<p>In 2024, Daniel Litt posed a simple coinflip game pitting Alice’s “Heads–Heads” versus Bob’s “Heads–Tails”: Who is more likely to win if they score 1 point per occurrence of their substring in a sequence of <i>n</i> fair coinflips? This attracted over 1 million views on X and quickly spawned several articles explaining the counterintuitive solution. We study the generalized game, where the set of coin outcomes, <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10959_2025_1452_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="107" /> </InlineMediaObject> <EquationSource Format="TEX">\(\{ \text {Heads}, \text {Tails} \}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">{</mo> <mtext>Heads</mtext> <mo>,</mo> <mtext>Tails</mtext> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation>, is generalized to an arbitrary finite alphabet <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10959_2025_1452_Article_IEq2.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="18" /> </InlineMediaObject> <EquationSource Format="TEX">\({\mathcal {A}}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">A</mi> </math></EquationSource> </InlineEquation>, and where Alice’s and Bob’s substrings are any finite <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10959_2025_1452_Article_IEq2.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="18" /> </InlineMediaObject> <EquationSource Format="TEX">\({\mathcal {A}}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">A</mi> </math></EquationSource> </InlineEquation>-strings of the same length. We find that the winner of Litt’s game can be determined by a single quantity which measures the amount of prefix/suffix self-overlaps in each string; whoever’s string has more overlaps <i>loses</i>. For example, “Heads–Tails” beats “Heads–Heads” in the original problem because “Heads–Heads” has a prefix/suffix overlap of length 1 while “Heads–Tails” has none. The method of proof is to develop a precise Edgeworth expansion for discrete Markov chains and apply this to calculate Alice’s and Bob’s probability to win the game correct to order <i>O</i>(1/<i>n</i>). </p>

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The Generalized Alice HH Vs Bob HT Problem

  • Svante Janson,
  • Mihai Nica,
  • Simon Segert

摘要

In 2024, Daniel Litt posed a simple coinflip game pitting Alice’s “Heads–Heads” versus Bob’s “Heads–Tails”: Who is more likely to win if they score 1 point per occurrence of their substring in a sequence of n fair coinflips? This attracted over 1 million views on X and quickly spawned several articles explaining the counterintuitive solution. We study the generalized game, where the set of coin outcomes, \(\{ \text {Heads}, \text {Tails} \}\) { Heads , Tails } , is generalized to an arbitrary finite alphabet \({\mathcal {A}}\) A , and where Alice’s and Bob’s substrings are any finite \({\mathcal {A}}\) A -strings of the same length. We find that the winner of Litt’s game can be determined by a single quantity which measures the amount of prefix/suffix self-overlaps in each string; whoever’s string has more overlaps loses. For example, “Heads–Tails” beats “Heads–Heads” in the original problem because “Heads–Heads” has a prefix/suffix overlap of length 1 while “Heads–Tails” has none. The method of proof is to develop a precise Edgeworth expansion for discrete Markov chains and apply this to calculate Alice’s and Bob’s probability to win the game correct to order O(1/n).