Replacement and sampling protocols

Keep first-draw probabilities fixed while reconstructing the different joint laws created by replacement and removal.

15 minutes · intermediate · Original Poker4Quant lesson

What you will learn

Construct a two-draw probability tree under each replacement protocol and distinguish exactly-one from at-least-one events.

Prerequisites: Combining conditional branches · Conditional draws · Hypergeometric models

State what happens between draws

A sampling experiment is not fully specified by its initial bag or deck. You must also say whether a drawn item returns before another selection and how the next selection is made. Replacing the item and uniformly reshuffling restores the original population. Removing it changes the available population and creates conditional second-draw probabilities.

Use a toy deck of four distinct cards: three red and one black. The first red probability is 3/4 under either protocol. With replacement and a fresh uniform shuffle, the next red probability remains 3/4 regardless of the first color. Without replacement, a red first card leaves two reds and one black, while a black first card leaves three reds. The first-draw marginal alone cannot distinguish these joint laws.

Ordered branches make the event precise

Exactly one red in two draws occurs through red-then-black or black-then-red. With replacement, their probabilities are (3/4)(1/4)=3/16 and (1/4)(3/4)=3/16, so the event probability is 3/8. Red-then-red is not part of this event, even though it contains a red.

Without replacement, red-then-black has probability (3/4)(1/3)=1/4. Black-then-red has probability (1/4)(3/3)=1/4. Their sum is 1/2. The black-then-black path is still a meaningful ordered color description but has probability zero in this toy deck: no black remains after drawing its only black card.

Cross-check with physical outcomes

With replacement there are 4×4=16 equally likely ordered physical-card pairs, because a physical card may reappear after being returned. Without replacement there are 4×3=12 equally likely ordered distinct-card pairs. Exactly one red has six physical ordered outcomes under each protocol, but the denominators differ. The same favorable count therefore need not give the same probability.

At least one red is another event. With replacement its complement, two blacks, has probability 1/16, so at least one red is 15/16. Without replacement two blacks are impossible, so at least one red has probability one. Neither result means the red cards necessarily win a poker hand: these are color events in a toy sampling model, not showdown equity or a strategy recommendation.

Worked example

A bag has two distinct marked tokens and three unmarked tokens. Draw twice without replacement. What is the probability both are marked?

The first marked probability is 2/5.
After a marked draw, one marked token remains among four total tokens. The second conditional marked probability is 1/4.
Multiply down this ordered path: (2/5)(1/4)=1/10.
Check physical ordered pairs: 2×1 marked pairs out of 5×4 total pairs gives 2/20=1/10.

The result is 1/10 under removal. Squaring 2/5 would describe a different replacement protocol.

Hands-on: Rebuild the second draw after replacement or removal

Construct each branch population, select exactly-one-red ordered paths and compare the joint laws under two explicit sampling protocols.

Reading branch workshop…

Common mistakes

Using the original denominator after an item is removed.
Assuming equal first-draw distributions imply equal joint distributions.
Counting red-red as exactly one red.
Translating a target-color event directly into a poker win probability.

Keep these ideas

Specify replacement and the next-draw sampling law.
Use branch-specific populations after removal.
Multiply ordered-path probabilities and sum only the target paths.
Cross-check against the correct physical ordered sample space.