Under a dollar
After 100 rounds 54% of players have less than $1 left of their original $100, even though average wealth is $13,150.
You have $100. Each round you bet all your money on a fair coin: heads adds 50% to what you have, tails takes away 40%. On average a round gives +5%, so the game looks like a good deal. 10,000 people play it, each on their own, for 100 rounds. What share of them will finish with more than $100? Lock in your answer, then toss the coin yourself.
Average wealth after 100 rounds really does grow: $13,150 on average instead of $100. But that average is held up by a few lucky millionaires. The typical player, with 50 heads and 50 tails, has 52 cents: a heads-and-tails pair multiplies the money by 1.5 · 0.6 = 0.9. To come out ahead you need 56 heads out of 100, and only 13.6% of people get that. The average over the crowd and the fate of one person over time differ here: the game is non-ergodic.
The physicist Ole Peters analysed this game in "The ergodicity problem in economics" (Nature Physics, 2019). It is as simple as tossing a coin, yet it separates two ways of thinking about risk: as an average over many people, and as the path of one person through time.
You have $100. Each of the 10,000 players also has $100, and each plays with their own coin, independently of the others.
You stake all the money you have right now and toss a fair coin. Heads: the sum grows by 50%. Tails: it shrinks by 40%. From $100, heads gives $150 and tails $60.
The percentages apply to your current sum. If you have $150, heads gives $225 and tails $90. You can't leave the game halfway.
Each player plays 100 rounds. What share of the 10,000 players will finish with more than $100?
"The game pays: +5% a round"
Heads and tails come up equally often. The average change per round: ½ · (+50%) + ½ · (−40%) = +5%. Over 100 rounds at +5% the sum should grow 1.05¹⁰⁰ ≈ 131 times. So most people should get richer? Or at least half, since the coin is fair?
10,000 people, each with $100, each plays 100 rounds: heads +50%, tails −40%. What percentage of them will finish with more than they started with? Pick a zone or type a number.
The exact share and a comparison with your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Play two rounds: heads, then tails. What is left of $100? And the other way round, tails first?
Over 100 rounds a typical player gets about 50 heads and 50 tails. What do 50 heads-and-tails pairs do to the money?
How many heads out of 100 does it take for 1.5^k · 0.6^(100 − k) to exceed 1? How often does a fair coin give that many?
The next rounds show how many players get richer: your own game with the coin, 10,000 players and the explanation. Answer round 1 first, or go on?
You chose to always show all rounds.
The blue line is your money, round by round. The grey lines are other players playing alongside, each with their own coin. The scale is logarithmic: each gridline is 100 times the one below, so a dollar and a million fit on one chart. The gold line is the average of all players, the green one the median, the sum of the player in the middle of the queue from poorest to richest.
The computer plays for 10,000 people. Each bar of the histogram collects the players with the same number of heads, and hence the same sum: the order of heads and tails doesn't change the result. Magenta bars ended richer than they started. The circles show how many players the exact formula puts in each bar. Move the number of rounds.
Once you answer, the simulation code appears here so you can run it yourself.
The same simulation, short and with no libraries. Run the JavaScript in your browser console (F12) or in Node; the Python needs only the standard library.
// The coin game: 10,000 players, 100 rounds each, heads +50%, tails −40%
const N = 10000, T = 100;
const money = [];
for (let i = 0; i < N; i++) {
let w = 100; // everyone starts with $100
for (let t = 0; t < T; t++) w *= Math.random() < 0.5 ? 1.5 : 0.6;
money.push(w);
}
money.sort((a, b) => a - b);
const mean = money.reduce((s, w) => s + w, 0) / N;
// exact: it takes at least 56 heads out of 100
let exact = 0, c = 1; // c = C(100, k)
for (let k = 0; k <= T; k++) { if (k >= 56) exact += c / 2 ** T; c = c * (T - k) / (k + 1); }
console.log(`in profit: ${(100 * money.filter(w => w > 100).length / N).toFixed(1)}% exact ${(100 * exact).toFixed(2)}%`);
console.log(`median: $${money[N / 2].toFixed(2)} exact $${(100 * 0.9 ** 50).toFixed(2)}`);
console.log(`average: $${mean.toFixed(0)} expected $${(100 * 1.05 ** T).toFixed(0)} (the average of 10,000 players is usually lower: a few lucky ones hold it up)`);
# The coin game: 10,000 players, 100 rounds each, heads +50%, tails −40%
import random
from math import comb
N, T = 10_000, 100
money = []
for _ in range(N):
w = 100 # everyone starts with $100
for _ in range(T):
w *= 1.5 if random.random() < 0.5 else 0.6
money.append(w)
money.sort()
mean = sum(money) / N
# exact: it takes at least 56 heads out of 100
exact = sum(comb(T, k) for k in range(56, T + 1)) / 2 ** T
print(f"in profit: {100 * sum(w > 100 for w in money) / N:.1f}% exact {100 * exact:.2f}%")
print(f"median: ${money[N // 2]:.2f} exact ${100 * 0.9 ** 50:.2f}")
print(f"average: ${mean:.0f} expected ${100 * 1.05 ** T:.0f} (the average of 10,000 players is usually lower: a few lucky ones hold it up)")
The "+5% a round" calculation is correct. It just answers a different question: what happens to all the crowd's money taken together, not to the money of one person.
“In plain words” is hidden on all puzzles.
One heads and one tails, in either order, leave you 90% of your money: 1.5 · 0.6 = 0.9. In 100 rounds a typical player gets about 50 heads and 50 tails, that is 50 such pairs, losing 10% each time. After 50 pairs, $100 has become 100 · 0.9⁵⁰ ≈ 52 cents. Even 55 heads out of 100, more than half, give only $50. It takes at least 56 heads to come out ahead, and only 13.6% of players get that many.
That explains the typical player, but not yet where the average +5% comes from. The two cards below deal with that.
| Coins | Money after 2 rounds |
|---|---|
| heads, heads | $225 |
| heads, tails | $90 |
| tails, heads | $90 |
| tails, tails | $36 |
| Average | $110.25 |
Four equally likely outcomes. The average is $110.25, exactly +5% twice: 100 · 1.05². But three of the four lost money. The average is pulled up by one lucky player with $225: their gain is bigger than the other three's losses put together.
Thirty people sit in a café, each with a few thousand dollars of savings on average. A billionaire walks in, and the visitors' average wealth jumps to tens of millions. None of them got richer; the average now just describes one person.
The coin game is the same. With every round the crowd's money gathers more and more in a few record holders who threw lots of heads. In a simulation of 10,000 players, the richest 1% (100 people) usually hold over 90% of all the money after 100 rounds. The average grows because of them, not because of you.
“Rigorous” is hidden on all puzzles.
Each round, money is multiplied by 1.5 or by 0.6. The arithmetic mean of these factors, (1.5 + 0.6)/2 = 1.05, is the expected value: the average over countless parallel players after one round.
But for one person the factors aren't added, they are multiplied. So the typical factor per round is the geometric mean: √(1.5 · 0.6) = √0.9 ≈ 0.949, about −5.1% a round. The geometric mean is never above the arithmetic one, and the more the factors swing, the wider the gap. Here it even flips the sign: +5% against −5.1%.
With k heads in 100 rounds you have 100 · 1.5k · 0.6100−k dollars. That is more than $100 when 1.5k · 0.6100−k > 1, that is k > 100 · ln(1/0.6) / ln(2.5) ≈ 55.75. So you need at least 56 heads.
The number of heads in 100 tosses of a fair coin follows a binomial distribution with mean 50 and standard deviation 5. The probability of 56 or more is 13.56%. The more rounds, the further the threshold sits from the middle and the smaller the share in profit.
| Rounds | Heads needed | Share in profit | Average | Median |
|---|
The average grows like 100 · 1.05n, the median falls roughly like 100 · 0.949n. The share in profit falls overall, but not smoothly: the threshold is always a whole number of heads, and for some n it sits just above the middle, for others well above it. So after 20 rounds 25.2% are in profit, after 30 again 29.2%, and after 50 already 24.0%. After 1,000 rounds about one player in 7,000 is left in profit.
W₀ = $100 is the starting sum, n the number of rounds, K the number of heads, K ~ Bin(n, ½). The sum after n rounds:
The expectation is multiplied by 1.05 every round, because rounds are independent:
One person's growth rate: the average logarithm of the factor. By the law of large numbers ln Wn/n tends to it almost surely:
The share in profit after 100 rounds:
The expectation E[Wn] describes the average over an infinite crowd at one moment. The rate g describes what happens to one person over a long time. For additive games the two numbers agree; for multiplicative ones they don't.
“Deeper” is hidden on all puzzles.
A process is called ergodic if the average over many parallel copies equals the time average of a single copy. Then it makes no difference whether 10,000 people play once each or one person plays 10,000 times.
Picture the same coin with fixed sums: heads +$50, tails −$40, and enough reserve not to go broke. Then both the average over people and one person's long-run average come to +$5 a round: the game is ergodic. With percentages of the current sum, the average over people gives +5%, while one person over time loses about 5.1% a round. Here the expectation doesn't tell you what is in store for you.
$13,150 is the average over an infinite crowd. Half of it comes from players with 72 or more heads, and they are one in 160,000. Among 10,000 players there is usually none, so the average of those particular 10,000 after 100 rounds is mostly lower: over 300 repeats of the simulation the median of that average was about $5,000, and 80% of repeats fell between $2,700 and $16,000.
The smaller the crowd, the sooner its average breaks away from the expectation and starts falling with the typical player. In the long run even the average of a large but finite crowd ends up behaving like one person.
In 1738 Daniel Bernoulli, analysing the St. Petersburg paradox, proposed judging a game not by its average win but by the average logarithm of wealth: that explained why nobody pays a fortune for a game with infinite expectation. He justified the logarithm by psychology: each extra dollar pleases less.
Ole Peters, together with Murray Gell-Mann ("Evaluating gambles using dynamics", 2016) and in his 2019 Nature Physics article, offered another reading: the average logarithm is simply one person's growth rate over time, and for multiplicative bets it can be derived with no psychology at all. Many economists dispute that this changes anything in utility theory, and the debate goes on. But the coin game itself doesn't depend on that debate: the numbers on this page are the same whichever view you take.
What makes the game a loser is not the coin but staking all your money every round. Stake only a fraction and see how the typical player's fate changes. Then from the numbers to the idea behind them, and to where it shows up in life.
Staking a fraction f, heads multiplies the money by 1 + 0.5f, tails by 1 − 0.4f. The typical growth rate is g(f) = ½ ln(1 + 0.5f) + ½ ln(1 − 0.4f). Its derivative is zero when 0.25/(1 + 0.5f) = 0.2/(1 − 0.4f), that is at f = 0.25. This is the Kelly criterion: for a bet that wins a fraction b with probability p and loses a fraction a with probability q, f* = p/a − q/b = 0.5/0.4 − 0.5/0.5 = 0.25.
With a 25% stake a heads-and-tails pair multiplies the money by 1.125 · 0.9 = 1.0125, and after 100 rounds the typical player has $186 instead of 52 cents, with 69% of players in profit. With a 50% stake the pair gives exactly 1.25 · 0.8 = 1: the typical player marks time at $100. A bigger stake raises the average but sinks the typical player. The average is highest at f = 100%, exactly where a person gets poorer fastest.
In "Skin in the Game" (2018), in the chapter "The Logic of Risk Taking", Nassim Taleb compares two pictures. A hundred people go to a casino once each: some lose everything, but the group's share of wins and losses can be worked out, and it says nothing about what happens to one person who plays every day. The second picture: one person, his cousin Theodorus, goes to the casino a hundred days in a row. On day 28 he is bust, and there will be no day 29 for him.
The idea Taleb borrows from Peters and Gell-Mann: when a loss takes a share of your wealth or knocks you out of the game, risk must be judged along one person's path through time, not by the average over a group. Ruin is absorbing: after it, the average no longer matters.
Peters and Alexander Adamou ("An evolutionary advantage of cooperation") showed another way out. Let several players, after every round, pour all their money into a common pot and split it equally. Each still stakes everything, but the pot swings less than each player alone, so the typical growth rate moves towards the average +5%.
| Players pooling | Rate g per round | Typical over 100 rounds |
|---|
Two players still get slightly poorer, three already get richer, and with a hundred the pot grows almost like the expectation: the limit is g = ln 1.05 ≈ 0.0488.
A multiplicative game hides wherever gains and losses are counted as a percentage of what you already have: savings, a portfolio, a business. The check is simple: multiply the factors instead of adding the percentages.
| Situation | What the average says | What happens over time |
|---|---|---|
| A share falls 50%, then rises 50% | Average change 0% | 0.5 · 1.5 = 0.75: minus 25%. The bigger the swings, the further the geometric mean lags the arithmetic one. This is called volatility drag. |
| A fund with triple leverage, reset daily | Three times the index's daily result | In an SEC and FINRA alert (August 2009), from 1 December 2008 to 30 April 2009 the Russell 1000 Financial Services Index rose about 8%, a triple-leveraged fund fell 53%, and a triple inverse fund fell 90%. |
| Bet size in a game with an edge | The bigger the bet, the bigger the average win | In 1956 John Kelly derived the stake at which capital grows fastest in the long run. Edward Thorp used the idea at blackjack and later in investing: a stake above Kelly adds risk and lowers growth. |
| Diversification, mutual insurance | The expectation doesn't change | The swings shrink, so each member's growth rate rises, as in the pot from the table above. That is why sensible risk pooling can benefit everyone at once, even though nobody raised the average. |
Other puzzles about bets, averages and time: the St. Petersburg paradox, gambler's ruin, Parrondo's paradox, where two losing games win together, and the longer you wait, the longer you wait.
After 100 rounds 54% of players have less than $1 left of their original $100, even though average wealth is $13,150.
Daniel Bernoulli proposed judging risk by the average logarithm of wealth as early as 1738. Almost three centuries later Peters pointed out that in a multiplicative game this is exactly one person's growth rate.
John Kelly, a physicist at Bell Labs, derived the optimal stake in "A New Interpretation of Information Rate", building on Claude Shannon's information theory. William Poundstone's book "Fortune's Formula" tells the formula's story.
This exact game, +50% for heads and −40% for tails, is the main example in Ole Peters's article "The ergodicity problem in economics" in Nature Physics in 2019.
Over five months in the winter and spring of 2008–09 a financial-stocks index rose about 8%, while a fund delivering three times its daily result fell 53%. The SEC and FINRA used the example in an investor alert.
If 10 players stake everything but split their money equally after every round, the typical player multiplies their wealth about 51 times in 100 rounds. Alone, they would have 52 cents.
Share your own explanation, an objection or a variant of the puzzle. You need a GitHub account to comment.
Is “Advanced” the right level for this puzzle?
Thank you! This helps set the levels right.
I keep adding new puzzles, new ways to explain them and write-ups of the discussions they start. Leave your email and I'll write when something new is out. No spam, one-click unsubscribe.
The sign-up form will appear here soon.