Module 3 · Lesson 17 of 39 · Free
At least once: the complement rule
In 100 hands you are dealt aces at least once only 36.5% of the time. Adding 100 chances of 1 in 221 would say 45%, but the chances of separate hands do not add up like that. Find the chance that it never happens and take it from 1: that is the complement rule.
The hand
Hand 100 of your session, and still no aces: this time K♣ 7♦. Every deal is a fresh shuffle, with aces 1 time in 221.
Predict first
How many hands do you need for an even chance of being dealt aces at least once?
- About 111: half of 221
- About 153: the misses multiply
- 221: aces come once in 221 hands
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Answer: About 153: the misses multiply
The chance of no aces in n hands is (220/221)ⁿ, because the misses multiply. It first falls below one half at 153 hands: 1 − (220/221)¹⁵³ ≈ 50.04%. Even after 221 hands the chance of at least one pair of aces is only 63.3%.
Working it out
A hand without aces comes 220 times in 221. Two such hands in a row come 220/221 × 220/221 of the time, because the deals are independent: the first tells you nothing about the second. Over 100 hands the misses multiply 100 times: (220/221)¹⁰⁰ ≈ 0.635, so 63.5% of 100-hand sessions see no aces at all. The rest, 1 − 0.635 = 36.5%, see aces at least once. The same rule works for anything: a pocket pair comes 1 hand in 17, so in 10 hands you see at least one 1 − (16/17)¹⁰ ≈ 1 − 0.545 = 45.5% of the time.
The complement rule fits every “at least once” question: at least one pocket pair in the next ten hands, at least one set from your next five pocket pairs. Find the chance that it never happens, multiply it once for each try, and take the result from 1. When the tries are cards from one deck, like the three cards of a flop, the miss chance changes card by card, so multiply those instead. It also tells you how long to wait: aces at least once becomes more likely than not only after 153 hands, and even 500 hands leave a 10.4% chance of none.
Formula
P(at least once in n tries) = 1 − (1 − p)ⁿ: aces in 100 hands, 1 − (220/221)¹⁰⁰ ≈ 1 − 0.635 = 36.5%
Check your understanding
Your flush draw comes in 1 time in 5. You hold one on 4 different hands. What is the chance that it comes in at least once, in %?
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Answer: 59%
Each time it misses 4 times in 5: 0.8⁴ = 0.4096 is the chance of four misses, so 1 − 0.4096 = 59.0%.
You are all in three times, each time a coin flip that you win half the time. What is the chance that you lose all three, in %?
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Answer: 12.5%
Independent chances multiply: 0.5 × 0.5 × 0.5 = 0.125, or 12.5%.
You play 30 hands a day. Why is the chance of at least one pair of aces in a day less than 30 ÷ 221 = 13.6%?
- Adding counts a day with two pairs of aces as two lucky days
- Aces are less likely right after you have had them
- Aces come less often at a full table
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Answer: Adding counts a day with two pairs of aces as two lucky days
Chances of separate hands do not add. The right number is 1 − (220/221)³⁰ ≈ 12.7%; the sum counts every day with two or more pairs of aces more than once.