Card Betting: Higher or Lower
You draw one card from a shuffled standard 52-card deck and see its value. You are offered a bet at even odds (win $\
- For each possible rank of your first card, what is the probability that the second card is strictly higher?
- For which first-card ranks should you take the bet, and for which should you pass (or bet on "lower" instead)?
- If you can choose how much to bet (with a fixed bankroll), how should the optimal bet size depend on the first card's rank? What is the breakeven rank?
- Does the order of draws matter? Why or why not?
Hints
- After seeing rank $r$, count the remaining cards: each of the 13 ranks has 4 cards, minus the one you drew. How many of the remaining 51 cards beat rank $r$?
- At even odds, you need $P > 0.5$ for positive EV. Solve $4(13 - r)/51 > 0.5$ for the cutoff rank.
- For optimal sizing, the Kelly criterion at even odds says bet fraction $f = 2p - 1$. Plug in the probability for each rank.
Worked Solution
How to Think About It: This is a classic trading-floor interview question. You see one card and need to quickly assess your edge. The key insight: after seeing a card of rank $r$, you know exactly how many of the remaining 51 cards are higher, lower, or equal. There is no randomness left in the edge calculation -- it is pure counting. The practical analogy is a market maker seeing one piece of information and deciding how aggressively to quote.
Quick Estimate: If you draw an Ace (rank 1), almost everything is higher: $48/51 \approx 94\%$ -- huge edge, bet big. If you draw a 7, there are 24 cards higher and 24 lower (and 3 equal), so $P(\text{higher}) = 24/51 \approx 47\%$ -- slight negative edge, pass. The breakeven is somewhere around rank 6-7.
Formal Solution:
Part 1 -- Probability for each rank:
After drawing a card of rank $r$, there are 51 remaining cards. Cards strictly higher than rank $r$: there are
$P(\text{next card higher} \mid \text{rank } r) = \frac{4(13 - r)}{51}$
| Rank $r$ | Higher cards | $P(\text{higher})$ | Edge at even odds | |----------|-------------|---------------------|--------------------| | 1 (Ace) | 48 | 48/51 = 0.941 | +0.882 | | 2 | 44 | 44/51 = 0.863 | +0.725 | | 3 | 40 | 40/51 = 0.784 | +0.569 | | 4 | 36 | 36/51 = 0.706 | +0.412 | | 5 | 32 | 32/51 = 0.627 | +0.255 | | 6 | 28 | 28/51 = 0.549 | +0.098 | | 7 | 24 | 24/51 = 0.471 | -0.059 | | 8 | 20 | 20/51 = 0.392 | -0.216 | | 9 | 16 | 16/51 = 0.314 | -0.373 | | 10 | 12 | 12/51 = 0.235 | -0.529 | | 11 (J) | 8 | 8/51 = 0.157 | -0.686 | | 12 (Q) | 4 | 4/51 = 0.078 | -0.843 | | 13 (K) | 0 | 0/51 = 0 | -1.000 |
Part 2 -- When to bet:
The bet on "higher" has positive EV when $P(\text{higher}) > 0.5$, i.e., $4(13 - r)/51 > 0.5$, giving
For $r = 7$ specifically: $P(\text{higher}) = 24/51$, $P(\text{lower}) = 24/51$, $P(\text{equal}) = 3/51$. Both bets have negative EV. Pass.
Part 3 -- Optimal bet sizing (Kelly criterion):
At even odds ($b = 1$), the Kelly fraction is $f^{*} = 2p - 1$ when $p > 0.5$, where $p$ is your probability of winning.
For betting on "higher" with rank $r \leq 6$: $f^{*}(r) = 2 \cdot \frac{4(13-r)}{51} - 1 = \frac{8(13-r) - 51}{51} = \frac{53 - 8r}{51}$
- $r = 1$: $f^{*} = 45/51 \approx 88\%$ of bankroll
- $r = 6$: $f^{*} = 5/51 \approx 10\%$ of bankroll
- $r = 7$: $f^{*} < 0$, do not bet on "higher"
For high ranks, bet on "lower" with analogous Kelly sizing.
Part 4 -- Does order matter?
Yes, order matters because you are drawing without replacement. After the first card is removed, the composition of the remaining deck changes. If you drew a low card, the remaining deck is enriched in high cards (and vice versa). This is fundamentally different from drawing with replacement, where the two draws would be independent and the conditional distribution would not change.
However, if the question asks whether the EV changes if you bet before seeing the first card, then the symmetry of the deck makes the unconditional probability of the second card being higher exactly the same as it being lower (each is $(52 \times 51/2 - 52 \times 3/2)/(52 \times 51)$, accounting for ties). The bet has zero unconditional EV -- all the edge comes from conditioning on the first card.
Answer: $P(\text{higher} \mid \text{rank } r) = 4(13-r)/51$. Bet on "higher" for ranks 1-6, pass on rank 7, bet on "lower" for ranks 8-13. Kelly optimal bet size is $f^{*} = (53 - 8r)/51$ for the "higher" bet. Order matters because drawing without replacement changes the conditional distribution.
Intuition
This problem is a microcosm of market making. You observe a signal (the first card), update your beliefs about the next outcome, and decide how much risk to take. The key lesson: your edge is entirely in the conditioning. Before seeing any card, the game is fair (by symmetry). After seeing the card, you have a precise edge that you can size accordingly. This is exactly how a market maker operates -- the unconditional expected P&L of providing liquidity might be zero, but conditional on each piece of information (order flow, inventory, volatility), there is a well-defined edge that determines quote width and position size.
The Kelly criterion application is also instructive: even with a 94% chance of winning (Ace drawn), Kelly says bet only 88% of your bankroll, not everything. Aggressive sizing in the face of certainty is tempting but suboptimal for long-run wealth growth. This discipline -- sizing bets proportional to edge, not conviction -- is what separates professional risk-takers from gamblers.