Card Game: Redraw vs. Switch Option
Two players each draw one card from a well-shuffled deck of 100 cards numbered 1 through 100. The player with the higher card wins the difference -- if you hold card $c$ and your opponent holds card $d$, you collect $c - d$ dollars (negative means you pay).
Before the game is resolved, you are offered one of two options (you must choose exactly one):
- Redraw option: You may put your card back, the deck is reshuffled, and you draw a fresh card.
- Switch option: You may swap cards with your opponent.
Neither you nor your opponent can see each other's card when you decide.
(a) Which option should you prefer?
(b) How much would you pay for each option separately, before drawing your own card?
Hints
- You see your own card before deciding to exercise each option. Think about what information you have and when you would choose to use each option -- the value comes from selective exercise, not from forced use.
- For the redraw option: your expected payoff holding card $c$ is proportional to $c - 50.5$. You redraw when this is negative (i.e., $c \leq 50$). The option value is $E[\max(c - 50.5, 0)]$ averaged over your initial draw.
- For the switch option: switching flips your expected payoff from $V_{\text{keep}}(c)$ to $-V_{\text{keep}}(c)$. You switch when $c < 50.5$ -- the same threshold. Compute $\frac{1}{100}\sum_{c=1}^{50}|V_{\text{keep}}(c)|$ and compare it to the redraw value.
Worked Solution
How to Think About It: At first glance both options seem equivalent: they look like symmetric reshufflings. But they are not. The key is that you see your own card before deciding whether to exercise the option, and that private information makes *both* options valuable: you exercise only when your card is bad. The tempting intuition that the switch is worthless ("you cannot see your opponent's card, so swapping is just trading one unknown for another") is wrong -- your own card tells you whether you are likely ahead or behind. The redraw *flattens* a bad position (back to roughly average); the switch *reverses* it -- you hand your bad card to your opponent and take their better-on-average card. Reversing a losing position swings the payoff by twice as much as flattening it.
Start by pinning down the baseline: with no options, your expected payoff is zero (by symmetry, $E[c - d] = 0$). Any option value comes purely from the ability to act on private information.
Quick Estimate:
For the redraw option: you see your card $c$. Your current payoff is $c - E[d] = c - 50.5$. If you redraw, you get a fresh card with expected value $\approx 50.5$, for expected payoff $\approx 0$. So you exercise the option when $c < 50.5$, i.e., when your card is 50 or below. The gain from exercising is $50.5 - c$ for each such $c$.
Before drawing, your expected payoff with the option is:
$$E[\text{payoff with option}] = \frac{1}{100}\sum_{c=1}^{100} \max(c - 50.5,\; 0)$$
$$= \frac{1}{100}\sum_{c=51}^{100}(c - 50.5) = \frac{1}{100}(0.5 + 1.5 + \cdots + 49.5)$$
$$= \frac{1}{100} \cdot 50 \cdot 25 = 12.5$$
So the redraw option is worth roughly $\$12.50$ before drawing.
For the switch option: you see your card $c$, but you do not see $d$. Switching gives you $d - c$ instead of $c - d$: it flips the sign of your conditional edge $c - 50.5$. You switch exactly when $c \leq 50$, and the swing relative to keeping is $2(50.5 - c)$. The value with the option is roughly $\frac{1}{100}\sum_{c=1}^{100} |c - 50.5| = 25$. So the switch option is worth about $\$25$ -- roughly twice the redraw.
Approach: Formal derivation using conditional expected value and the optional exercise structure.
Formal Solution:
*Redraw option:*
You observe your card $c$ (uniform on $\{1, \ldots, 100\}$). Opponent's card $d$ is uniform on the remaining 99 values, with $E[d \mid c] = (5050 - c)/99$. For large decks this is $\approx 50.5$. Your expected payoff holding $c$ is:
$$V_{\text{keep}}(c) = c - E[d \mid c] = c - \frac{5050 - c}{99} = \frac{100c - 5050}{99}$$
If you redraw, your card is returned and the deck is reshuffled: the fresh card $c'$ is uniform on the 99 cards excluding only the opponent's card $d$, so $E[c' \mid d] = (5050 - d)/99$. Averaging over $d \mid c$ (you decide knowing only $c$):
$$V_{\text{redraw}}(c) = E_{d \mid c}\!\left[\frac{5050 - d}{99} - d\right] = \frac{5050 - E[d \mid c]}{99} - E[d \mid c] = \frac{100c - 5050}{9801} = \frac{V_{\text{keep}}(c)}{99}$$
Note $V_{\text{redraw}}(c)$ is *not* exactly zero: your own card goes back into the deck, so a low $c$ slightly depresses the fresh draw. You exercise the option (redraw) when $V_{\text{redraw}}(c) > V_{\text{keep}}(c)$, i.e., when $\frac{98}{99} V_{\text{keep}}(c) < 0$, i.e., $c \leq 50$, with net gain $V_{\text{redraw}}(c) - V_{\text{keep}}(c) = \frac{494900 - 9800c}{9801}$.
The value of the option before drawing (the no-option baseline is 0):
$$W_{\text{redraw}} = \frac{1}{100}\sum_{c=1}^{100} \max\!\left(V_{\text{keep}}(c),\; V_{\text{redraw}}(c)\right) = \frac{1}{100}\left[\sum_{c=51}^{100} \frac{100c - 5050}{99} + \sum_{c=1}^{50} \frac{100c - 5050}{9801}\right]$$
$$= \frac{125000}{9900} - \frac{125000}{980100} = \frac{1250}{99} - \frac{1250}{9801} = \frac{122500}{9801} = \left(\frac{350}{99}\right)^{2} \approx 12.50$$
(confirmed by exact enumeration over all $(c, d)$ pairs). A common variant instead draws the fresh card from the 98 cards excluding *both* drawn cards; in that model $V_{\text{redraw}}(c) = 0$ exactly and $W_{\text{redraw}} = \frac{1250}{99} \approx 12.63$ -- slightly more, because your bad card can no longer be redrawn.
*Switch option:*
You observe $c$ but not $d$. If you switch, your payoff becomes $d - c$. Conditioned on $c$:
$$E[d - c \mid c] = E[d \mid c] - c = \frac{5050 - c}{99} - c = -\frac{100c - 5050}{99} = -V_{\text{keep}}(c)$$
So switching always flips the sign of your expected payoff. You benefit from switching when $V_{\text{keep}}(c) < 0$, i.e., when $c < 50.5$ (the same threshold as redrawing). When $c \leq 50$, switching converts a negative expected payoff into a positive one; when $c \geq 51$, you hold.
The value of the switch option:
$$W_{\text{switch}} = \frac{1}{100}\sum_{c=1}^{100}\max\!\left(V_{\text{switch}}(c),\; V_{\text{keep}}(c)\right) - \underbrace{\frac{1}{100}\sum_{c=1}^{100} V_{\text{keep}}(c)}_{=\;0}$$
Since $V_{\text{switch}} = -V_{\text{keep}}$, whenever $V_{\text{keep}}(c) < 0$ you switch -- and the gain *relative to keeping* is $-2V_{\text{keep}}(c)$, not $-V_{\text{keep}}(c)$: switching moves your expected payoff from $V_{\text{keep}}(c)$ all the way to $-V_{\text{keep}}(c)$, a swing of twice the edge. Equivalently, $\max(V_{\text{keep}}, -V_{\text{keep}}) = |V_{\text{keep}}|$:
$$W_{\text{switch}} = \frac{1}{100}\sum_{c=1}^{100}\left|V_{\text{keep}}(c)\right| = \frac{2}{100}\sum_{c=1}^{50}\frac{5050 - 100c}{99}$$
$$= \frac{2}{9900}\left(5050 \times 50 - 100\sum_{c=1}^{50}c\right) = \frac{2\,(252500 - 127500)}{9900} = \frac{250000}{9900} = \frac{2500}{99} \approx 25.25$$
The switch option is worth about twice the redraw option: $\frac{2500}{99} \approx 25.25$ versus $\frac{122500}{9801} \approx 12.50$. A tempting slip is to count only $-V_{\text{keep}}(c)$ as the switching gain -- that drops the factor of 2 and wrongly suggests the two options are equivalent.
Answer:
- (a) Prefer the switch option. It is worth $W_{\text{switch}} = \frac{2500}{99} \approx \$25.25$, roughly twice the redraw option's $W_{\text{redraw}} = \frac{122500}{9801} \approx \$12.50$.
- (b) You would pay up to $\frac{122500}{9801} = \left(\frac{350}{99}\right)^{2} \approx \$12.50$ for the redraw option (fresh card drawn from the 99-card deck with only the opponent's card removed) and up to $\frac{2500}{99} \approx \$25.25$ for the switch option.
- The intuition: redrawing replaces your below-average card with a roughly average one (gain $\approx |V_{\text{keep}}|$), while switching hands your bad card to your opponent *and* takes their better-on-average card (gain $= 2|V_{\text{keep}}|$). Both options are exercised on the same set $c \leq 50$, so reversing a bad position is worth about twice flattening it.
Intuition
The punchline is that the switch option is worth about twice the redraw option -- $\frac{2500}{99} \approx \$25.25$ versus $\frac{122500}{9801} \approx \$12.50$. Many people wrongly conclude that switching is worthless (you cannot see $d$, so swapping "cannot help"), or that the two options must be equivalent by symmetry. Both intuitions fail because you see your own card before deciding. Conditional on a bad card, your expected payoff is $V_{\text{keep}}(c) < 0$; redrawing lifts you back to roughly zero (a gain of about $|V_{\text{keep}}|$), while switching carries you all the way to $-V_{\text{keep}}(c) > 0$ (a gain of $2|V_{\text{keep}}|$). Reversing a losing position is worth twice as much as exiting it.
In practice, problems like this show up whenever you have an option to abandon versus an option to reverse a position. A trader long a losing stock can sell (flatten: the P&L stops here) or go short (reverse: profit if the adverse move continues) -- and the payoff swing from reversing is double that of flattening. The key discipline is to compute conditional expected values given what you actually know -- not to reason qualitatively about whether "switching seems better." Writing down $V_{\text{keep}}(c)$, $V_{\text{redraw}}(c)$, and $V_{\text{switch}}(c) = -V_{\text{keep}}(c)$ explicitly is what exposes the factor of two.