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"A and B throw a die in turn, A first. The first to throw a six wins. Find the probability that A wins." It looks like it needs an endless tree diagram. It needs one geometric series. This post sets up the method, works through four standard versions and shows the slips that turn a two-line answer into a wrong one.
Where is this in the syllabus?
The multiplication rule and independent events are in sections 13.3 and 13.4 of the NCERT Class 12 chapter Probability (2026-27 reprint). The JEE Advanced 2026 brochure lists "independence of events" under probability and "infinite geometric series" under algebra. The JEE (Main) 2026 syllabus lists "addition and multiplication theorems of probability" and geometric progressions. Check the syllabus for your exam year.
How do you set up a turn-taking game?
Let each turn succeed with probability p and fail with q = 1 − p, independently of other turns.
A can win on turn 1, turn 3, turn 5, and so on. To win on turn 3, A and B must both fail first, then A succeeds: q × q × p. To win on turn 5, four failures come first. So
P(A wins) = p + q²p + q⁴p + ... = p/(1 − q²).
This is an infinite GP with first term p and ratio q², and it converges because q² < 1. B's chance is q times A's, because B's sequence is A's shifted by one failed turn: qp/(1 − q²).
Worked example 1: the first six
p = 1/6, q = 5/6, so q² = 25/36 and 1 − q² = 11/36.
P(A wins) = (1/6)/(11/36) = (1/6) × (36/11) = 6/11.
P(B wins) = (5/6) × (6/11) = 5/11.
They add to 1, so the game ends with probability 1. A's edge comes only from going first.
Worked example 2: three players
A, B and C throw in that order until someone gets a six. Now a full round has three failures, so the ratio is q³ = 125/216 and 1 − q³ = 91/216.
P(A) = (1/6)/(91/216) = 36/91.
P(B) = (5/6) × 36/91 = 30/91.
P(C) = (5/6) × 30/91 = 25/91.
Check: 36 + 30 + 25 = 91. The chances fall in the ratio 1 : 5/6 : 25/36, one failure factor for each place in the order.
Worked example 3: players with different skills
A hits a target with probability 1/3 per shot and B with probability 1/2 (sample numbers for this exercise). They shoot alternately, A first, until someone hits.
One full round with no hit has probability (2/3) × (1/2) = 1/3. So
P(A wins) = (1/3)/(1 − 1/3) = (1/3)/(2/3) = 1/2,
and B also has 1/2. Going first exactly makes up for A's lower skill here. A coin version is quicker still: first head wins, A first, gives (1/2)/(1 − 1/4) = 2/3 for A.
Worked example 4: at least one success in n trials
What is the chance of at least one six in four throws? Work through the complement: no six in four throws has probability (5/6)⁴ = 625/1296. So the answer is 1 − 625/1296 = 671/1296, about 0.518.
How many throws make at least one six 90% likely? You need (5/6)ⁿ ≤ 0.1. Since (5/6)¹² ≈ 0.112 and (5/6)¹³ ≈ 0.093, the answer is 13 throws. Taking logarithms gives the same: n ≥ log 0.1 / log(5/6) ≈ 12.6.
Which mistakes cost marks?
Using q instead of q² as the ratio. Between A's turns, both players fail.
Adding p for every turn. p + p + p ... exceeds 1. Each later turn needs all earlier failures in front of it.
Using n × p for "at least one". Six throws do not make a six certain: the true chance is 1 − (5/6)⁶ ≈ 0.665, not 6 × 1/6 = 1.
Forgetting independence. The GP works only when each turn's chance stays the same. Drawing balls without replacement changes p each turn, so you must multiply term by term instead.
Not checking the total. All players' chances must add to 1.
The rules underneath all of this are in our guide to independent and mutually exclusive events. For the series side, our post on the binomial theorem covers the expansions that often appear in the same papers.
If you cannot see which terms go into the series, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
For each game, write A's first three winning paths in words ("fail, fail, succeed") before writing any formula.
Name the ratio as "one full round of failures" and compute it separately.
Check that all players' probabilities add to 1.
Do one "at least one" question with the complement each session.
Change one number in a solved problem and predict the direction of the change before recomputing.
Try this now: A and B toss a coin in turn, B first, and the first to get a head wins. Find A's chance. (Answer: 1/3.)
