What "one in six" really promises
Tonight's idea
A fair die has six faces, all equally likely. So the chance of a six is \(\tfrac16\), "one in six". That does not mean one six in every six rolls. Roll six times and you get no six at all about one time in three.
The promise is about the long run:
fraction of sixes = number of sixes ÷ number of rolls. Over more and more rolls, this fraction settles close to \(\tfrac16\), which is about 16.7%.
Why does it settle? Each roll is a fresh start. The die has no memory, so it never "owes" you a six. Early luck, good or bad, is not paid back. It is swamped. Three extra sixes matter a lot in 60 rolls and hardly at all in 6,000.
This also works backwards. To estimate an unknown chance, do the thing many times and use the fraction you see.
Watch one: the run in the picture
Here is the score at three moments.
| Rolls | Sixes expected | Sixes rolled | Off by | Fraction of sixes |
|---|---|---|---|---|
| 60 | 10 | 13 | 3 | \(13 \div 60 \approx 21.7\%\) |
| 600 | 100 | 106 | 6 | \(106 \div 600 \approx 17.7\%\) |
| 6,000 | 1,000 | 1,033 | 33 | \(1033 \div 6000 \approx 17.2\%\) |
"Sixes expected" is \(\tfrac16\) of the rolls. The count drifts further from that: off by 3, then 6, then 33. But the fraction closes in on 16.7%. The promise is about the fraction, not the count.
Roll 10,000 times in a second and watch the fraction close in: mathdetours.com/demos/longrun.html.
Your turn
- A fair spinner has 5 equal sections. One is red. (a) What is the chance of red? (b) About how many reds would you expect in 50 spins? In 2,000 spins? (c) Is exactly that many reds certain?
- Drop a drawing pin and it lands point up or point down. You cannot find the chance by counting faces. Maya drops one 200 times and it lands point up 124 times. (a) Estimate the chance of point up, as a percentage. (b) About how many times would it land point up in 1,500 drops?
- Ana flips a coin 10 times and gets 7 heads. Ben flips a coin 1,000 times and gets 540 heads. (a) What percentage of heads did each get? (b) How many heads should a fair coin give each of them? How far off is each, counted in flips? (c) One of these coins might be unfair. Whose would you suspect, and why?
- Leo rolls a fair die 20 times and gets no six. "A six is due," he says. (a) What is the chance of a six on his next roll? (b) Suppose his next 600 rolls contain exactly the expected number of sixes. What fraction of all 620 rolls are sixes? Give a percentage. (c) Did the die have to make up for the bad start?
Stretch: no six in six rolls
The chance of "not a six" on one roll is \(\tfrac56\). For two rolls in a row it is \(\tfrac56 \times \tfrac56 = \tfrac{25}{36}\). Carry on to six rolls in a row, using a calculator. Write the answer as a decimal. Is "no six in six rolls" rare?
Answers
- (a) \(\tfrac15\), or 20%. (b) \(50 \div 5 = 10\) and \(2000 \div 5 = 400\). (c) No. These are the numbers to expect, give or take. Only the long-run fraction is promised.
- (a) \(124 \div 200 = 0.62\), so about 62%. (b) \(0.62 \times 1500 = 930\), so about 930.
- (a) Ana 70%, Ben 54%. (b) Ana: 5 expected, off by 2. Ben: 500 expected, off by 40. (c) Ben's. With only 10 flips the fraction is still jumpy, and a fair coin gives 7 or more heads about one time in six. After 1,000 flips the fraction should have settled near 50%. A fair coin gives 540 or more heads less than one time in a hundred.
- (a) Still \(\tfrac16\). The die has no memory. (b) The 600 rolls contain 100 sixes, so \(100 \div 620 \approx 16.1\%\). (c) No. With no making up at all, the fraction is already close to 16.7%. The bad start is swamped by the later rolls, not paid back.
Stretch. \(\tfrac56\) multiplied by itself six times is \(\tfrac{15625}{46656} \approx 0.335\). Not rare at all: about one time in three, a set of six rolls has no six.