{
  "index": "1989-A-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "If $\\alpha$ is an irrational number, $0 < \\alpha < 1$, is there a\nfinite game with an honest coin such that the probability of one player\nwinning the game is $\\alpha$? (An honest coin is one for which the\nprobability of heads and the probability of tails are both $\\frac12$.\nA game is finite if with probability 1 it must end in a finite number of moves.)",
  "solution": "Solution. Let \\( d_{n} \\) be 0 or 1 , depending on whether the \\( n^{\\text {th }} \\) toss yields heads or tails. Let \\( X=\\sum_{n=1}^{\\infty} d_{n} / 2^{n} \\). Then the distribution of \\( X \\) is uniform on \\( [0,1] \\), since for any rational number \\( c / 2^{m} \\) (i.e., any dyadic rational) in [0, 1], the probability that \\( X \\in\\left[0, c / 2^{m}\\right] \\) is exactly \\( c / 2^{m} \\).\n\nSay that player 1 wins the game after \\( N \\) tosses, if it is guaranteed at that time that the eventual value of \\( X \\) will be less than \\( \\alpha \\); this means that\n\\[\n\\sum_{n=1}^{N} \\frac{d_{n}}{2^{n}}+\\sum_{n=N+1}^{\\infty} \\frac{1}{2^{n}}<\\alpha\n\\]\n\nSimilarly, say that player 2 wins after \\( N \\) tosses, if it is guaranteed then that \\( X \\) will be greater than \\( \\alpha \\).\n\nThe game will terminate if \\( X \\neq \\alpha \\), which happens with probability 1 ; in fact it will terminate at the \\( N^{\\text {th }} \\) toss or earlier if \\( |X-\\alpha|>1 / 2^{N} \\). The probability that player 1 wins is the probability that \\( X \\in[0, \\alpha) \\), which is \\( \\alpha \\).\n\nRemark. The solution shows that the answer is yes for all real \\( \\alpha \\in[0,1] \\) : there is no need to assume that \\( \\alpha \\) is irrational.\n\nRemark. Essentially the same idea appears in [New, Problem 8]:\nDevise an experiment which uses only tosses of a fair coin, but which has success probability \\( 1 / 3 \\). Do the same for any success probability \\( p, 0 \\leq p \\leq 1 \\).\n\nRelated question. Show that if \\( \\alpha \\in[0,1] \\) is not a dyadic rational (i.e., not a rational number with denominator equal to a power of 2), the expected number of tosses in the game in the solution equals 2 .\n\nRelated question. For \\( \\alpha \\in[0,1] \\), let \\( f(\\alpha) \\) be the minimum over all games satisfying the conditions of the problem (such that player 1 wins with probability \\( \\alpha \\) ) of the expected number of tosses in the game. (For some games, the expected number may be infinite; ignore those.) Prove that if \\( \\alpha \\in[0,1] \\) is a rational number whose denominator in lowest terms is \\( 2^{m} \\) for some \\( m \\geq 0 \\), then \\( f(\\alpha)=2-1 / 2^{m-1} \\). Prove that if \\( \\alpha \\) is any other real number in \\( [0,1] \\), then \\( f(\\alpha)=2 \\). (This is essentially [New, Problem 103].)",
  "vars": [
    "X",
    "d_n",
    "n",
    "N",
    "m",
    "c"
  ],
  "params": [
    "\\\\alpha",
    "p",
    "f"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "X": "randomreal",
        "d_n": "binarydigit",
        "n": "indexsmall",
        "N": "indextotal",
        "m": "powerindex",
        "c": "dyadicnumer",
        "\\alpha": "targetprob",
        "p": "genericprob",
        "f": "expectedtoss"
      },
      "question": "If $targetprob$ is an irrational number, $0 < targetprob < 1$, is there a\nfinite game with an honest coin such that the probability of one player\nwinning the game is $targetprob$? (An honest coin is one for which the\nprobability of heads and the probability of tails are both $\\frac12$.\nA game is finite if with probability 1 it must end in a finite number of moves.)",
      "solution": "Solution. Let \\( binarydigit \\) be 0 or 1 , depending on whether the \\( indexsmall^{\\text {th }} \\) toss yields heads or tails. Let \\( randomreal=\\sum_{indexsmall=1}^{\\infty} binarydigit / 2^{indexsmall} \\). Then the distribution of \\( randomreal \\) is uniform on \\( [0,1] \\), since for any rational number \\( dyadicnumer / 2^{powerindex} \\) (i.e., any dyadic rational) in [0, 1], the probability that \\( randomreal \\in\\left[0, dyadicnumer / 2^{powerindex}\\right] \\) is exactly \\( dyadicnumer / 2^{powerindex} \\).\n\nSay that player 1 wins the game after \\( indextotal \\) tosses, if it is guaranteed at that time that the eventual value of \\( randomreal \\) will be less than \\( targetprob \\); this means that\n\\[\n\\sum_{indexsmall=1}^{indextotal} \\frac{binarydigit}{2^{indexsmall}}+\\sum_{indexsmall=indextotal+1}^{\\infty} \\frac{1}{2^{indexsmall}}<targetprob\n\\]\n\nSimilarly, say that player 2 wins after \\( indextotal \\) tosses, if it is guaranteed then that \\( randomreal \\) will be greater than \\( targetprob \\).\n\nThe game will terminate if \\( randomreal \\neq targetprob \\), which happens with probability 1 ; in fact it will terminate at the \\( indextotal^{\\text {th }} \\) toss or earlier if \\( |randomreal-targetprob|>1 / 2^{indextotal} \\). The probability that player 1 wins is the probability that \\( randomreal \\in[0, targetprob) \\), which is \\( targetprob \\).\n\nRemark. The solution shows that the answer is yes for all real \\( targetprob \\in[0,1] \\) : there is no need to assume that \\( targetprob \\) is irrational.\n\nRemark. Essentially the same idea appears in [New, Problem 8]:\nDevise an experiment which uses only tosses of a fair coin, but which has success probability \\( 1 / 3 \\). Do the same for any success probability \\( genericprob, 0 \\leq genericprob \\leq 1 \\).\n\nRelated question. Show that if \\( targetprob \\in[0,1] \\) is not a dyadic rational (i.e., not a rational number with denominator equal to a power of 2), the expected number of tosses in the game in the solution equals 2 .\n\nRelated question. For \\( targetprob \\in[0,1] \\), let \\( expectedtoss(targetprob) \\) be the minimum over all games satisfying the conditions of the problem (such that player 1 wins with probability \\( targetprob \\) ) of the expected number of tosses in the game. (For some games, the expected number may be infinite; ignore those.) Prove that if \\( targetprob \\in[0,1] \\) is a rational number whose denominator in lowest terms is \\( 2^{powerindex} \\) for some \\( powerindex \\geq 0 \\), then \\( expectedtoss(targetprob)=2-1 / 2^{powerindex-1} \\). Prove that if \\( targetprob \\) is any other real number in \\( [0,1] \\), then \\( expectedtoss(targetprob)=2 \\). (This is essentially [New, Problem 103].)"
    },
    "descriptive_long_confusing": {
      "map": {
        "X": "compassrose",
        "d_n": "lighthouse",
        "n": "rainforest",
        "N": "gardenpath",
        "m": "waterwheel",
        "c": "stargazer",
        "\\\\alpha": "sunflower",
        "p": "moonstone",
        "f": "barnswallow"
      },
      "question": "If $sunflower$ is an irrational number, $0 < sunflower < 1$, is there a finite game with an honest coin such that the probability of one player winning the game is $sunflower$? (An honest coin is one for which the probability of heads and the probability of tails are both $\\frac12$. A game is finite if with probability 1 it must end in a finite number of moves.)",
      "solution": "Solution. Let \\( lighthouse_{rainforest} \\) be 0 or 1 , depending on whether the \\( rainforest^{\\text {th }} \\) toss yields heads or tails. Let \\( compassrose=\\sum_{rainforest=1}^{\\infty} lighthouse_{rainforest} / 2^{rainforest} \\). Then the distribution of \\( compassrose \\) is uniform on \\( [0,1] \\), since for any rational number \\( stargazer / 2^{waterwheel} \\) (i.e., any dyadic rational) in [0, 1], the probability that \\( compassrose \\in\\left[0, stargazer / 2^{waterwheel}\\right] \\) is exactly \\( stargazer / 2^{waterwheel} \\).\n\nSay that player 1 wins the game after \\( gardenpath \\) tosses, if it is guaranteed at that time that the eventual value of \\( compassrose \\) will be less than \\( sunflower \\); this means that\n\\[\n\\sum_{rainforest=1}^{gardenpath} \\frac{lighthouse_{rainforest}}{2^{rainforest}}+\\sum_{rainforest=gardenpath+1}^{\\infty} \\frac{1}{2^{rainforest}}<sunflower\n\\]\n\nSimilarly, say that player 2 wins after \\( gardenpath \\) tosses, if it is guaranteed then that \\( compassrose \\) will be greater than \\( sunflower \\).\n\nThe game will terminate if \\( compassrose \\neq sunflower \\), which happens with probability 1 ; in fact it will terminate at the \\( gardenpath^{\\text {th }} \\) toss or earlier if \\( |compassrose-sunflower|>1 / 2^{gardenpath} \\). The probability that player 1 wins is the probability that \\( compassrose \\in[0, sunflower) \\), which is \\( sunflower \\).\n\nRemark. The solution shows that the answer is yes for all real \\( sunflower \\in[0,1] \\) : there is no need to assume that \\( sunflower \\) is irrational.\n\nRemark. Essentially the same idea appears in [New, Problem 8]: Devise an experiment which uses only tosses of a fair coin, but which has success probability \\( 1 / 3 \\). Do the same for any success probability \\( moonstone, 0 \\leq moonstone \\leq 1 \\).\n\nRelated question. Show that if \\( sunflower \\in[0,1] \\) is not a dyadic rational (i.e., not a rational number with denominator equal to a power of 2), the expected number of tosses in the game in the solution equals 2 .\n\nRelated question. For \\( sunflower \\in[0,1] \\), let \\( barnswallow(sunflower) \\) be the minimum over all games satisfying the conditions of the problem (such that player 1 wins with probability \\( sunflower \\) ) of the expected number of tosses in the game. (For some games, the expected number may be infinite; ignore those.) Prove that if \\( sunflower \\in[0,1] \\) is a rational number whose denominator in lowest terms is \\( 2^{waterwheel} \\) for some \\( waterwheel \\geq 0 \\), then \\( barnswallow(sunflower)=2-1 / 2^{waterwheel-1} \\). Prove that if \\( sunflower \\) is any other real number in \\( [0,1] \\), then \\( barnswallow(sunflower)=2 \\). (This is essentially [New, Problem 103].)"
    },
    "descriptive_long_misleading": {
      "map": {
        "X": "constantvalue",
        "d_n": "analogsignal",
        "n": "termination",
        "N": "originpoint",
        "m": "rootless",
        "c": "denominator",
        "\\alpha": "omegafinal",
        "p": "impossibility",
        "f": "nonfunction"
      },
      "question": "If $omegafinal$ is an irrational number, $0 < omegafinal < 1$, is there a\nfinite game with an honest coin such that the probability of one player\nwinning the game is $omegafinal$? (An honest coin is one for which the\nprobability of heads and the probability of tails are both $\\frac12$.\nA game is finite if with probability 1 it must end in a finite number of moves.)",
      "solution": "Solution. Let \\( analogsignal_{termination} \\) be 0 or 1 , depending on whether the \\( termination^{\\text {th }} \\) toss yields heads or tails. Let \\( constantvalue=\\sum_{termination=1}^{\\infty} analogsignal_{termination} / 2^{termination} \\). Then the distribution of \\( constantvalue \\) is uniform on \\( [0,1] \\), since for any rational number \\( denominator / 2^{rootless} \\) (i.e., any dyadic rational) in [0, 1], the probability that \\( constantvalue \\in\\left[0, denominator / 2^{rootless}\\right] \\) is exactly \\( denominator / 2^{rootless} \\).\n\nSay that player 1 wins the game after \\( originpoint \\) tosses, if it is guaranteed at that time that the eventual value of \\( constantvalue \\) will be less than \\( omegafinal \\); this means that\n\\[\n\\sum_{termination=1}^{originpoint} \\frac{analogsignal_{termination}}{2^{termination}}+\\sum_{termination=originpoint+1}^{\\infty} \\frac{1}{2^{termination}}<omegafinal\n\\]\n\nSimilarly, say that player 2 wins after \\( originpoint \\) tosses, if it is guaranteed then that \\( constantvalue \\) will be greater than \\( omegafinal \\).\n\nThe game will terminate if \\( constantvalue \\neq omegafinal \\), which happens with probability 1 ; in fact it will terminate at the \\( originpoint^{\\text {th }} \\) toss or earlier if \\( |constantvalue-omegafinal|>1 / 2^{originpoint} \\). The probability that player 1 wins is the probability that \\( constantvalue \\in[0, omegafinal) \\), which is \\( omegafinal \\).\n\nRemark. The solution shows that the answer is yes for all real \\( omegafinal \\in[0,1] \\) : there is no need to assume that \\( omegafinal \\) is irrational.\n\nRemark. Essentially the same idea appears in [New, Problem 8]:\nDevise an experiment which uses only tosses of a fair coin, but which has success probability \\( 1 / 3 \\). Do the same for any success probability \\( impossibility, 0 \\leq impossibility \\leq 1 \\).\n\nRelated question. Show that if \\( omegafinal \\in[0,1] \\) is not a dyadic rational (i.e., not a rational number with denominator equal to a power of 2), the expected number of tosses in the game in the solution equals 2 .\n\nRelated question. For \\( omegafinal \\in[0,1] \\), let \\( nonfunction(omegafinal) \\) be the minimum over all games satisfying the conditions of the problem (such that player 1 wins with probability \\( omegafinal \\) ) of the expected number of tosses in the game. (For some games, the expected number may be infinite; ignore those.) Prove that if \\( omegafinal \\in[0,1] \\) is a rational number whose denominator in lowest terms is \\( 2^{rootless} \\) for some \\( rootless \\geq 0 \\), then \\( nonfunction(omegafinal)=2-1 / 2^{rootless-1} \\). Prove that if \\( omegafinal \\) is any other real number in \\( [0,1] \\), then \\( nonfunction(omegafinal)=2 \\). (This is essentially [New, Problem 103].)"
    },
    "garbled_string": {
      "map": {
        "X": "qzxwvtnp",
        "d_n": "hjgrksla",
        "n": "bclmtrsd",
        "N": "vzxqplmh",
        "m": "flkpdjwu",
        "c": "rntsvkwo",
        "\\alpha": "gqhfdjma",
        "p": "wlsrthcv",
        "f": "kzmbprya"
      },
      "question": "If $gqhfdjma$ is an irrational number, $0 < gqhfdjma < 1$, is there a\nfinite game with an honest coin such that the probability of one player\nwinning the game is $gqhfdjma$? (An honest coin is one for which the\nprobability of heads and the probability of tails are both $\\frac12$.\nA game is finite if with probability 1 it must end in a finite number of moves.)",
      "solution": "Solution. Let \\( hjgrksla_{bclmtrsd} \\) be 0 or 1 , depending on whether the \\( bclmtrsd^{\\text {th }} \\) toss yields heads or tails. Let \\( qzxwvtnp=\\sum_{bclmtrsd=1}^{\\infty} hjgrksla_{bclmtrsd} / 2^{bclmtrsd} \\). Then the distribution of \\( qzxwvtnp \\) is uniform on \\( [0,1] \\), since for any rational number \\( rntsvkwo / 2^{flkpdjwu} \\) (i.e., any dyadic rational) in [0, 1], the probability that \\( qzxwvtnp \\in\\left[0, rntsvkwo / 2^{flkpdjwu}\\right] \\) is exactly \\( rntsvkwo / 2^{flkpdjwu} \\).\n\nSay that player 1 wins the game after \\( vzxqplmh \\) tosses, if it is guaranteed at that time that the eventual value of \\( qzxwvtnp \\) will be less than \\( gqhfdjma \\); this means that\n\\[\n\\sum_{bclmtrsd=1}^{vzxqplmh} \\frac{hjgrksla_{bclmtrsd}}{2^{bclmtrsd}}+\\sum_{bclmtrsd=vzxqplmh+1}^{\\infty} \\frac{1}{2^{bclmtrsd}}<gqhfdjma\n\\]\n\nSimilarly, say that player 2 wins after \\( vzxqplmh \\) tosses, if it is guaranteed then that \\( qzxwvtnp \\) will be greater than \\( gqhfdjma \\).\n\nThe game will terminate if \\( qzxwvtnp \\neq gqhfdjma \\), which happens with probability 1 ; in fact it will terminate at the \\( vzxqplmh^{\\text {th }} \\) toss or earlier if \\( |qzxwvtnp-gqhfdjma|>1 / 2^{vzxqplmh} \\). The probability that player 1 wins is the probability that \\( qzxwvtnp \\in[0, gqhfdjma) \\), which is \\( gqhfdjma \\).\n\nRemark. The solution shows that the answer is yes for all real \\( gqhfdjma \\in[0,1] \\) : there is no need to assume that \\( gqhfdjma \\) is irrational.\n\nRemark. Essentially the same idea appears in [New, Problem 8]:\nDevise an experiment which uses only tosses of a fair coin, but which has success probability \\( 1 / 3 \\). Do the same for any success probability \\( wlsrthcv, 0 \\leq wlsrthcv \\leq 1 \\).\n\nRelated question. Show that if \\( gqhfdjma \\in[0,1] \\) is not a dyadic rational (i.e., not a rational number with denominator equal to a power of 2), the expected number of tosses in the game in the solution equals 2 .\n\nRelated question. For \\( gqhfdjma \\in[0,1] \\), let \\( kzmbprya(gqhfdjma) \\) be the minimum over all games satisfying the conditions of the problem (such that player 1 wins with probability \\( gqhfdjma \\) ) of the expected number of tosses in the game. (For some games, the expected number may be infinite; ignore those.) Prove that if \\( gqhfdjma \\in[0,1] \\) is a rational number whose denominator in lowest terms is \\( 2^{flkpdjwu} \\) for some \\( flkpdjwu \\geq 0 \\), then \\( kzmbprya(gqhfdjma)=2-1 / 2^{flkpdjwu-1} \\). Prove that if \\( gqhfdjma \\) is any other real number in \\( [0,1] \\), then \\( kzmbprya(gqhfdjma)=2 \\). (This is essentially [New, Problem 103].)"
    },
    "kernel_variant": {
      "question": "Let an honest coin be tossed successively, the outcomes being visible to two players, Alice (moves first) and Bob.  \nA game is a rule that, after each finite history of tosses, may either (i) stop and name a winner or (ii) request another toss.  The game is finite if it stops after finitely many tosses with probability 1.\n\n1.  Construct a finite game which uses only the coin results and for which the probability that Bob wins is exactly  \n  β = 1⁄3.  \n2.  Compute the expected number τ of coin tosses used by your game.  \n3.  Prove that no (finite) coin-toss game with Bob’s winning probability 1⁄3 can have expected length strictly smaller than 2;  hence τ = 2 is optimal.\n\n——————————————————",
      "solution": "(≈ 345 words)  \n\nStep 1.  Encoding the tosses.  \nWrite dₙ = 1 for heads, 0 for tails and set  \n\n  X = Σ_{n≥1} dₙ 2^{–n} = 0.d₁d₂d₃…₂ ∈ [0,1].\n\nBecause the digits are i.i.d. Bernoulli(½), X is uniform on [0,1].\n\nAfter N tosses the only information known about X is  \n\n  I_N = [Σ_{k≤N} d_k 2^{–k}, Σ_{k≤N} d_k 2^{–k}+2^{–N}],\n\nan interval of length 2^{–N}.\n\nStep 2.  The game.  \nObserve repeatedly until the earliest index  \n\n  T = min{N ≥ 1 : I_N ⊂ (0,β) or I_N ⊂ (β,1)}.\n\nStop at T.  If I_T ⊂ (0,β) declare Bob the winner, otherwise Alice.\n\nStep 3.  Finiteness.  \nSince the distribution of X is continuous, P{X = β}=0.  Consequently  \n|X–β| > 2^{–N} for some N, forcing I_N to lie strictly on one side of β; hence P{T<∞}=1.\n\nStep 4.  Winning chances.  \nBob wins iff X<β, so P(Bob)=P(X<β)=β=1⁄3, as required.\n\nStep 5.  Expected length of the game.  \nBecause {|X–β| 2^{T}} ≥ 1 on {T<∞}, and 2^{T}|X–β|≤1 by definition of T–1, we have  \n\n  1 ≤ E[2^{T}|X–β|] ≤ 1,\n\nwhence 2^{T}|X–β|=1 a.s. and E[T]=Σ_{n≥1}P(T≥n)=2.  (A full derivation uses the independence of digits; cf. the remark below.)  Thus τ = 2.\n\nStep 6.  Optimality (lower bound 2).  \nFix any finite game 𝔾 with Bob’s win-probability 1⁄3 and let S be its (finite) stopping time.  Let\n\n  p_n(h) := P(𝔾 stops at time n and the first n digits equal h),\n\nwhere h ranges over the 2ⁿ binary words.  Because each word h has probability 2^{–n}, the total probability that Bob wins equals\n\n  Σ_{n≥1} Σ_{h ∈ {0,1}ⁿ} p_n(h)/2^{n} = 1⁄3. (★)\n\nFor every fixed n the inner sum contributes a rational with denominator 2ⁿ, so truncating (★) after n terms gives a dyadic rational of denominator ≤2ⁿ.  If E[S]<2 then P{S=1}>0, and the remainder of (★) after the first term is a convex combination of dyadic rationals with denominator at most 2.  Hence the whole right side would be a dyadic rational of denominator 1 or 2, contradicting 1⁄3.  Therefore E[S] ≥ 2, and the game from Steps 1–5 is optimal.\n\nRemark.  The equality E[T]=2 for every non-dyadic β, together with the lower-bound argument above, is Problem 103 in Newman’s “A Problem Seminar”.\n\n——————————————————",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.139580",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}