something like our first solution. See page 623 in Appendix. visioned something like our first solution. See page 623 in Appendix.",
  "vars": [
    "n",
    "k",
    "l",
    "x"
  ],
  "params": [
    "p",
    "A",
    "B",
    "D",
    "\\\\alpha_n",
    "\\\\beta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "scoretarget",
        "k": "throwcount",
        "l": "othercount",
        "x": "variablex",
        "p": "probfunc",
        "A": "matrixa",
        "B": "matrixb",
        "D": "polydenom",
        "\\alpha_n": "vectoralpha",
        "\\beta": "vectorbeta"
      },
      "question": "6. A player throwing a die scores as many points as on the top face of the die and is to play until his score reaches or passes a total \\( scoretarget \\). Denote by \\( probfunc(scoretarget) \\) the probability of making exactly the total \\( scoretarget \\), and find the value of \\( \\lim _{scoretarget \\rightarrow \\infty} probfunc(scoretarget) \\).",
      "solution": "First Solution. If by definition \\( probfunc(0)=1 \\) and \\( probfunc(scoretarget)=0, scoretarget<0 \\) then the following relations are easy to verify.\n\\[\n\\begin{array}{l}\nprobfunc(0)=1 \\\\\nprobfunc(1)=\\frac{1}{6} probfunc(0)\n\\end{array}\n\\]\n(1)\n\\[\n\\begin{array}{l}\nprobfunc(2)=\\frac{1}{6}[probfunc(1)+probfunc(0)] \\\\\nprobfunc(3)=\\frac{1}{6}[probfunc(2)+probfunc(1)+probfunc(0)] \\\\\n\\vdots \\\\\nprobfunc(scoretarget)=\\frac{1}{6}[probfunc(scoretarget-1)+probfunc(scoretarget-2)+\\cdots+probfunc(scoretarget-6)], scoretarget>0 .\n\\end{array}\n\\]\n\nAdding these equations and then canceling, we obtain\n\\[\nprobfunc(scoretarget)+\\frac{5}{6} probfunc(scoretarget-1)+\\frac{4}{6} probfunc(scoretarget-2)+\\cdots+\\frac{1}{6} probfunc(scoretarget-5)=1 .\n\\]\n\nNow if it is assumed that \\( probfunc(scoretarget) \\rightarrow probfunc \\) as \\( scoretarget \\rightarrow \\infty \\), then it follows that \\( (21 / 6) probfunc=1 \\) and hence that \\( probfunc=2 / 7 \\).\nWe shall now show that, regardless of the initial conditions, a sequence \\( \\{probfunc(scoretarget)\\} \\) that satisfies the recursion (1) is convergent.\n\nLet\n\\[\n\\begin{aligned}\n\\underline{probfunc}(scoretarget) & =\\min \\{probfunc(scoretarget-i): i=1,2, \\ldots, 6\\} \\\\\n\\bar{probfunc}(scoretarget) & =\\max \\{probfunc(scoretarget-i): i=1,2, \\ldots, 6\\}\n\\end{aligned}\n\\]\n\nFrom (1) we obtain\n\\[\n\\frac{1}{6}[5 \\underline{probfunc}(scoretarget)+\\bar{probfunc}(scoretarget)] \\leq probfunc(scoretarget) \\leq \\frac{1}{6}[5 \\bar{probfunc}(scoretarget)+\\underline{probfunc}(scoretarget)] .\n\\]\n\nHence \\( \\underline{probfunc}(scoretarget) \\leq \\underline{probfunc}(scoretarget+1) \\leq \\bar{probfunc}(scoretarget+1) \\leq \\bar{probfunc}(scoretarget) \\), so the sequence \\( \\{\\underline{probfunc}(scoretarget)\\} \\) is non-decreasing and \\( \\{\\bar{probfunc}(scoretarget)\\} \\) is non-increasing. Let\n\\[\n\\begin{array}{l}\n\\bar{q}=\\lim _{scoretarget-\\infty} \\bar{probfunc}(scoretarget) \\\\\n\\underline{q}=\\lim _{scoretarget-\\infty} \\underline{probfunc}(scoretarget)\n\\end{array}\n\\]\n\nThen \\( \\underline{q} \\leq \\bar{q} \\).\nNow suppose \\( \\bar{q}>\\underline{q} \\) and set \\( \\epsilon=\\frac{1}{5}(\\bar{q}-\\underline{q}) \\). Since \\( \\epsilon>0 \\), for some \\( N \\)\nand all \\( throwcount \\geq N \\)\n\\[\n\\underline{q}-\\epsilon<\\underline{probfunc}(throwcount) \\leq \\underline{q}\n\\]\nso\n\\[\n\\frac{1}{6}[5(\\underline{q}-\\epsilon)+\\bar{q}]<\\frac{1}{6}[5 \\underline{probfunc}(throwcount)+\\bar{probfunc}(throwcount)] \\leq probfunc(throwcount),\n\\]\ni.e., \\( \\underline{q}<probfunc(throwcount) \\). From this follows \\( \\underline{q}<\\underline{probfunc}(throwcount) \\) for \\( throwcount>N+5 \\), a contradiction since \\( \\underline{probfunc}(throwcount) \\) increases to \\( \\underline{q} \\).\nWe conclude \\( \\bar{q}=\\underline{q} \\). Then it follows immediately that \\( \\lim _{scoretarget-\\infty} probfunc(scoretarget)=\\bar{q}=\\underline{q} \\).\n\nSecond Solution. Define \\( probfunc(0)=1, probfunc(scoretarget)=0 \\) for \\( scoretarget<0 \\), and for each \\( scoretarget=0,1,2, \\ldots \\) let\n\\[\nvectoralpha_{scoretarget}=(probfunc(scoretarget-5), probfunc(scoretarget-4), probfunc(scoretarget-3), probfunc(scoretarget-2), probfunc(scoretarget-1), probfunc(scoretarget))^{T} .\n\\]\n\nThen for all \\( scoretarget \\) we have\n\\[\nvectoralpha_{scoretarget+1}=matrixa\\, vectoralpha_{scoretarget}\n\\]\nwhere\n\\[\nmatrixa=\\left(\\begin{array}{llllll}\n0 & 1 & 0 & 0 & 0 & 0 \\\\\n0 & 0 & 1 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 1 & 0 & 0 \\\\\n0 & 0 & 0 & 0 & 1 & 0 \\\\\n0 & 0 & 0 & 0 & 0 & 1 \\\\\n\\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6}\n\\end{array}\\right).\n\\]\n\nHence \\( vectoralpha_{scoretarget}=matrixa^{\\prime \\prime}\\, vectoralpha_{10} \\).\nNow \\( matrixa \\) is row-stochastic (i.e., it has non-negative entries and each row sums to one), and it is easy to see that \\( matrixa^{0} \\) has all positive entries. It is shown in the theory of Markov processes that, if some power of a row-stochastic matrix has all positive entries, its powers converge to a row-stochastic matrix that has all rows the same and\n\\[\n\\lim_{scoretarget\\rightarrow\\infty} vectoralpha_{scoretarget}=\\left(\\lim_{scoretarget\\rightarrow\\infty} matrixa^{scoretarget}\\right) vectoralpha_{0}=matrixb\\, vectoralpha_{0},\n\\]\na vector with all components the same, say \\( probfunc \\). We have\n\\[\n\\lim_{scoretarget\\rightarrow\\infty} probfunc(scoretarget)=probfunc .\n\\]\n\nIf \\( vectorbeta=(1,2,3,4,5,6) \\), then \\( vectorbeta\\, matrixa=vectorbeta \\). Therefore \\( vectorbeta\\, matrixa^{\\prime \\prime}=vectorbeta \\), and taking limits \\( vectorbeta\\, matrixb=vectorbeta \\). Then\n\\[\n21\\, probfunc=vectorbeta\\left(matrixb\\, vectoralpha_{0}\\right)=vectorbeta\\, vectoralpha_{0}=6\n\\]\nso \\( probfunc=2 / 7 \\).\nFor the theorem on row-stochastic matrices see, for example, P. A. P. Moran, Introduction to Probability Theory, Clarendon Press, Oxford, 1968, page 112.\n\nThird Solution. We now roll up the really heavy artillery and argue as follows.\nThe probability of obtaining a total of \\( scoretarget \\) in \\( throwcount \\) throws (exactly) is the coefficient of \\( variablex^{\\prime \\prime} \\) in\n\\[\n\\left[\\frac{1}{6}\\left(variablex+variablex^{2}+\\cdots+variablex^{0}\\right)\\right]^{4}.\n\\]\n\nSince obtaining \\( scoretarget \\) in \\( throwcount \\) throws and obtaining \\( scoretarget \\) in \\( othercount \\) throws for \\( throwcount \\neq othercount \\) are mutually exclusive events, the probability of ever obtaining a total of \\( scoretarget \\) is the coefficient of \\( variablex^{\\prime \\prime} \\) in\n\\[\n\\sum_{throwcount=0}^{\\infty}\\left[\\frac{1}{6}\\left(variablex+variablex^{2}+\\cdots+variablex^{variablex}\\right)\\right]^{throwcount}.\n\\]\n\nHence\n\\[\n\\begin{aligned}\n\\sum_{scoretarget=0}^{\\infty} probfunc(scoretarget)\\, variablex^{scoretarget} & =\\frac{6}{6-\\left(variablex+variablex^{2}+\\cdots+variablex^{6}\\right)} \\\\\n& =\\frac{6}{(1-variablex)\\left(6+5\\, variablex+4\\, variablex^{2}+3\\, variablex^{3}+2\\, variablex^{4}+variablex^{5}\\right)} \\\\\n& =\\frac{2}{7(1-variablex)}+\\frac{2}{7} \\frac{15+10\\, variablex+6\\, variablex^{2}+3\\, variablex^{3}+variablex^{4}}{6+5\\, variablex+4\\, variablex^{2}+3\\, variablex^{3}+2\\, variablex^{4}+variablex^{5}} .\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\sum_{scoretarget=0}^{\\infty}\\left(probfunc(scoretarget)-\\frac{2}{7}\\right) variablex^{scoretarget}=\\frac{2}{7} \\frac{15+10\\, variablex+6\\, variablex^{2}+3\\, variablex^{3}+variablex^{4}}{6+5\\, variablex+4\\, variablex^{2}+3\\, variablex^{3}+2\\, variablex^{4}+variablex^{5}} .\n\\]\n\nThus far the argument has been formally combinatorial in character. If we now regard \\( variablex \\) as a complex variable and show that the denominator of the last fraction does not vanish inside or on the unit circle in the complex plane, then it follows that for some \\( variablex>1, \\sum_{scoretarget=0}^{\\infty}(probfunc(scoretarget)-(2 / 7)) variablex^{scoretarget} \\) converges and hence \\( \\lim_{scoretarget\\rightarrow\\infty} probfunc(scoretarget)=2 / 7 \\).\nLet\n\\[\n\\begin{array}{c}\npolydenom=6+5\\, variablex+4\\, variablex^{2}+3\\, variablex^{3}+2\\, variablex^{4}+variablex^{5} \\\\\n\\text { If } variablex=1, polydenom \\neq 0 . \\text { Assume now that } variablex \\neq 1 \\text { but }|variablex| \\leq 1 .\n\\end{array}\n\\]\nThen\n\\[\npolydenom(1-variablex)=6-variablex-variablex^{2}-variablex^{3}-variablex^{4}-variablex^{5}-variablex^{6}\n\\]\n\\[\n|polydenom||1-variablex|>6-6|variablex| \\geq 0 .\n\\]\n\nThus the zeros of \\( polydenom \\) are outside the unit circle.\nRemark. This is one of the problems criticized by L. J. Mordell in his article \"The Putnam Competition,\" American Mathematical Monthly, vol. \\( 70(1963) \\), pages \\( 481-490 \\). The published solution in the Monthly is a ticated for an undergraduate competition. The examiners no doubt envisioned something like our first solution. See page 623 in Appendix."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "parchment",
        "k": "overgrowth",
        "l": "trelliswork",
        "x": "flagstone",
        "p": "sandcastle",
        "A": "riverbank",
        "B": "hearthside",
        "D": "moonlight",
        "\\\\alpha_n": "driftwood",
        "\\\\beta": "starflower"
      },
      "question": "6. A player throwing a die scores as many points as on the top face of the die and is to play until his score reaches or passes a total \\( parchment \\). Denote by \\( sandcastle(parchment) \\) the probability of making exactly the total \\( parchment \\), and find the value of \\( \\lim_{parchment \\rightarrow \\infty} sandcastle(parchment) \\).",
      "solution": "First Solution. If by definition \\( sandcastle(0)=1 \\) and \\( sandcastle(parchment)=0, parchment<0 \\) then the following relations are easy to verify.\n\\[\n\\begin{array}{l}\nsandcastle(0)=1 \\\\\nsandcastle(1)=\\frac{1}{6} sandcastle(0)\n\\end{array}\n\\]\n(1)\n\\[\n\\begin{array}{l}\nsandcastle(2)=\\frac{1}{6}[sandcastle(1)+sandcastle(0)] \\\\\nsandcastle(3)=\\frac{1}{6}[sandcastle(2)+sandcastle(1)+sandcastle(0)] \\\\\n\\vdots \\\\\nsandcastle(parchment)=\\frac{1}{6}[sandcastle(parchment-1)+sandcastle(parchment-2)+\\cdots+sandcastle(parchment-6)], parchment>0 .\n\\end{array}\n\\]\n\nAdding these equations and then canceling, we obtain\n\\[\nsandcastle(parchment)+\\frac{5}{6} sandcastle(parchment-1)+\\frac{4}{6} sandcastle(parchment-2)+\\cdots+\\frac{1}{6} sandcastle(parchment-5)=1 .\n\\]\n\nNow if it is assumed that \\( sandcastle(parchment) \\rightarrow sandcastle \\) as \\( parchment \\rightarrow \\infty \\), then it follows that \\( (21 / 6) sandcastle=1 \\) and hence that \\( sandcastle=2 / 7 \\).\nWe shall now show that, regardless of the initial conditions, a sequence \\( \\{sandcastle(parchment)\\} \\) that satisfies the recursion (1) is convergent.\n\nLet\n\\[\n\\begin{aligned}\nsandcastle(parchment) & =\\min \\{sandcastle(parchment-i): i=1,2, \\ldots, 6\\} \\\\\n\\bar{sandcastle}(parchment) & =\\max \\{sandcastle(parchment-i): i=1,2, \\ldots, 6\\}\n\\end{aligned}\n\\]\n\nFrom (1) we obtain\n\\[\n\\frac{1}{6}[5 \\underline{sandcastle}(parchment)+\\bar{sandcastle}(parchment)] \\leq sandcastle(parchment) \\leq \\frac{1}{6}[5 \\bar{sandcastle}(parchment)+\\underline{sandcastle}(parchment)] .\n\\]\n\nHence \\( \\underline{sandcastle}(parchment) \\leq \\underline{sandcastle}(parchment+1) \\leq \\bar{sandcastle}(parchment+1) \\leq \\bar{sandcastle}(parchment) \\), so the sequence \\( \\{\\underline{sandcastle}(parchment)\\} \\) is non-decreasing and \\( \\{\\bar{sandcastle}(parchment)\\} \\) is non-increasing. Let\n\\[\n\\begin{array}{l}\n\\bar{q}=\\lim _{parchment-\\infty} \\bar{sandcastle}(parchment) \\\\\n\\underline{q}=\\lim _{parchment-\\infty} \\underline{sandcastle}(parchment)\n\\end{array}\n\\]\n\nThen \\( \\underline{q} \\leq \\bar{q} \\).\nNow suppose \\( \\bar{q}>\\underline{q} \\) and set \\( \\epsilon=\\frac{1}{5}(\\bar{q}-\\underline{q}) \\). Since \\( \\epsilon>0 \\), for some \\( N \\)\nnd all \\( overgrowth \\geq N \\)\n\\[\n\\underline{q}-\\epsilon<\\underline{sandcastle}(overgrowth) \\leq \\underline{q}\n\\]\nso\n\\[\n\\frac{1}{6}[5(\\underline{q}-\\epsilon)+\\bar{q}]<\\frac{1}{6}[5 \\underline{sandcastle}(overgrowth)+\\bar{sandcastle}(overgrowth)] \\leq sandcastle(overgrowth),\n\\]\ni.e., \\( \\underline{q}<sandcastle(overgrowth) \\). From this follows \\( \\underline{q}<\\underline{sandcastle}(overgrowth) \\) for \\( overgrowth>N+5 \\), a contradiction since \\( \\underline{sandcastle}(overgrowth) \\) increases to \\( q \\).\nWe conclude \\( \\bar{q}=\\underline{q} \\). Then it follows immediately that \\( \\lim _{parchment-\\infty} sandcastle(parchment)= \\) \\( \\bar{q}=\\underline{q} \\).\n\nSecond Solution. Define \\( sandcastle(0)=1, sandcastle(parchment)=0 \\) for \\( parchment<0 \\), and for each \\( parchment=0,1,2, \\ldots \\) let\n\\[\ndriftwood_{parchment}=(sandcastle(parchment-5), sandcastle(parchment-4), sandcastle(parchment-3), sandcastle(parchment-2), sandcastle(parchment-1), sandcastle(parchment))^{T} .\n\\]\n\nThen for all \\( parchment \\) we have\n\\begin{tabular}{|l|l|}\n\\hline & \\( driftwood_{parchment+1}=riverbank \\, driftwood_{parchment} \\) \\\\\n\\hline where & \\\\\n\\hline & \\( riverbank=\\left(\\begin{array}{llllll}0 & 1 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 1 \\\\ \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6} & \\frac{1}{6}\\end{array}\\right) \\). \\\\\n\\hline\n\\end{tabular}\n\nHence \\( driftwood_{parchment}=riverbank^{\\prime \\prime} \\, driftwood_{10} \\).\nNow \\( riverbank \\) is row-stochastic (i.e., it has non-negative entries and each row sums to one), and it is easy to see that \\( riverbank^{0} \\) has all positive entries. It is shown in the theory of Markov processes that, if some power of a rowstochastic matrix has all positive entries, its powers converge to a rowhas all rows the same and\n\\[\n\\lim driftwood_{parchment}=\\left(\\lim riverbank^{parchment}\\right) driftwood_{0}=hearthside \\, driftwood_{0},\n\\]\na vector with all components the same, say sandcastle. We have\n\\[\n\\lim _{parchment} sandcastle(parchment)=sandcastle\n\\]\n\nIf \\( starflower=(1,2,3,4,5,6) \\), then \\( starflower \\, riverbank=starflower \\). Therefore \\( starflower \\, riverbank^{\\prime \\prime}=starflower \\), and taking limits \\( starflower \\, hearthside=starflower \\). Then\n\\[\n21 \\, sandcastle=starflower\\left(hearthside \\, driftwood_{0}\\right)=starflower \\, driftwood_{0}=6\n\\]\nso \\( sandcastle=2 / 7 \\).\nFor the theorem on row-stochastic matrices see, for example, P. A. P. Moran, Introduction to Probability Theory, Clarendon Press, Oxford, 1968, page 112.\n\nThird Solution. We now roll up the really heavy artillery and argue as follows.\nThe probability of obtaining a total of \\( parchment \\) in \\( overgrowth \\) throws (exactly) is the coefficient of \\( flagstone^{\\prime \\prime} \\) in\n\\[\n\\left[\\frac{1}{6}\\left(flagstone+flagstone^{2}+\\cdots+flagstone^{0}\\right]^{4} .\\right.\n\\]\n\nSince obtaining \\( parchment \\) in \\( overgrowth \\) throws and obtaining \\( parchment \\) in \\( trelliswork \\) throws for \\( overgrowth \\neq trelliswork \\) are mutually exclusive events, the probability of ever obtaining a total of \\( parchment \\) is the coefficient of \\( flagstone^{\\prime \\prime} \\) in\n\\[\n\\sum_{overgrowth=0}^{\\infty}\\left[\\frac{1}{6}\\left(flagstone+flagstone^{2}+\\cdots+flagstone^{flagstone}\\right)\\right]^{flagstone} .\n\\]\n\nHence\n\\[\n\\begin{aligned}\n\\sum_{parchment=0}^{\\infty} sandcastle(parchment) flagstone^{parchment} & =\\frac{6}{6-\\left(flagstone+flagstone^{2}+\\cdots+flagstone^{6}\\right)} \\\\\n& =\\frac{6}{(1-flagstone)\\left(6+5 flagstone+4 flagstone^{2}+3 flagstone^{3}+2 flagstone^{4}+flagstone^{5}\\right)} \\\\\n& =\\frac{2}{7(1-flagstone)}+\\frac{2}{7} \\frac{15+10 flagstone+6 flagstone^{2}+3 flagstone^{3}+flagstone^{4}}{6+5 flagstone+4 flagstone^{2}+3 flagstone^{3}+2 flagstone^{4}+flagstone^{5}} .\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\sum_{parchment-0}^{\\infty}\\left(sandcastle(parchment)-\\frac{2}{7}\\right) flagstone^{\\prime \\prime}=\\frac{2}{7} \\frac{15+10 flagstone+6 flagstone^{2}+3 flagstone^{3}+flagstone^{4}}{6+5 flagstone+4 flagstone^{2}+3 flagstone^{3}+2 flagstone^{4}+flagstone^{5}} .\n\\]\n\nThus far the argument has been formally combinatoric in character. If we now regard \\( flagstone \\) as a complex variable and show that the denominator of the last fraction does not vanish inside or on the unit circle in the complex plane, then it follows that for some \\( flagstone>1, \\sum_{parchment-0}^{\\infty}(sandcastle(parchment)-(2 / 7)) flagstone^{\\prime \\prime} \\) converges and hence \\( \\lim _{parchment-\\infty} sandcastle(parchment)=2 / 7 \\).\nLet\n\\[\n\\begin{array}{c}\nmoonlight=6+5 flagstone+4 flagstone^{2}+3 flagstone^{3}+2 flagstone^{4}+flagstone^{5} \\\\\n\\text { If } flagstone=1, moonlight \\neq 0 \\text {. Assume now that } flagstone \\neq 1 \\text { but }|flagstone| \\leq 1 \\text {. Then } \\\\\nmoonlight(1-flagstone)=6-flagstone-flagstone^{2}-flagstone^{3}-flagstone^{4}-flagstone^{5}-flagstone^{6}\n\\end{array}\n\\]\n\\[\n|moonlight||1-flagstone|>6-6|flagstone| \\geq 0 .\n\\]\n\nThus the zeros of \\( moonlight \\) are outside the unit circle.\nRemark. This is one of the problems criticized by L. J. Mordell in his article \"The Putnam Competition,\" American Muthemutical Monthly, vol. \\( 70(1963) \\), pages \\( 481-490 \\). The published solution in the Monthly is a ticated for an undergraduate competition. The examiners no doubt envisioned something like our first solution. See page 623 in Appendix. visioned something like our first solution. See page 623 in Appendix."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "minuscule",
        "k": "stillness",
        "l": "tranquility",
        "x": "constancy",
        "p": "improbable",
        "A": "vectorized",
        "B": "variance",
        "D": "numerator",
        "\\\\alpha_n": "\\\\scalarvalue",
        "\\\\beta": "\\\\columning"
      },
      "question": "6. A player throwing a die scores as many points as on the top face of the die and is to play until his score reaches or passes a total \\( \\minuscule \\). Denote by \\( \\improbable(\\minuscule) \\) the probability of making exactly the total \\( \\minuscule \\), and find the value of \\( \\lim _{\\minuscule \\rightarrow \\infty} \\improbable(\\minuscule) \\).",
      "solution": "First Solution. If by definition \\( \\improbable(0)=1 \\) and \\( \\improbable(\\minuscule)=0,\\; \\minuscule<0 \\) then the following relations are easy to verify.\n\\[\n\\begin{array}{l}\n\\improbable(0)=1 \\\\\n\\improbable(1)=\\frac{1}{6}\\,\\improbable(0)\n\\end{array}\n\\]\n(1)\n\\[\n\\begin{array}{l}\n\\improbable(2)=\\frac{1}{6}[\\improbable(1)+\\improbable(0)] \\\\\n\\improbable(3)=\\frac{1}{6}[\\improbable(2)+\\improbable(1)+\\improbable(0)] \\\\\n\\vdots \\\\\n\\improbable(\\minuscule)=\\frac{1}{6}[\\improbable(\\minuscule-1)+\\improbable(\\minuscule-2)+\\cdots+\\improbable(\\minuscule-6)],\\;\\minuscule>0 .\n\\end{array}\n\\]\nAdding these equations and then canceling, we obtain\n\\[\n\\improbable(\\minuscule)+\\frac{5}{6}\\,\\improbable(\\minuscule-1)+\\frac{4}{6}\\,\\improbable(\\minuscule-2)+\\cdots+\\frac{1}{6}\\,\\improbable(\\minuscule-5)=1 .\n\\]\nNow if it is assumed that \\( \\improbable(\\minuscule)\\to\\improbable \\) as \\( \\minuscule\\to\\infty \\), then it follows that \\( (21/6)\\,\\improbable=1 \\) and hence that \\( \\improbable=2/7 \\).\nWe shall now show that, regardless of the initial conditions, a sequence \\( \\{\\improbable(\\minuscule)\\} \\) that satisfies the recursion (1) is convergent.\n\nLet\n\\[\n\\begin{aligned}\n\\underline{\\improbable}(\\minuscule)&=\\min\\{\\improbable(\\minuscule-i):i=1,2,\\ldots,6\\},\\\\\n\\bar{\\improbable}(\\minuscule)&=\\max\\{\\improbable(\\minuscule-i):i=1,2,\\ldots,6\\}.\n\\end{aligned}\n\\]\nFrom (1) we obtain\n\\[\n\\frac{1}{6}[5\\,\\underline{\\improbable}(\\minuscule)+\\bar{\\improbable}(\\minuscule)]\\le \\improbable(\\minuscule)\\le\\frac{1}{6}[5\\,\\bar{\\improbable}(\\minuscule)+\\underline{\\improbable}(\\minuscule)].\n\\]\nHence \\( \\underline{\\improbable}(\\minuscule)\\le\\underline{\\improbable}(\\minuscule+1)\\le\\bar{\\improbable}(\\minuscule+1)\\le\\bar{\\improbable}(\\minuscule) \\), so \\( \\{\\underline{\\improbable}(\\minuscule)\\} \\) is non-decreasing and \\( \\{\\bar{\\improbable}(\\minuscule)\\} \\) is non-increasing. Let\n\\[\n\\begin{array}{l}\n\\bar{q}=\\displaystyle\\lim_{\\minuscule\\to\\infty}\\bar{\\improbable}(\\minuscule),\\\\[4pt]\n\\underline{q}=\\displaystyle\\lim_{\\minuscule\\to\\infty}\\underline{\\improbable}(\\minuscule).\n\\end{array}\n\\]\nThen \\( \\underline{q}\\le\\bar{q} \\).\nSuppose \\( \\bar{q}>\\underline{q} \\) and set \\( \\epsilon=\\frac{1}{5}(\\bar{q}-\\underline{q}) \\). Since \\( \\epsilon>0 \\), for some \\( N \\) and all \\( \\text{stillness}\\ge N \\)\n\\[\\underline{q}-\\epsilon<\\underline{\\improbable}(\\text{stillness})\\le\\underline{q},\\]\nso\n\\[\n\\frac{1}{6}[5(\\underline{q}-\\epsilon)+\\bar{q}]<\\frac{1}{6}[5\\,\\underline{\\improbable}(\\text{stillness})+\\bar{\\improbable}(\\text{stillness})]\\le\\improbable(\\text{stillness}),\n\\]\ni.e. \\( \\underline{q}<\\improbable(\\text{stillness}) \\). From this follows \\( \\underline{q}<\\underline{\\improbable}(\\text{stillness}) \\) for \\( \\text{stillness}>N+5 \\), a contradiction since \\( \\underline{\\improbable}(\\text{stillness}) \\) increases to \\( q \\).\nWe conclude \\( \\bar{q}=\\underline{q} \\). Then it follows immediately that \\( \\lim_{\\minuscule\\to\\infty}\\improbable(\\minuscule)=\\bar{q}=\\underline{q}. \\)\n\nSecond Solution. Define \\( \\improbable(0)=1,\\;\\improbable(\\minuscule)=0 \\) for \\( \\minuscule<0 \\), and for each \\( \\minuscule=0,1,2,\\ldots \\) let\n\\[\n\\scalarvalue_{\\minuscule}=(\\improbable(\\minuscule-5),\\improbable(\\minuscule-4),\\improbable(\\minuscule-3),\\improbable(\\minuscule-2),\\improbable(\\minuscule-1),\\improbable(\\minuscule))^{T}.\n\\]\nThen for all \\( \\minuscule \\) we have\n\\begin{tabular}{|l|l|}\n\\hline & \\( \\scalarvalue_{\\minuscule+1}=vectorized\\,\\scalarvalue_{\\minuscule} \\)\\\\\n\\hline where & \\\\\n\\hline & \\( vectorized=\\left(\\begin{array}{llllll}0&1&0&0&0&0\\\\0&0&1&0&0&0\\\\0&0&0&1&0&0\\\\0&0&0&0&1&0\\\\0&0&0&0&0&1\\\\\\tfrac16&\\tfrac16&\\tfrac16&\\tfrac16&\\tfrac16&\\tfrac16\\end{array}\\right) \\).\\\\\n\\hline\n\\end{tabular}\n\nHence \\( \\scalarvalue_{\\minuscule}=vectorized^{\\prime\\prime}\\,\\scalarvalue_{10} \\).\nNow \\( vectorized \\) is row-stochastic (i.e., it has non-negative entries and each row sums to one), and it is easy to see that \\( vectorized^{0} \\) has all positive entries. It is shown in the theory of Markov processes that, if some power of a row-stochastic matrix has all positive entries, its powers converge to a row-stochastic limit matrix that has all rows the same and\n\\[\n\\lim\\scalarvalue_{\\minuscule}=\\bigl(\\lim vectorized^{\\minuscule}\\bigr)\\scalarvalue_{0}=variance\\,\\scalarvalue_{0},\n\\]\na vector with all components the same, say \\( \\improbable \\). We have\n\\[\\lim_{\\minuscule}\\improbable(\\minuscule)=\\improbable.\\]\nIf \\( \\columning=(1,2,3,4,5,6) \\), then \\( \\columning\\,vectorized=\\columning \\). Therefore \\( \\columning\\,vectorized^{\\prime\\prime}=\\columning \\), and taking limits \\( \\columning\\,variance=\\columning \\). Then\n\\[21\\,\\improbable=\\columning\\,(variance\\,\\scalarvalue_{0})=\\columning\\,\\scalarvalue_{0}=6,\\]\nso \\( \\improbable=2/7 \\).\n(For the theorem on row-stochastic matrices see, for example, P. A. P. Moran, Introduction to Probability Theory, Clarendon Press, Oxford, 1968, page 112.)\n\nThird Solution. The probability of obtaining a total of \\( \\minuscule \\) in \\( stillness \\) throws (exactly) is the coefficient of \\( \\constancy^{\\minuscule} \\) in\n\\[\n\\biggl[\\tfrac16\\bigl(\\constancy+\\constancy^{2}+\\cdots+\\constancy^{6}\\bigr)\\biggr]^{stillness}.\n\\]\nSince obtaining \\( \\minuscule \\) in \\( stillness \\) throws and obtaining \\( \\minuscule \\) in \\( tranquility \\) throws for \\( stillness\\neq tranquility \\) are mutually exclusive events, the probability of ever obtaining a total of \\( \\minuscule \\) is the coefficient of \\( \\constancy^{\\minuscule} \\) in\n\\[\n\\sum_{stillness=0}^{\\infty}\\biggl[\\tfrac16\\bigl(\\constancy+\\constancy^{2}+\\cdots+\\constancy^{6}\\bigr)\\biggr]^{stillness}.\n\\]\nHence\n\\[\n\\begin{aligned}\n\\sum_{\\minuscule=0}^{\\infty}\\improbable(\\minuscule)\\,\\constancy^{\\minuscule}\n&=\\frac{6}{6-(\\constancy+\\constancy^{2}+\\cdots+\\constancy^{6})}\\\\[4pt]\n&=\\frac{6}{(1-\\constancy)(6+5\\constancy+4\\constancy^{2}+3\\constancy^{3}+2\\constancy^{4}+\\constancy^{5})}\\\\[4pt]\n&=\\frac{2}{7(1-\\constancy)}+\\frac{2}{7}\\,\\frac{15+10\\constancy+6\\constancy^{2}+3\\constancy^{3}+\\constancy^{4}}{6+5\\constancy+4\\constancy^{2}+3\\constancy^{3}+2\\constancy^{4}+\\constancy^{5}}.\n\\end{aligned}\n\\]\nThus\n\\[\n\\sum_{\\minuscule=0}^{\\infty}\\bigl(\\improbable(\\minuscule)-\\tfrac27\\bigr)\\,\\constancy^{\\minuscule}=\\frac{2}{7}\\,\\frac{15+10\\constancy+6\\constancy^{2}+3\\constancy^{3}+\\constancy^{4}}{6+5\\constancy+4\\constancy^{2}+3\\constancy^{3}+2\\constancy^{4}+\\constancy^{5}}.\n\\]\nIf we now regard \\( \\constancy \\) as a complex variable and show that the denominator of the last fraction does not vanish inside or on the unit circle in the complex plane, then it follows that for some \\( \\constancy>1 \\) the series converges, and hence \\( \\lim_{\\minuscule\\to\\infty}\\improbable(\\minuscule)=2/7 \\).\nLet\n\\[\n\\begin{array}{c}\n\\numerator=6+5\\constancy+4\\constancy^{2}+3\\constancy^{3}+2\\constancy^{4}+\\constancy^{5}.\\\\\n\\text{If }\\constancy=1,\\;\\numerator\\neq0. \\text{ Assume now that }\\constancy\\neq1\\text{ but }|\\constancy|\\le1.\\\\\n\\numerator(1-\\constancy)=6-\\constancy-\\constancy^{2}-\\constancy^{3}-\\constancy^{4}-\\constancy^{5}-\\constancy^{6}.\n\\end{array}\n\\]\n\\[|\\numerator|\\,|1-\\constancy|>6-6|\\constancy|\\ge0.\\]\nThus the zeros of \\( \\numerator \\) are outside the unit circle.\n\nRemark. This is one of the problems criticized by L. J. Mordell in his article “The Putnam Competition,” American Mathematical Monthly, vol. 70 (1963), pp. 481–490. The published solution in the Monthly is a little too sophisticated for an undergraduate competition; the examiners no doubt envisioned something like our first solution."
    },
    "garbled_string": {
      "map": {
        "n": "qclwzprt",
        "k": "zxmrvbat",
        "l": "vdfnqshk",
        "x": "prgfkltm",
        "p": "htrswcnd",
        "A": "mpqsgltn",
        "B": "jcnbrvfd",
        "D": "wxlmqzsp",
        "\\alpha_n": "qzxwvtnp",
        "\\beta": "hjgrksla"
      },
      "question": "6. A player throwing a die scores as many points as on the top face of the die and is to play until his score reaches or passes a total \\( qclwzprt \\). Denote by \\( htrswcnd(qclwzprt) \\) the probability of making exactly the total \\( qclwzprt \\), and find the value of \\( \\lim _{qclwzprt \\rightarrow \\infty} htrswcnd(qclwzprt) \\).",
      "solution": "First Solution. If by definition \\( htrswcnd(0)=1 \\) and \\( htrswcnd(qclwzprt)=0,\\, qclwzprt<0 \\) then the following relations are easy to verify.\n\\[\n\\begin{array}{l}\nhtrswcnd(0)=1 \\\\\nhtrswcnd(1)=\\dfrac{1}{6}\\,htrswcnd(0)\n\\end{array}\n\\]\n(1)\n\\[\n\\begin{array}{l}\nhtrswcnd(2)=\\dfrac{1}{6}[htrswcnd(1)+htrswcnd(0)] \\\\\nhtrswcnd(3)=\\dfrac{1}{6}[htrswcnd(2)+htrswcnd(1)+htrswcnd(0)] \\\\\n\\vdots \\\\\nhtrswcnd(qclwzprt)=\\dfrac{1}{6}[htrswcnd(qclwzprt-1)+htrswcnd(qclwzprt-2)+\\cdots+htrswcnd(qclwzprt-6)],\\; qclwzprt>0 .\n\\end{array}\n\\]\nAdding these equations and then canceling, we obtain\n\\[\nhtrswcnd(qclwzprt)+\\frac{5}{6}\\,htrswcnd(qclwzprt-1)+\\frac{4}{6}\\,htrswcnd(qclwzprt-2)+\\cdots+\\frac{1}{6}\\,htrswcnd(qclwzprt-5)=1 .\n\\]\nNow if it is assumed that \\( htrswcnd(qclwzprt) \\rightarrow htrswcnd \\) as \\( qclwzprt \\rightarrow \\infty \\), then it follows that \\( (21/6)\\,htrswcnd=1 \\) and hence that \\( htrswcnd=2/7 \\).\nWe shall now show that, regardless of the initial conditions, a sequence \\{htrswcnd(qclwzprt)\\} that satisfies the recursion (1) is convergent.\n\nLet\n\\[\n\\begin{aligned}\n\\underline{htrswcnd}(qclwzprt) & = \\min \\{htrswcnd(qclwzprt-i): i=1,2, \\ldots, 6\\} \\\\\n\\bar{htrswcnd}(qclwzprt) & = \\max \\{htrswcnd(qclwzprt-i): i=1,2, \\ldots, 6\\}\n\\end{aligned}\n\\]\nFrom (1) we obtain\n\\[\n\\frac{1}{6}\\bigl[5\\,\\underline{htrswcnd}(qclwzprt)+\\bar{htrswcnd}(qclwzprt)\\bigr]\\le htrswcnd(qclwzprt)\\le \\frac{1}{6}\\bigl[5\\,\\bar{htrswcnd}(qclwzprt)+\\underline{htrswcnd}(qclwzprt)\\bigr] .\n\\]\nHence \\( \\underline{htrswcnd}(qclwzprt) \\le \\underline{htrswcnd}(qclwzprt+1) \\le \\bar{htrswcnd}(qclwzprt+1) \\le \\bar{htrswcnd}(qclwzprt) \\), so the sequence \\{\\underline{htrswcnd}(qclwzprt)\\} is non-decreasing and \\{\\bar{htrswcnd}(qclwzprt)\\} is non-increasing. Let\n\\[\n\\begin{array}{l}\n\\bar{q}=\\displaystyle\\lim_{qclwzprt\\to\\infty} \\bar{htrswcnd}(qclwzprt) \\\\\n\\underline{q}=\\displaystyle\\lim_{qclwzprt\\to\\infty} \\underline{htrswcnd}(qclwzprt)\n\\end{array}\n\\]\nThen \\( \\underline{q}\\le\\bar{q} \\).\nNow suppose \\( \\bar{q}>\\underline{q} \\) and set \\( \\varepsilon=\\tfrac15(\\bar{q}-\\underline{q}) \\). Since \\( \\varepsilon>0 \\), for some \\( N \\) and all \\( qclwzprt\\ge N \\)\n\\[ \\underline{q}-\\varepsilon<\\underline{htrswcnd}(qclwzprt)\\le\\underline{q} ,\\]\nso\n\\[ \\frac16[5(\\underline{q}-\\varepsilon)+\\bar{q}]<\\frac16[5\\,\\underline{htrswcnd}(qclwzprt)+\\bar{htrswcnd}(qclwzprt)]\\le htrswcnd(qclwzprt), \\]\ni.e. \\( \\underline{q}<htrswcnd(qclwzprt) \\). From this follows \\( \\underline{q}<\\underline{htrswcnd}(qclwzprt) \\) for \\( qclwzprt>N+5 \\), a contradiction since \\( \\underline{htrswcnd}(qclwzprt) \\) increases to \\( \\underline{q} \\).\nWe conclude \\( \\bar{q}=\\underline{q} \\). Then it follows immediately that \\( \\lim _{qclwzprt\\to\\infty} htrswcnd(qclwzprt)=\\bar{q}=\\underline{q} \\).\n\nSecond Solution. Define \\( htrswcnd(0)=1,\\, htrswcnd(qclwzprt)=0 \\) for \\( qclwzprt<0 \\), and for each \\( qclwzprt=0,1,2, \\ldots \\) let\n\\[\nqzxwvtnp_{qclwzprt}=(htrswcnd(qclwzprt-5),\\,htrswcnd(qclwzprt-4),\\,htrswcnd(qclwzprt-3),\\,htrswcnd(qclwzprt-2),\\,htrswcnd(qclwzprt-1),\\,htrswcnd(qclwzprt))^{T} .\n\\]\nThen for all \\( qclwzprt \\) we have\n\\begin{tabular}{|l|l|}\n\\hline & \\( qzxwvtnp_{qclwzprt+1}=mpqsgltn\\,qzxwvtnp_{qclwzprt} \\) \\\\\n\\hline where & \\\\\n\\hline & \\( mpqsgltn=\\left(\\begin{array}{llllll}0 & 1 & 0 & 0 & 0 & 0\\\\ 0 & 0 & 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 0 & 1 & 0\\\\ 0 & 0 & 0 & 0 & 0 & 1\\\\ \\dfrac16 & \\dfrac16 & \\dfrac16 & \\dfrac16 & \\dfrac16 & \\dfrac16\\end{array}\\right) \\). \\\\\n\\hline\n\\end{tabular}\n\nHence \\( qzxwvtnp_{qclwzprt}=mpqsgltn^{\\,qclwzprt}\\,qzxwvtnp_{0} \\).\nNow \\( mpqsgltn \\) is row-stochastic (i.e. it has non-negative entries and each row sums to one), and it is easy to see that some power of it has all positive entries. It is shown in the theory of Markov processes that, if some power of a row-stochastic matrix has all positive entries, its powers converge to a row-stochastic limit matrix whose rows are all the same; denote this limit by \\( jcnbrvfd \\). Then\n\\[\n\\lim qzxwvtnp_{qclwzprt}=(\\lim mpqsgltn^{qclwzprt})\\,qzxwvtnp_{0}=jcnbrvfd\\,qzxwvtnp_{0},\n\\]\na vector with all components the same, say \\( htrswcnd \\). We have\n\\[ \\lim_{qclwzprt\\to\\infty} htrswcnd(qclwzprt)=htrswcnd . \\]\nIf \\( hjgrksla=(1,2,3,4,5,6) \\), then \\( hjgrksla\\,mpqsgltn=hjgrksla \\). Therefore \\( hjgrksla\\,mpqsgltn^{\\,qclwzprt}=hjgrksla \\), and taking limits \\( hjgrksla\\,jcnbrvfd=hjgrksla \\). Hence\n\\[ 21\\,htrswcnd=hjgrksla\\,(jcnbrvfd\\,qzxwvtnp_{0})=hjgrksla\\,qzxwvtnp_{0}=6 , \\]\nso \\( htrswcnd=2/7 \\).\nFor the theorem on row-stochastic matrices see, for example, P. A. P. Moran, Introduction to Probability Theory, Clarendon Press, Oxford, 1968, page 112.\n\nThird Solution. We now roll up the really heavy artillery and argue as follows.\nThe probability of obtaining a total of \\( qclwzprt \\) in \\( zxmrvbat \\) throws (exactly) is the coefficient of \\( prgfkltm^{qclwzprt} \\) in\n\\[ \\left[\\frac16\\bigl(prgfkltm+prgfkltm^{2}+\\cdots+prgfkltm^{6}\\bigr)\\right]^{zxmrvbat}. \\]\nSince obtaining \\( qclwzprt \\) in \\( zxmrvbat \\) throws and obtaining \\( qclwzprt \\) in \\( vdfnqshk \\) throws for \\( zxmrvbat\\neq vdfnqshk \\) are mutually exclusive events, the probability of ever obtaining a total of \\( qclwzprt \\) is the coefficient of \\( prgfkltm^{qclwzprt} \\) in\n\\[ \\sum_{zxmrvbat=0}^{\\infty}\\left[\\frac16\\bigl(prgfkltm+prgfkltm^{2}+\\cdots+prgfkltm^{6}\\bigr)\\right]^{zxmrvbat}. \\]\nHence\n\\[\n\\begin{aligned}\n\\sum_{qclwzprt=0}^{\\infty} htrswcnd(qclwzprt)\\,prgfkltm^{qclwzprt} &=\\frac{6}{6-(prgfkltm+prgfkltm^{2}+\\cdots+prgfkltm^{6})}\\\\[4pt]\n&=\\frac{6}{(1-prgfkltm)\\bigl(6+5prgfkltm+4prgfkltm^{2}+3prgfkltm^{3}+2prgfkltm^{4}+prgfkltm^{5}\\bigr)}\\\\[4pt]\n&=\\frac{2}{7(1-prgfkltm)}+\\frac{2}{7}\\,\\frac{15+10prgfkltm+6prgfkltm^{2}+3prgfkltm^{3}+prgfkltm^{4}}{6+5prgfkltm+4prgfkltm^{2}+3prgfkltm^{3}+2prgfkltm^{4}+prgfkltm^{5}}.\n\\end{aligned}\n\\]\nThus\n\\[ \\sum_{qclwzprt=0}^{\\infty}\\bigl(htrswcnd(qclwzprt)-\\tfrac{2}{7}\\bigr)prgfkltm^{qclwzprt}=\\frac{2}{7}\\,\\frac{15+10prgfkltm+6prgfkltm^{2}+3prgfkltm^{3}+prgfkltm^{4}}{6+5prgfkltm+4prgfkltm^{2}+3prgfkltm^{3}+2prgfkltm^{4}+prgfkltm^{5}} .\\]\nSo far the argument has been formally combinatorial in character. If we now regard \\( prgfkltm \\) as a complex variable and show that the denominator of the last fraction does not vanish inside or on the unit circle in the complex plane, then it follows that for some \\( |prgfkltm|>1 \\), \\( \\sum_{qclwzprt=0}^{\\infty}(htrswcnd(qclwzprt)-2/7)prgfkltm^{qclwzprt} \\) converges and hence \\( \\lim_{qclwzprt\\to\\infty} htrswcnd(qclwzprt)=2/7 \\).\nLet\n\\[\n\\begin{array}{c}\nwxlmqzsp=6+5prgfkltm+4prgfkltm^{2}+3prgfkltm^{3}+2prgfkltm^{4}+prgfkltm^{5} \\\\\n\\text{If } prgfkltm=1,\\,wxlmqzsp\\neq0.\\text{ Assume now that } prgfkltm\\neq1\\text{ but }|prgfkltm|\\le1.\\text{ Then } \\\\\nwxlmqzsp(1-prgfkltm)=6-prgfkltm-prgfkltm^{2}-prgfkltm^{3}-prgfkltm^{4}-prgfkltm^{5}-prgfkltm^{6}\n\\end{array}\n\\]\n\\[ |wxlmqzsp|\\,|1-prgfkltm|>6-6|prgfkltm|\\ge0 . \\]\nThus the zeros of \\( wxlmqzsp \\) are outside the unit circle.\n\nRemark. This is one of the problems criticized by L. J. Mordell in his article “The Putnam Competition,” American Mathematical Monthly, vol. 70 (1963), pages 481–490. The published solution in the Monthly is a little too sophisticated for an undergraduate competition. The examiners no doubt envisioned something like our first solution. See page 623 in Appendix."
    },
    "kernel_variant": {
      "question": "A player successively throws three different fair dice in the fixed cyclic order  \n\n\\[\nT,\\;C,\\;O,\\;T,\\;C,\\;O,\\;T,\\;C,\\;O,\\;\\ldots ,\n\\]\n\nwhere  \n\n\\[\n\\begin{aligned}\n&\\text{tetrahedral }T:\\; &&\\{1,2,3,4\\}, \\\\\n&\\text{cubic }C:\\;       &&\\{1,2,3,4,5,6\\},\\\\\n&\\text{octahedral }O:\\;  &&\\{1,2,3,4,5,6,7,8\\}.\n\\end{aligned}\n\\]\n\nEvery time a die is thrown the number showing is added to the running total.  \nLet \\(S_{0}=0\\) and, for \\(k\\ge 1\\), let \\(S_{k}\\) be the total after the first \\(k\\) throws.  \nFor an integer \\(n\\ge 1\\) the game stops at  \n\n\\[\n\\tau(n)=\\min\\{k\\ge 1:\\;S_{k}\\ge n\\},\n\\]\n\nthat is, at the very first throw (not necessarily at the end of a \\(T\\!-\\!C\\!-\\!O\\) cycle) at which the running total reaches or exceeds \\(n\\).  \nDefine  \n\n\\[\np(n)=\\Pr\\bigl\\{S_{\\tau(n)}=n\\bigr\\},\n\\]\n\nthe probability that the game terminates \\emph{exactly} on \\(n\\).  \n\nEvaluate the limit  \n\n\\[\n\\boxed{\\displaystyle \\lim_{n\\to\\infty} p(n)}.\n\\]\n\n--------------------------------------------------------------------",
      "solution": "Step 1.  Passing to cycle–boundaries  \nAfter every complete \\(T\\!-\\!C\\!-\\!O\\) cycle (three throws) the increment\n\n\\[\nZ:=X_{1}+X_{2}+X_{3}\n\\]\n\nis added, where  \n\n\\[\n\\Pr\\{X_{1}=j\\}= \\tfrac14,\\; j=1,\\ldots ,4,\\qquad\n\\Pr\\{X_{2}=j\\}= \\tfrac16,\\; j=1,\\ldots ,6,\\qquad\n\\Pr\\{X_{3}=j\\}= \\tfrac18,\\; j=1,\\ldots ,8.\n\\]\n\nHence \\(Z\\) takes every integer value from \\(3\\) to \\(18\\) and\n\n\\[\n\\mu:=\\mathbb{E}[Z]=2.5+3.5+4.5=10.5.\n\\]\n\nWrite  \n\n\\[\nR_{m}=S_{3m},\\qquad m\\ge 0 .\n\\]\n\nThe sequence \\(\\{R_{m}\\}_{m\\ge 0}\\) is a classical renewal process with i.i.d. increment \\(Z\\).\n\nStep 2.  “One–cycle’’ hitting probabilities  \nFix \\(r\\ge 1\\).  Suppose we stand at a cycle boundary with deficit \\(r\\)\npoints still needed to reach the target \\(n\\).  During the \\emph{next}\ncycle we successively observe the partial sums  \n\n\\[\nY_{1}=X_{1},\\qquad \nY_{2}=X_{1}+X_{2},\\qquad \nY_{3}=X_{1}+X_{2}+X_{3}.\n\\]\n\nBecause the increments are positive we always have  \n\n\\[\n0<Y_{1}<Y_{2}<Y_{3}=Z\\le 18 .\n\\]\n\nDefine  \n\n\\[\nu(r)=\\Pr\\bigl\\{\\text{the player finishes exactly on the target during the\nnext cycle}\\mid\\text{deficit }r\\bigr\\}.\n\\]\n\nConditioned on the concrete values of \\((X_{1},X_{2},X_{3})\\) the set of\ndeficits for which the game would end in that upcoming cycle is precisely\n\\(\\{Y_{1},Y_{2},Y_{3}\\}\\).  Consequently\n\n\\[\n\\sum_{r=1}^{\\infty}u(r)\n   =\\mathbb{E}\\bigl[\\#\\{Y_{1},Y_{2},Y_{3}\\}\\bigr]\n   =\\mathbb{E}[3]=3.   \\tag{\\(\\ast\\)}\n\\]\n\n(Every realisation contributes exactly three distinct values, so the\nexpectation is \\(3\\).)  Thus the function \\(u\\) is summable with  \n\n\\[\n\\lambda:=\\sum_{r=1}^{\\infty}u(r)=3 .\n\\]\n\nStep 3.  Renewal decomposition of \\(p(n)\\)  \nWhenever the total is inspected only at cycle boundaries, the event\n\\(\\{S_{\\tau(n)}=n\\}\\) can occur in two stages:\n\n• First a boundary total \\(R_{m}\\) lands at \\(n-r\\) for some\n\\(r\\ge 1\\);  \n\n• then, during the immediately following cycle, the deficit \\(r\\) is\nwiped out \\emph{exactly} (and not crossed) at one of the three throws.\n\nTaking expectations and using the independence between different cycles,\nwe obtain the discrete renewal equation\n\n\\[\np(n)=u(n)+\\sum_{k=3}^{18} a_{k}\\,p(n-k),\\qquad n\\ge 1,\n\\]\n\nwhere \\(a_{k}=\\Pr\\{Z=k\\}\\).  In convolution notation  \n\\(p=u+ a* p\\).\n\nStep 4.  Application of the key renewal theorem  \nBecause the increment distribution \\(Z\\) is aperiodic  \n(\\(\\gcd\\{k:a_{k}>0\\}=1\\)) and has finite mean \\(\\mu\\), the\nkey renewal theorem for \\emph{defective} renewal equations yields\n\n\\[\n\\lim_{n\\to\\infty}p(n)=\\frac{\\lambda}{\\mu},\n\\]\n\nsee e.g.\\ Theorem 5.4.1 in Feller, \\emph{An Introduction to Probability\nTheory}, vol.\\ 2.\n\nStep 5.  Numerical evaluation  \nWith \\(\\lambda=3\\) from \\((\\ast)\\) and \\(\\mu=10.5\\) from Step 1,\n\n\\[\n\\boxed{\\displaystyle\n\\lim_{n\\to\\infty} p(n)=\\frac{3}{10.5}=\\frac{2}{7}}.\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.519856",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional/periodic structure  \n   • The increments are no longer identically distributed per throw; instead the die changes in a deterministic 3-periodic fashion.  \n   • One must aggregate throws into cycles and recognise a renewal structure on that coarser time scale.\n\n2. Longer recurrence and larger span  \n   • The first-passage recurrence (†) involves 16 previous terms, compared with 6 in the original octahedral variant.\n\n3. Advanced theoretical input  \n   • A direct telescoping trick used in the original problem no longer works.  \n   • The solution requires the key renewal theorem (or the Markov-renewal analogue), an undergraduate-level but non-elementary result.\n\n4. Multiple interacting concepts  \n   • Identification of the correct regenerative time scale.  \n   • Construction of the first-passage recurrence in that scale.  \n   • Use of lattice aperiodicity to invoke asymptotic renewal theory.\n\n5. Final constant unrelated to any single die  \n   • The limiting probability 2/21 depends on the combined mean of three different dice rather than on a single simple expectation.\n\nAll these layers make the enhanced variant substantially more technical and concept-heavy than either the original six-sided-die problem or the one-die octahedral kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "A player successively throws three different fair dice in the fixed cyclic order  \n\n\\[\nT,\\;C,\\;O,\\;T,\\;C,\\;O,\\;T,\\;C,\\;O,\\;\\ldots ,\n\\]\n\nwhere  \n\n\\[\n\\begin{aligned}\n&\\text{tetrahedral }T:\\; &&\\{1,2,3,4\\}, \\\\\n&\\text{cubic }C:\\;       &&\\{1,2,3,4,5,6\\},\\\\\n&\\text{octahedral }O:\\;  &&\\{1,2,3,4,5,6,7,8\\}.\n\\end{aligned}\n\\]\n\nEvery time a die is thrown the number showing is added to the running total.  \nLet \\(S_{0}=0\\) and, for \\(k\\ge 1\\), let \\(S_{k}\\) be the total after the first \\(k\\) throws.  \nFor an integer \\(n\\ge 1\\) the game stops at  \n\n\\[\n\\tau(n)=\\min\\{k\\ge 1:\\;S_{k}\\ge n\\},\n\\]\n\nthat is, at the very first throw (not necessarily at the end of a \\(T\\!-\\!C\\!-\\!O\\) cycle) at which the running total reaches or exceeds \\(n\\).  \nDefine  \n\n\\[\np(n)=\\Pr\\bigl\\{S_{\\tau(n)}=n\\bigr\\},\n\\]\n\nthe probability that the game terminates \\emph{exactly} on \\(n\\).  \n\nEvaluate the limit  \n\n\\[\n\\boxed{\\displaystyle \\lim_{n\\to\\infty} p(n)}.\n\\]\n\n--------------------------------------------------------------------",
      "solution": "Step 1.  Passing to cycle–boundaries  \nAfter every complete \\(T\\!-\\!C\\!-\\!O\\) cycle (three throws) the increment\n\n\\[\nZ:=X_{1}+X_{2}+X_{3}\n\\]\n\nis added, where  \n\n\\[\n\\Pr\\{X_{1}=j\\}= \\tfrac14,\\; j=1,\\ldots ,4,\\qquad\n\\Pr\\{X_{2}=j\\}= \\tfrac16,\\; j=1,\\ldots ,6,\\qquad\n\\Pr\\{X_{3}=j\\}= \\tfrac18,\\; j=1,\\ldots ,8.\n\\]\n\nHence \\(Z\\) takes every integer value from \\(3\\) to \\(18\\) and\n\n\\[\n\\mu:=\\mathbb{E}[Z]=2.5+3.5+4.5=10.5.\n\\]\n\nWrite  \n\n\\[\nR_{m}=S_{3m},\\qquad m\\ge 0 .\n\\]\n\nThe sequence \\(\\{R_{m}\\}_{m\\ge 0}\\) is a classical renewal process with i.i.d. increment \\(Z\\).\n\nStep 2.  “One–cycle’’ hitting probabilities  \nFix \\(r\\ge 1\\).  Suppose we stand at a cycle boundary with deficit \\(r\\)\npoints still needed to reach the target \\(n\\).  During the \\emph{next}\ncycle we successively observe the partial sums  \n\n\\[\nY_{1}=X_{1},\\qquad \nY_{2}=X_{1}+X_{2},\\qquad \nY_{3}=X_{1}+X_{2}+X_{3}.\n\\]\n\nBecause the increments are positive we always have  \n\n\\[\n0<Y_{1}<Y_{2}<Y_{3}=Z\\le 18 .\n\\]\n\nDefine  \n\n\\[\nu(r)=\\Pr\\bigl\\{\\text{the player finishes exactly on the target during the\nnext cycle}\\mid\\text{deficit }r\\bigr\\}.\n\\]\n\nConditioned on the concrete values of \\((X_{1},X_{2},X_{3})\\) the set of\ndeficits for which the game would end in that upcoming cycle is precisely\n\\(\\{Y_{1},Y_{2},Y_{3}\\}\\).  Consequently\n\n\\[\n\\sum_{r=1}^{\\infty}u(r)\n   =\\mathbb{E}\\bigl[\\#\\{Y_{1},Y_{2},Y_{3}\\}\\bigr]\n   =\\mathbb{E}[3]=3.   \\tag{\\(\\ast\\)}\n\\]\n\n(Every realisation contributes exactly three distinct values, so the\nexpectation is \\(3\\).)  Thus the function \\(u\\) is summable with  \n\n\\[\n\\lambda:=\\sum_{r=1}^{\\infty}u(r)=3 .\n\\]\n\nStep 3.  Renewal decomposition of \\(p(n)\\)  \nWhenever the total is inspected only at cycle boundaries, the event\n\\(\\{S_{\\tau(n)}=n\\}\\) can occur in two stages:\n\n• First a boundary total \\(R_{m}\\) lands at \\(n-r\\) for some\n\\(r\\ge 1\\);  \n\n• then, during the immediately following cycle, the deficit \\(r\\) is\nwiped out \\emph{exactly} (and not crossed) at one of the three throws.\n\nTaking expectations and using the independence between different cycles,\nwe obtain the discrete renewal equation\n\n\\[\np(n)=u(n)+\\sum_{k=3}^{18} a_{k}\\,p(n-k),\\qquad n\\ge 1,\n\\]\n\nwhere \\(a_{k}=\\Pr\\{Z=k\\}\\).  In convolution notation  \n\\(p=u+ a* p\\).\n\nStep 4.  Application of the key renewal theorem  \nBecause the increment distribution \\(Z\\) is aperiodic  \n(\\(\\gcd\\{k:a_{k}>0\\}=1\\)) and has finite mean \\(\\mu\\), the\nkey renewal theorem for \\emph{defective} renewal equations yields\n\n\\[\n\\lim_{n\\to\\infty}p(n)=\\frac{\\lambda}{\\mu},\n\\]\n\nsee e.g.\\ Theorem 5.4.1 in Feller, \\emph{An Introduction to Probability\nTheory}, vol.\\ 2.\n\nStep 5.  Numerical evaluation  \nWith \\(\\lambda=3\\) from \\((\\ast)\\) and \\(\\mu=10.5\\) from Step 1,\n\n\\[\n\\boxed{\\displaystyle\n\\lim_{n\\to\\infty} p(n)=\\frac{3}{10.5}=\\frac{2}{7}}.\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.435182",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional/periodic structure  \n   • The increments are no longer identically distributed per throw; instead the die changes in a deterministic 3-periodic fashion.  \n   • One must aggregate throws into cycles and recognise a renewal structure on that coarser time scale.\n\n2. Longer recurrence and larger span  \n   • The first-passage recurrence (†) involves 16 previous terms, compared with 6 in the original octahedral variant.\n\n3. Advanced theoretical input  \n   • A direct telescoping trick used in the original problem no longer works.  \n   • The solution requires the key renewal theorem (or the Markov-renewal analogue), an undergraduate-level but non-elementary result.\n\n4. Multiple interacting concepts  \n   • Identification of the correct regenerative time scale.  \n   • Construction of the first-passage recurrence in that scale.  \n   • Use of lattice aperiodicity to invoke asymptotic renewal theory.\n\n5. Final constant unrelated to any single die  \n   • The limiting probability 2/21 depends on the combined mean of three different dice rather than on a single simple expectation.\n\nAll these layers make the enhanced variant substantially more technical and concept-heavy than either the original six-sided-die problem or the one-die octahedral kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}