A shuffled collection contains two green and three blue tokens. You want to take one green token. Before seeing the next token, either take it and end, or discard it, reveal its color and continue. You must take the last token if you get that far. Does observing discarded colors give a strategy better than taking immediately?
These rules distinguish information available before a decision from information learned afterward. You may remember the entire revealed history and randomize decisions using independent private randomness. You may not inspect an unseen token before choosing to take or skip it, and you cannot recover a discarded token. The initial order is a uniform permutation of five distinct tokens with the stated colors.
Predict
Taking immediately succeeds with probability 2/5. Try this adaptive policy: skip the first token; if it is green, take the next unseen token; if it is blue, skip one more and then take the next unseen token. Predict the success chance before drawing its tree.
After a green reveal the remaining collection is less favorable, with one green among four tokens. After a blue reveal it is more favorable, with two green among four. The observations really do change conditional probabilities. A proof claiming that every remaining collection stays at the initial proportion would be false.
Experiment
Draw three leaves for the revealed prefixes G, BG and BB under this policy. Each leaf says that the next unseen token will be taken. Fill in the remaining composition and success contribution before checking the table.
Revealed prefix
Prefix probability
Remaining green/total
Success contribution
G
2/5
1/4
1/10
BG
(3/5)(2/4)=3/10
1/3
1/10
BB
(3/5)(2/4)=3/10
2/3
1/5
The prefix probabilities sum to one. The success contributions sum to 2/5, exactly the immediate-take chance. The policy skips one token always and a second with probability 3/5, giving expected skips 8/5. Equality of success does not imply equality of work.
Change the policy on one branch and recompute. For example, after the first blue, take immediately instead. Its two contributions become (2/5)(1/4) and (3/5)(2/4), again totaling 2/5. These examples suggest a general strategy bound, but neither checks every possible history-dependent rule.
The explorer below starts with a separate two-green, one-blue case. Predict both successor fractions before revealing its exact skip calculation. Change the counts, including a zero-color boundary, and distinguish the finite calculation from the general induction that follows.
Choose before seeing the next token
A uniformly shuffled population contains green and blue tokens. At each decision, either take the next unseen token and finish, or skip it, reveal its color and remove it. If only one token remains, you must take it. Success means taking green. Skipping has no cost; no token can be recalled after it is revealed.
The total must be between 1 and 8. Counts describe the currently unseen population. After a revealed skip, the remaining order is still uniformly shuffled conditional on that history.
Current state: 2 green, 1 blue, 3 total. Taking now succeeds with probability 2/3 = 0.666667. Predict whether skipping once and then making decisions from the remaining population can improve it.
Write a prediction first. Revealing the calculation is an explanation, not a simulated random draw.
Policy comparison: taking immediately has success probability 2/3. Skipping until only one token remains has the same probability, because each original token is equally likely to be last. The lesson proves the stronger claim that every admissible history-dependent policy has this value, using induction on the number remaining and the branch calculation. One inspected state alone does not prove that universal statement.
Change the information: peek before deciding
If the next color can be seen before deciding whether to take it, take the first green and skip blues, with the same forced-last rule. For these counts that succeeds with probability 1: at least one green exists, and the policy will find it. This changes the information available at a decision. It is not a counterexample to the unseen-choice theorem. The ordinary protocol reveals a skipped token only after that token is no longer available.
Try 0 green with 3 blue, then 3 green with 0 blue. Predict why each has just one skip branch. Next choose a one-token population and explain why the calculation stops. Changing counts hides any earlier revealed calculation. This temporary explorer saves no work and awards no completion.
Understand
Let g and b be the remaining green and blue counts, with n=g+b≥1. We prove the stronger statement: every admissible strategy, after any legal revealed history leaving a uniform remaining order, succeeds with probability g/n. Proving only a balanced starting case would not be enough, because one reveal generally leaves an unbalanced state.
First justify uniformity. Given an exact revealed prefix of a uniformly shuffled collection of distinct tokens, every order of the remaining tokens completes that prefix in exactly one way and had equal probability. Thus their conditional order is uniform. If only colors were recorded, average over compatible labeled prefixes: they give the same remaining color counts and hence the same color probabilities. A decision based only on the revealed prefix does not supply extra information about the unseen order. This is sampling without replacement, not independent successive draws.
Proceed by induction on n. If n=1, the last token must be taken, so success is g, equal to g/n. No value at the empty state (0,0) is needed. For n>1, taking now succeeds with probability g/n because the decision precedes seeing the token.
If instead the strategy skips, a green reveal has probability g/n and leaves counts (g−1,b); a blue reveal has probability b/n and leaves (g,b−1). On each branch with positive probability, the induction hypothesis applies to whatever continuation the strategy chooses, including a different continuation based on the full history. Its success chances are respectively (g−1)/(n−1) and g/(n−1). Therefore the skip branch has value
[g(g−1)+bg]/[n(n−1)]=g(n−1)/[n(n−1)]=g/n.
If a color is absent, omit that probability-zero branch rather than invoking an induction hypothesis on a negative count. All-green and all-blue states consequently have values one and zero. Both available decisions have the same value, completing the induction for deterministic strategies.
Independent private randomization cannot improve the result. Conditional on a random seed independent of the shuffled order, the rule becomes a deterministic nonanticipating strategy, already covered by the proof. Averaging its same value g/n over seeds preserves that value. Equivalently, randomizing between the equal-valued take and skip options changes nothing. This does not assert independence between the final color and the number of skips.
The theorem gives equality for every admissible strategy, not merely an upper bound for a particular proposed optimizer. Forced taking at the last token ensures that every such strategy ends after at most n−1 skips. A strategy allowed to refuse forever would require a different contract.
Apply
Change the information rule: reveal the next token before deciding whether to take it. With at least one green token, accept the first green and discard preceding blues. Success is then certain. Identify the failed step in the old proof: a decision to take is now based on the token’s own color, so the take branch no longer has success probability g/n. This is a different experiment, not a contradiction.
Now restore the original rule and charge c≥0 per skipped token. Let K count skips. Expected net payoff is g/n−cE[K], because the expected terminal reward remains g/n. Immediate taking attains g/n with zero cost. If c=0 all admissible strategies tie. If c is positive, every strategy that skips with positive probability is strictly worse; a nominal skip branch on an impossible history does not incur cost.
Check a numerical cost
The five-token worked policy has expected skips 8/5. At cost c=1/8 per skip its expected net payoff is 2/5−(1/8)(8/5)=1/5. Immediate taking gives 2/5. The reward and skip count may be dependent; linearity of expectation still justifies subtracting their expectations.
Continue
In the practice, state when the decision occurs, what history is visible, and whether randomness is independent of the hidden order. Use actual remaining counts in every conditional branch. The stronger induction handles histories that produce unequal color proportions, while separate counterexamples explain why peeking, nontermination or a changed payoff would require a new proof.
Apply · explain · revise
Practice and reasoning
Drafts save locally as you type. Reveal worked solutions when useful; the site remembers that exposure. Automatic matches check the requested answer, while written arguments require review against the rubric.
Start with four green and two blue tokens. Skip the first token. If it is blue, take the next unseen token; if it is green, skip one more token and then take the next unseen token. Draw the three relevant revealed-prefix branches, compute their probabilities and conditional success chances, and add the weighted contributions. Compute expected skips.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Prove for every g,b with n=g+b≥1 that every admissible strategy starting from a uniformly shuffled remaining collection succeeds with probability g/n. Use induction on n, not just balanced initial states. Include a history-dependent strategy and explain what happens at the forced final take.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Justify the uniform-remainder premise after a revealed prefix of a uniform permutation of distinct tokens. Explain why independent random choices at decision times do not improve success. Contrast a mixture of strategies with independence of successive revealed colors.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Analyze all-green, all-blue and one-token states. A proposed proof for an initially balanced collection says every later decision still has success chance 1/2. Refute that claim with a two-green/two-blue initial collection, then explain why the correct unconditional strategy theorem can nevertheless hold.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
There are two green and three blue tokens. Change the rule so the next token is revealed before deciding whether to take it; a rejected token is discarded and the last token must still be taken. Construct a strategy with success 1. Identify exactly why it is not an admissible strategy for the original theorem. Treat the all-blue boundary too.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Under the original information rule, charge c≥0 for each skipped token and define net payoff as the terminal green indicator minus c times the number K of skips. Prove the optimal expected value is g/(g+b). Characterize equality for c=0 and c>0. Compute the net value for the first task’s policy when c=1/10, and compare immediate taking.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
