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Recognize that the payoff is the sum of four independent Bernoulli trials, each with success probability 0.5. Use linearity of expectation to compute the expected number of heads as 4 * 0.5 = 2, so the expected payoff is $2. Optionally, verify by computing the full binomial distribution.
Pro tip: Mention that linearity of expectation holds even without independence, but here independence makes it straightforward. Also, note that the expected value is not necessarily the most likely outcome; the most likely outcome is 2 heads, but the expected value is also 2, which is a nice coincidence.
Let X be the total payoff, which is the sum of four indicator variables for each coin landing heads.
Since expectation is linear, E[X] = sum of E[each indicator] = 4 * 0.5 = 2.
Compute the binomial probabilities for k heads (k=0..4) and sum k * P(X=k) to confirm the expected value is 2.
Conclude that the expected payoff is $2.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the rules: the wand allows flipping any two coins after seeing the initial toss, repeatedly for free. Then, determine the maximum expected value of the best possible outcome by analyzing the optimal strategy, and compute the fair price as the difference between that maximum expected value and the expected value of the initial toss.
Pro tip: Frame the problem as an optimal stopping or control problem, and explicitly state that the fair price is the expected gain from the wand, not the total expected value. This shows you understand pricing in terms of incremental value.
Confirm that the wand can be used any number of times for free, that you can choose which two coins to flip after seeing the initial toss, and that the goal is to maximize expected value. Assume fair coins unless stated otherwise.
Compute the expected value of the initial toss (e.g., for n coins, expected number of heads is n/2). This serves as the baseline without the wand.
Analyze how to use the wand to maximize expected value. For example, if the goal is to maximize heads, you can flip two tails to heads if at least two tails exist, otherwise flip one tail and one head, etc. Consider the best achievable outcome.
Calculate the expected value under the optimal strategy. This may involve considering the distribution of the initial toss and the effect of repeated flips. For large n, the maximum expected value approaches n (all heads) if you can flip any two coins repeatedly.
Subtract the initial expected value from the maximum expected value to get the expected gain from the wand. This difference is the fair price.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the game: you start with some number of coins (likely 100) each showing heads or tails, and the wand randomly re-tosses any two chosen coins, meaning each selected coin becomes heads or tails with equal probability. The goal is to maximize the expected number of heads when you stop, and the fair price is the expected maximum value achievable under optimal stopping. Model the state as the number of heads, derive the optimal stopping rule via dynamic programming, and compute the value function.
Pro tip: Emphasize that the wand's randomness means you can only improve the expected number of heads if you have fewer than half heads; otherwise, using the wand is neutral or harmful. This threshold insight simplifies the optimal strategy and shows you understand the stochastic nature.
Let the state be the current number of heads h out of N coins. The objective is to maximize the expected number of heads at stopping, and the fair price is the expected value under the optimal stopping policy.
When you choose two coins, each is independently re-tossed to heads with probability 1/2. The expected change in the number of heads depends on the current states of the chosen coins: if both are tails, expected gain is +1; if one head and one tail, expected change is 0; if both heads, expected loss is -1. Thus, to maximize expected gain, always choose two tails if available; otherwise, the expected change is non-positive.
Since using the wand on two tails increases expected heads by 1, it is beneficial to continue as long as there are at least two tails (i.e., h ≤ N-2). If there is at most one tail (h ≥ N-1), using the wand cannot improve the expected value (it may decrease it), so stop. Thus, the optimal strategy is: stop when h ≥ N-1, otherwise use the wand on two tails.
Starting from an initial configuration (e.g., all tails, h=0), repeatedly apply the wand to two tails until h = N-1 or N. The process is a Markov chain where each successful use increases h by 1 with probability 1/2 (if both tails become heads) or leaves h unchanged with probability 1/2 (if one or both remain tails). The expected number of steps to reach N-1 or N, and the final expected heads, can be computed via dynamic programming or by solving the recurrence.
The fair price is the expected number of heads at stopping under the optimal policy. For large N, this value approaches N - 1 + something? Actually, we need to compute exactly. For N coins, the expected value is N - 1 + (1/2)^{N-1}? Let's derive: Let E(h) be the expected final heads starting from h heads. For h ≤ N-2, E(h) = 0.5 * E(h+2) + 0.5 * E(h+1)? Wait, careful: when you choose two tails, each becomes heads with prob 1/2. So the number of new heads from those two is Binomial(2, 1/2): 0,1,2 with probs 1/4,1/2,1/4. So from h, you go to h+2 with prob 1/4, h+1 with prob 1/2, h with prob 1/4. But if you always choose two tails, you need at least two tails. So for h ≤ N-2, you can choose two tails. The recurrence: E(h) = 1/4 E(h+2) + 1/2 E(h+1) + 1/4 E(h). Rearranging: 3/4 E(h) = 1/4 E(h+2) + 1/2 E(h+1) => 3 E(h) = E(h+2) + 2 E(h+1). Boundary: E(N-1) = N-1 (stop), E(N) = N. Solve this recurrence. The solution is E(h) = N - 1 + (1/3)^{N-1-h}? Let's check: For h = N-1, E = N-1. For h = N-2, 3 E(N-2) = E(N) + 2 E(N-1) = N + 2(N-1) = 3N - 2 => E(N-2) = N - 2/3. But N-1 + (1/3)^{1} = N - 1 + 1/3 = N - 2/3. Yes. For h = N-3, 3 E = E(N-1) + 2 E(N-2) = (N-1) + 2(N - 2/3) = 3N - 7/3 => E = N - 7/9. And N-1 + (1/3)^2 = N - 1 + 1/9 = N - 8/9? Not matching. Let's recompute: N-1 + (1/3)^2 = N - 1 + 1/9 = N - 8/9. But we got N - 7/9. So maybe the solution is E(h) = N - 1 + (2/3)^{N-1-h}? Check h=N-2: N-1 + 2/3 = N - 1/3, not N-2/3. So not. Let's solve properly. Recurrence: 3E_h = E_{h+2} + 2E_{h+1}. Characteristic equation: 3 = r^2 + 2r => r^2 + 2r - 3 = 0 => (r+3)(r-1)=0 => r=1, r=-3. So general solution: E_h = A + B(-3)^h. But h is number of heads, and we have boundary at h=N-1 and h=N. Actually, the recurrence holds for h ≤ N-2. So for h from 0 to N-2, E_h = A + B(-3)^h. Use boundaries: E_{N-1} = N-1, E_N = N. But E_N is not in the recurrence domain? Actually, for h=N-1, we stop, so E_{N-1}=N-1. For h=N, E_N=N. But the recurrence for h=N-2 uses E_N and E_{N-1}. So we can use E_{N-1} and E_N to solve for A and B. We have: A + B(-3)^{N-1} = N-1, A + B(-3)^N = N. Subtract: B(-3)^N - B(-3)^{N-1} = 1 => B(-3)^{N-1}(-3 - 1) = 1 => B(-3)^{N-1}(-4) = 1 => B = -1/(4(-3)^{N-1}) = -(-3)^{-(N-1)}/4. Then A = N-1 - B(-3)^{N-1} = N-1 - (-1/4) = N - 3/4. So E_h = N - 3/4 - (1/4)(-3)^{h-(N-1)}? Actually B(-3)^h = -1/(4(-3)^{N-1}) * (-3)^h = -1/4 * (-3)^{h-(N-1)}. So E_h = N - 3/4 - 1/4 * (-3)^{h-(N-1)}. For h=N-1: E = N - 3/4 - 1/4 = N-1. Good. For h=N-2: E = N - 3/4 - 1/4 * (-3)^{-1} = N - 3/4 + 1/12 = N - 9/12 + 1/12 = N - 8/12 = N - 2/3. Good. For h=N-3: E = N - 3/4 - 1/4 * (-3)^{-2} = N - 3/4 - 1/4 * 1/9 = N - 3/4 - 1/36 = N - 27/36 - 1/36 = N - 28/36 = N - 7/9. Matches earlier. So the formula is E_h = N - 3/4 - (1/4)(-3)^{h-(N-1)}. For large N and starting from h=0 (all tails), the term (1/4)(-3)^{0-(N-1)} = (1/4)(-3)^{-(N-1)} = (1/4)(-
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem setup: define the two wand prices (e.g., optimal and suboptimal) and the rules for n coins. Then derive or reason about the asymptotic behavior of each price as n grows, focusing on scaling laws and convergence. Finally, compare the two prices to highlight the gap and its implications.
Pro tip: Emphasize that the key is not just the final formula but the reasoning process—demonstrate how you'd validate the result with small n and edge cases, showing rigor and intuition.
Restate the wand pricing problem for n coins, ensuring you understand the rules and what the two prices represent (e.g., optimal strategy vs. a fixed strategy).
Determine the stochastic process or combinatorial structure governing the coin flips and wand prices, such as expected values or dynamic programming recurrences.
For each wand price, derive a closed form or asymptotic expression as n increases, using techniques like recurrences, generating functions, or approximations.
Compare the growth rates of the two prices: do they converge, diverge, or maintain a constant ratio? Discuss the implications for large n.
Check your results with small n (e.g., n=4) and consider edge cases; interpret the findings in the context of the original problem.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Frame the problem as an expected value optimization: compare the expected gain from re-tossing (given the current state) against the cost c. Determine the threshold state where the expected benefit of re-tossing equals c, and only re-toss if the expected benefit exceeds c. Then assess whether the wand is worth buying by comparing the total expected value with the wand (including all re-toss costs) to the expected value without the wand.
Pro tip: Quantify the value of information: the wand is essentially an option to re-toss, so its value increases with volatility of outcomes and decreases with cost. Also, consider that the optimal strategy may involve multiple re-tosses, so think recursively or use dynamic programming.
Clarify the underlying game: what are the possible outcomes, their probabilities, and the payoff structure? Without the wand, what is the expected value?
For a given current outcome, compute the expected value of re-tossing (which may involve future re-tosses) minus the cost c. Compare this to the value of stopping. The optimal policy is to re-toss if the expected net gain is positive.
Determine the threshold outcome below which you should re-toss. This threshold depends on c and the distribution of outcomes. Use dynamic programming or backward induction if multiple re-tosses are allowed.
Calculate the expected value of the game when following the optimal re-toss policy, including the expected costs of re-tossing. This is the value of having the wand.
Compare the value with the wand (minus any purchase price) to the value without the wand. The wand is worth buying if the net increase in expected value exceeds its cost.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
They just wanted the DP formulation, not the solution.
Define a state that captures the remaining number of re-tosses and the current outcome, then write a Bellman equation that maximizes expected value by choosing between accepting the current value or using a re-toss. Emphasize the base cases and the recursive structure without solving the DP.
Pro tip: Clearly state the state variables and the decision at each state; this shows you understand the core of dynamic programming and can communicate complex ideas simply.
Let V(i, k) be the maximum expected value when the current outcome is i and you have k re-tosses remaining. Specify the range of i and k.
When k = 0, no re-tosses remain, so V(i, 0) = i (the current outcome). Also consider if i is the maximum possible outcome, you might not re-toss even if k > 0.
For k > 0, V(i, k) = max(i, E[V(X, k-1)]) where X is a new random toss. The expectation is over the distribution of the next toss.
If the toss is uniform over {1,...,n}, then E[V(X, k-1)] = (1/n) * sum_{j=1}^n V(j, k-1). This makes the recursion explicit.
Mention that the DP can be solved by iterating k from 0 to K, computing V(i, k) for all i. Note that the optimal policy is a threshold: re-toss if i < threshold(k).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.