Why the Ball-Drawing Convention Matters
In an urn allocation rule, a ball’s color determines the treatment assigned to the next participant. The probability of that assignment depends on the urn’s composition at the instant the ball is drawn. If the rule changes the composition after a draw, the next probability can change too. The key is to specify exactly what happens to the drawn ball and whether any other balls are added.
“Replace the ball” can mean putting the original ball back, putting a new ball in its place, or returning the original and adding extra balls. These operations do not necessarily produce the same urn state. For example, returning a drawn A ball leaves the number of A balls unchanged; returning it and adding a B ball increases the total number of balls by one. A correct probability calculation starts by writing the resulting counts, not by assuming that the urn total stays fixed.
Three Operations That Must Not Be Confused
First, returning the drawn ball means putting that same ball back. If an A ball is drawn from state \((a,b)\), the state remains \((a,b)\). Its next-draw A probability is still \(a/(a+b)\). The same is true after drawing B. If no other balls are added, repeated draws have the same treatment probabilities at every draw.
Second, replacing a drawn ball with one ball of the same color produces the same counts as returning the original ball. For purposes of treatment probabilities, the two operations are equivalent when the replacement has the same color and no additional balls are added.
Third, adding extra balls changes the counts in addition to whatever happens to the drawn ball. In the convention used below, the drawn ball is returned and then one ball of the opposite color is added: after drawing A, add one B; after drawing B, add one A. Thus an A draw changes \((a,b)\) to \((a,b+1)\), while a B draw changes it to \((a+1,b)\). The total becomes \(N+1\). Other addition rules are possible, but they must be stated explicitly.
A different operation is to remove the drawn ball and put a ball of the opposite color in its place. That is a one-for-one color swap, not an extra-ball addition: after A, the state is \((a-1,b+1)\), and the total remains \(N\). Saying only “replace the drawn ball” does not distinguish these cases.
| Rule after the draw | State after drawing A | Total after drawing A |
|---|---|---|
| Return the drawn ball only | \((a,b)\) | \(N\) |
| Return it and add one B ball | \((a,b+1)\) | \(N+1\) |
| Remove it and put one B ball in its place | \((a-1,b+1)\) | \(N\) |
Calculating the Next Assignment Probability
For the return-and-add-opposite-color rule, the next A probability depends on the first draw. If A is drawn first, the updated state is \((a,b+1)\), so there are still \(a\) A balls, but now \(N+1\) balls in total. If B is drawn first, the updated state is \((a+1,b)\), so there are \(a+1\) A balls among \(N+1\) total balls. The overall probability of A on the second draw is found by weighting each conditional probability by the probability of the first draw that led to it.
The first factor in each product is a first-draw probability. The second factor is conditional on the resulting updated urn state. In particular, after an A draw the next-A probability is \(a/(N+1)\), not \(a/N\): the state has \(N+1\) balls. This denominator check is essential.
Worked Example: Returning the Ball Without Adding Any
Suppose an urn contains 5 A balls and 3 B balls. A ball is drawn uniformly, its original ball is returned, and no other balls are added. Find the probability that the second draw assigns A.
The initial total is \(N=5+3=8\). After either possible first draw, the urn returns to state \((5,3)\). Therefore the conditional probability of A on the second draw is \(5/8\) on either path. Averaging over the first draw gives
Returning the ball without adding any others preserves the A proportion. The second assignment probability is therefore 0.625, or 62.5%, just as it was for the first draw.
Worked Example: Returning the Ball and Adding the Opposite Color
Now start with the same urn, \((5,3)\), but return the drawn ball and add one ball of the opposite color. Find the probability of A on the second draw.
If the first draw is A, which has probability \(5/8\), the updated state is \((5,4)\). The next-A probability is \(5/9\), because the updated total is 9. If the first draw is B, which has probability \(3/8\), the updated state is \((6,3)\), and the next-A probability is \(6/9\). Thus
The second-draw probability is about 59.72%, rather than the original 62.5%. The calculation averages over the two possible first draws; it does not assume that the second probability is the same after both first-draw outcomes.
A Short Derivation by Conditioning
The averaging rule above follows by splitting the event “the second draw is A” according to whether the first draw was A or B. These two first-draw events are mutually exclusive and cover all possibilities. For the return-and-add-opposite-color rule, the conditional probabilities are determined by the updated states, so the law of total probability gives
This derivation also shows why the state update and denominator must be checked together. After an A draw, the A count remains \(a\) but the total is \(N+1\); after a B draw, the A count becomes \(a+1\) and the total is also \(N+1\). Using \(N\) as either conditional denominator would not match the updated urn.
Worked Example: Comparing Three Replacement Conventions
Suppose the current state is \((4,2)\). An A ball is drawn. Compare the probability of A on the next draw under three clearly specified rules.
Return the drawn A ball only. The state stays \((4,2)\), with 6 balls, so the next-A probability is \(4/6=2/3\approx0.6667\).
Return the A ball and add two B balls. The state becomes \((4,4)\), with 8 balls. The next-A probability is \(4/8=0.5000\).
Remove the A ball and replace it with one B ball. The state becomes \((3,3)\), with 6 balls. The next-A probability is \(3/6=0.5000\).
The last two rules happen to give the same probability here, but they do not give the same state: one urn has 8 balls and the other has 6. Their later probabilities need not continue to agree. A numerical match at one step is not evidence that two update rules are equivalent.
Checking an Update and Comparing Alternatives
A reliable calculation records both color counts and the total after every possible draw. For the return-and-add-opposite-color rule, the total increases by one after each draw, regardless of which color was drawn. The A count stays fixed after A and increases by one after B. These checks catch common bookkeeping errors before they enter a probability calculation.
Record \((a,b)\) and verify that \(N=a+b\).
For each possible color drawn, say whether the original ball is returned, removed, or accompanied by additions.
Add the updated color counts. Do not carry over \(N\) automatically.
Use updated counts for a conditional probability; weight each path by its probability when combining paths.
These conventions lead to different dependence patterns. With return-only drawing from a fixed urn, each draw has the same treatment probabilities, and the draws are independent. With extra balls added after a draw, the next probability depends on the earlier draw because the state has changed. Removing a ball and replacing it with a different color also changes the state, but keeps the total fixed. These are distinct from complete randomization, where treatment assignments are generated under a specified allocation scheme rather than by repeatedly drawing from an evolving urn.
The comparison is useful when interpreting an urn rule: its treatment probabilities are not determined solely by the starting counts. They also depend on the update convention. However, a simple urn calculation describes assignment probabilities under the stated rule; it does not by itself establish that a particular rule is clinically appropriate or that treatment outcomes are independent of assignments.
Common Mistake and Limitations
A frequent error is to use the pre-draw total in a conditional probability after adding a ball. For example, after an A draw under the return-and-add-one-B rule, the state is \((a,b+1)\), so the next-A probability is \(a/(N+1)\), not \(a/N\). The numerator counts A balls; the denominator counts all balls now in the urn. Another error is to treat an extra ball as though it replaced the drawn ball. Returning the drawn ball and adding a ball increases the total; removing it and adding one replacement does not.
These calculations assume that each draw is uniform over the balls currently in the urn, that the color-to-treatment mapping is fixed, and that the stated update is carried out before the next draw. They also presume that the urn has the stated composition and that no balls are lost or misclassified. If the draw is not uniform, if the update depends on information beyond the draw, or if the ball-handling rule is ambiguous, the probabilities derived here do not apply without modification.
Check Your Understanding
Answer each question by writing the updated state before calculating a probability.
- An urn has 6 A balls and 4 B balls. The drawn ball is returned, with no additions. What is the probability of A on the next draw?
- From state \((6,4)\), an A ball is drawn, returned, and followed by one added B ball. What is the new state and the next-A probability?
- From state \((6,4)\), a B ball is drawn and removed, then one A ball is added in its place. What is the new state and the next-A probability?
- Why is it incorrect to use the original total as the denominator after returning a drawn ball and adding an extra ball?
Solutions. (1) The state remains \((6,4)\), so the next-A probability is \(6/10=0.6000\). (2) The state is \((6,5)\), with 11 balls, so the probability is \(6/11\approx0.5455\). (3) Removing B and adding A gives state \((7,3)\), with 10 balls, so the probability is \(7/10=0.7000\). (4) The added ball increases the total by one; the denominator must equal the number of balls in the updated state, not the original total.