What Is a Game Worth on Average?

An endless game is hard to think about, so cut it short. Suppose Prabin allows only three tosses: if all three are tails, the game ends and you get nothing.

There are four ways the game can go. Each toss is heads or tails with chance ½, and the tosses don’t affect one another, so the chances multiply (Grade 10’s multiplication law):

Tosses	Chance	Prize
H	½	Rs 2
T H	¼	Rs 4
T T H	⅛	Rs 8
T T T	⅛	Rs 0

Play this game eight times “on paper”, with each way coming up exactly as often as its chance says: H four times, TH twice, TTH once, TTT once. The prizes are 4 × Rs 2 + 2 × Rs 4 + 1 × Rs 8 + 1 × Rs 0 = Rs 24, which is Rs 3 a game.

Look at where those Rs 24 came from: Rs 8 from the H games, Rs 8 from the TH games, Rs 8 from the one TTH game. Each row of the table paid exactly the same. The prize doubles from row to row, and the chance halves, so (chance × prize) is ½ × 2 = ¼ × 4 = ⅛ × 8 = 1 every time.

Now take the cap away. Prabin’s real game has a row for every number of tosses, without end, and each row adds another Rs 1. Ten rows make Rs 10, a hundred rows Rs 100, a thousand rows Rs 1,000. No amount is ever enough: whatever price you name, enough rows add up to more.

What you have been working out has a name. A quantity that a chance event decides, like the prize in Prabin’s game, is a random variable, usually written with a capital letter, X. The table of its values with their chances is its probability distribution: every value it can take, and P(X = x), the chance it takes the value x. The chances add up to 1, because something has to happen.

The expected value of X is each value times its chance, all added up:

$$E(X) = \sum x \, P(X = x) = x_1 p_1 + x_2 p_2 + x_3 p_3 + \cdots$$

For the three-toss game, $E(X) = 2 \times \tfrac{1}{2} + 4 \times \tfrac{1}{4} + 8 \times \tfrac{1}{8} + 0 \times \tfrac{1}{8} = 3$. That is the Rs 3 of the eight games on paper, and no accident.

Why it is a mean

You already know this formula, in disguise. Grade 10 finds the mean of a frequency table as $\bar{x} = \frac{\sum f x}{\sum f}$: each value times how often it came, added up, divided by how many there were. Divide each frequency by the total first and you get $\sum x \times \frac{f}{\sum f}$: each value times the share of the games it came up in. In a long run of games those shares settle down near the chances (that is what a probability means), so the mean prize settles down near $\sum x \, P(X = x)$. The expected value is the mean of the probability distribution: the mean you would get from an endless run of games, worked out without playing any.

That is why a fair price for a game is its expected value. If you pay Rs 3 for each three-toss game, then over many games you and Prabin come out about even. Your gain in a game is the prize minus the price, and a game is fair when the expected gain is zero. The word “expected” is a trap, though: nobody expects to win Rs 3 in one three-toss game, which can only pay 0, 2, 4 or 8. The expected value is what many games average out to. (Is This Game Fair? in Grades 6–8 watches a fairground stall’s average settle as the games pile up.)

An average that never settles

Prabin’s real game has an expected value too, but the sum never stops growing:

$$E(X) = 2 \times \tfrac{1}{2} + 4 \times \tfrac{1}{4} + 8 \times \tfrac{1}{8} + 16 \times \tfrac{1}{16} + \cdots = 1 + 1 + 1 + 1 + \cdots$$

Mathematicians say the expected value is infinite: the sum passes every number. So what does an actual run of games do? In a run of n games, the rows whose chance is bigger than about $\frac{1}{n}$ come up often enough to pay their Rs 1 each; the rows beyond that mostly don’t come up at all, and once in a while one does and pays a fortune. A thousand games fill the first ten rows or so ($2^{10} = 1{,}024$); a million games fill about twenty. So the average of n games is usually close to the number of rows filled, about $\log_2 n$, plus a few rupees, and every so often a huge win throws it far above that.

Played by computer, thousands of runs for each length:

Games played	10	100	1,000	10,000	100,000
Typical average prize (the median run)	Rs 6	Rs 9.20	Rs 12.60	Rs 16	Rs 19.20
One run in ten averages more than	Rs 19	Rs 25	Rs 28	Rs 30	Rs 35

Ten times as many games add only about Rs 3 to the typical average; each further Rs 10 takes about a thousand times as many games (2¹⁰ = 1,024). The average does climb without end, as the expected value says it must, but so slowly that nobody will ever see a typical run reach Rs 100: that would take about $2^{97}$ games, roughly 1.6 × 10²⁹.

That is the first surprise of the game. An expected value is a long-run average, and for most games the long run arrives quickly. For Prabin’s game it never arrives.

Expected values answer other questions too. How many tosses does one of Prabin’s games take, on average? The table of tosses: 1 toss with chance ½, 2 with chance ¼, 3 with chance ⅛, and so on, so

$$E(\text{tosses}) = 1 \times \tfrac{1}{2} + 2 \times \tfrac{1}{4} + 3 \times \tfrac{1}{8} + 4 \times \tfrac{1}{16} + \cdots$$

That sum can be done, but there is a lovelier way. Call the answer E. Every game takes one toss to start with. Half the time it lands heads and the game is over. The other half, it lands tails, and from then on the game is exactly like a fresh one: it still needs E more tosses on average. So

$$E = 1 + \tfrac{1}{2} \times E, \qquad \text{so} \qquad \tfrac{1}{2} E = 1, \qquad E = 2.$$

A game lasts 2 tosses on average. The same trick finds how long you wait for a six on a die: one throw, then with chance $\frac{5}{6}$ no six and a fresh start, so $E = 1 + \frac{5}{6} E$, and E = 6. The answer you would guess, but now for a reason. And the reason warns you that 6 is only an average: about a third of the time (33.5%) six throws pass with no six at all, and one wait in nine is longer than twelve throws.

Expected values add

One more rule makes many hard questions easy: the expected value of a sum is the sum of the expected values, E(X + Y) = E(X) + E(Y), whenever both are ordinary numbers, even when X and Y depend on each other. The mean of a total is the total of the means, as it is for any list of numbers. The total of two dice has expected value 3.5 + 3.5 = 7, with no table of 36 throws needed. Ten throws of a die show 10 × $\frac{1}{6} = \frac{5}{3}$ sixes on average. (The birthday topic used this rule to find the 0.69 matching pairs to expect in a class of 23.)

To weigh a decision, give each outcome its chance and its payoff, and compare the expected values of the choices. Two real decisions:

Should you guess?

Nepal’s entrance examination for medical and health courses, MECEE-BL, has 200 multiple-choice questions with four options each. A right answer earns 1 mark and each wrong answer loses 0.25, the Medical Education Commission says, “in order to discourage the examinee from answering only on the basis of conjecture” (as of 2026; read the current notice). Does it?

Suppose you have no idea and guess at random. The mark for that question is a random variable: +1 with chance ¼, −0.25 with chance ¾. Its expected value is

$$1 \times \tfrac{1}{4} + (-0.25) \times \tfrac{3}{4} = \tfrac{4}{16} - \tfrac{3}{16} = +\tfrac{1}{16} \text{ mark}.$$

A blind guess is worth +0.0625 marks on average, more than leaving the question blank (0). If you can rule out one option and guess between three, it is worth $\frac{1}{3} - \frac{2}{3} \times 0.25 = +\frac{1}{6}$; between two, $+\frac{3}{8}$. For the deduction to make a blind guess worth exactly nothing, it would have to be $\frac{1}{3}$ of a mark. So on this scheme, a student who wants the most marks on average should answer every question, guessing where they must: the quarter-mark penalty doesn’t make a blind guess a bad bet, it only shrinks its reward from ¼ of a mark to $\frac{1}{16}$. (A single guess still loses 0.25 three times in four: guessing raises the average, not the certainty. Whether a test-taker should play for the average is the question of “Why nobody pays much”, below.)

Is a lottery ticket worth buying?

The American Powerball lottery costs $2 a ticket. You pick five numbers from 1 to 69 and one “Powerball” from 1 to 26; the prizes depend on how many you match. The chances come from counting, as in Grade 12’s combinations: there are $\binom{69}{5} \times 26 = 292{,}201{,}338$ possible tickets, and exactly one wins the jackpot. The fixed prizes (as of 2026), and what each adds to the expected value of one ticket:

Match	Chance (1 in …)	Prize	Adds to E(X)
Powerball only	38.32	$4	$0.104
1 number + Powerball	91.98	$4	$0.043
2 + Powerball	701.33	$7	$0.010
3 numbers	579.76	$7	$0.012
3 + Powerball	14,494.11	$100	$0.007
4 numbers	36,525.17	$100	$0.003
4 + Powerball	913,129.18	$50,000	$0.055
5 numbers	11,688,053.52	$1,000,000	$0.086
5 + Powerball	292,201,338	the jackpot	jackpot ÷ 292,201,338

The fixed prizes add up to about $0.32 of each $2 ticket. The jackpot starts at $20 million, which adds about 7 cents more: a ticket is then worth about 39 cents of its $2. For the ticket to be worth its price, the jackpot would have to pass about $491 million, and it does sometimes, because when nobody wins it rolls over and grows. The record, $2.04 billion in November 2022, would make a ticket worth about $7.30 by this sum.

But the sum leaves things out. The advertised jackpot is paid over 29 years; the winner of the $2.04 billion took the lump sum instead, which came to $628.5 million after taxes, under a third of it, and that alone brings the ticket down to about $2.50. A record jackpot also sells far more tickets, so a winner is more likely to have to share. Those record draws are rare: on almost every draw the jackpot is far below $491 million, and a ticket is worth much less than its $2, which is how a lottery keeps more than it pays out. And even with the most generous sum, a ticket loses everything 23 times in 24 (any prize at all comes up once in about 24.9 tickets), and the jackpot comes once in 292 million: buying a ticket every day, you would expect to wait about 800,000 years. This is where the expected value stops being the whole story.

Back to Prabin. His game’s expected value is infinite, so by the rule “a fair price is the expected value”, you should give him anything he asks: your phone, your savings, your house. No one would. Three reasons why, each of them a real idea.

No bank is endless

Prabin doesn’t have a crore. Suppose he has Rs 1,000 and the rule is: if the prize would be more than he has, you get everything he has. Then rows 1 to 9 pay their Rs 1 each (the ninth prize is Rs 512), and every longer game pays Rs 1,000, which happens with chance 1 in 2⁹ = 512, adding 1,000 ÷ 512 ≈ Rs 1.95. The expected value is 9 + 1.95, about Rs 10.95. The infinity was always in the rows beyond any real bank.

Each time the bank doubles, one more row fits and the game is worth only Rs 1 more. Rs 1 crore makes it worth about Rs 24; a lakh crore rupees, Rs 41; a crore crore rupees (10¹⁴), only about Rs 47. The expected value grows like the logarithm of the bank: multiply the bank by ten and the game is worth only about Rs 3.32 more.

A rupee means less to the rich

In 1738 Daniel Bernoulli suggested that what a person should average is not money but how much the money is worth to them, and that a gain matters less the richer you already are: a thousand rupees means a great deal to a poor student and little to a millionaire. He proposed measuring it by the logarithm of a person’s whole wealth, so that doubling your money is worth the same to anyone, rich or poor. (Economists now call this worth utility.)

On Bernoulli’s scale, the most you should pay for Prabin’s game depends on how much you have. With Rs 1,000 to your name, about Rs 11; with a lakh, about Rs 17.50; with a crore, about Rs 24. The rich can afford to pay more for a long shot, and even they not much.

Most games pay little

The third reason is the one your stomach gives you. Half the games pay Rs 2; three in four pay Rs 4 or less; only one game in 32 pays Rs 64 or more. The infinite average is carried entirely by long runs of tails that almost never come. Paying Rs 100 for one game means losing money unless the coin shows tails six times running first, a chance of 1 in 64. An expected value tells you how a game averages over many plays. For a single play, it can be a poor guide to what will actually happen, and the less likely the outcomes that carry it, the poorer.

• Is This Game Fair? (Grades 6–8) Expected value first met at a fairground stall, watched settling over a thousand games, with insurance for a buffalo.
• Should You Switch Doors? A game-show choice where working out the chances properly changes your mind.
• Is a Positive Test Result Bad News? Chances that depend on what you already know, in medical tests.
• Can You Add Halves Forever and Still Get One? (Grades 6–8) Endless sums that do settle, like question f.’s.
• Will Two People in Your Class Share a Birthday? Another chance everyone’s gut gets wrong, and the expected number of matching pairs.
• Why Do Snakes and Ladders Games Take So Long? (Computing, Going Further) The average length of a game, found by letting a program play thousands.
• Grade 10 Math: Probability The rules the chances on this page are built from.