The bet you must not take twice

The coin from the last lesson has been improved. It now comes up heads 55 times in a hundred. Even money, and you may play as long as you like.

You have 1,000. You decide to stake 500 a throw, because the edge is large and the arithmetic says every throw is worth 50 to you.

Before reading on, commit to an answer.

Predict

Betting half your money each time on a coin with a genuine 55-in-100 edge, and playing on: what is the chance you eventually end up with nothing?

Choose one answer

Almost none — the edge is enormous and it compounds in your favourAround one in tenAround one in threeAround two in three

Lock in answer

Why zero behaves differently from every other number

Between 900 and 1,100 there is nothing special. You can go up from either, down from either, and the arithmetic is the same in both directions.

Zero is not like that. Every other balance is a place you pass through; zero is a place you stop. Nothing multiplies back up from it, no edge helps you from it, and no amount of subsequent good luck reaches you there.

This makes the whole problem multiplicative rather than additive. A run of results is not a sum you can average — it is a product, and a single zero in a product sets the entire thing to zero however large the other terms were.

The same bet, sized differently

The model below runs the exact bet from the prediction. The odds cannot be changed. The only control is how much you put on each throw.

Probability of ruin

≈ Hypothetical model, not a forecast

The chance a bankroll reaches zero over a run of even-money bets, even when the bets are in your favour.

It assumes

• The same amount is staked on every bet, whatever the balance has done.
• Betting stops permanently the first time the balance reaches zero.
• Each bet is independent and pays even money at the stated odds.
• No money can be added to the bankroll part way through the run.

It ignores

• Bet sizing that scales with the bankroll, which changes everything
• Any way to add money to the bankroll mid-run
• A target at which the player would stop and walk away
• Payouts other than even money, and any commission on winnings

Without touching the odds, bring the chance of ruin inside this run below one in twenty. Then read what happened to the expected result per bet.

Staked on each bet500

The same amount every bet. The bankroll is counted in units of this.

Bankroll, counted in bets

2

Edge on each bet

1,000%

Expected result per bet

50

Chance of ruin inside this run

0.661

Chance of surviving this run

0.339

Chance of ruin if the betting never stops

0.669

Expected bankroll at the end, in bets

9.86

Expected bankroll at the end

4,930

• Chance of having reached zero
• Chance of still being in

◷ Not there yet — keep moving the controls.

A stake of 70 — a bankroll fourteen bets deep — brings the chance of ruin just under one in twenty. At a stake of 50 it is closer to one in ninety.

Now look at what that cost. The expected result per bet fell from 50 to 5. The edge is unchanged; you are collecting less of it per throw, in exchange for being extremely likely to still be throwing.

The two averages

There is a name for the split, and it is worth having, because once you have it you see it constantly.

Not the same thing

The average across players

Take a thousand people making this bet once. Add their results and divide. This is what expected value computes.

Measured in

Amount per player, averaged over the population at one moment

For example

A thousand people each stake half of 1,000 on the coin. The pile of money afterwards is larger than the pile before.

Chase it alone and

It answers a question nobody asked. You are not the population, and the population's average is propped up by winners whose money never reaches you.

The average along one path

Take one person making this bet repeatedly. Follow their balance through time. This is what actually happens to you.

Measured in

Amount per bet, averaged over the sequence one person lives

For example

One person stakes half of 1,000 repeatedly. Two thirds of the time the sequence ends at zero and stays there.

Chase it alone and

It is far harder to compute, and it depends on position size and order, so it resists being reduced to one comparable number.

Which is which? Put each one on a side.

• An insurer writing two hundred thousand small policies this year

  The average across playersThe average along one path

• One household putting most of its savings into one venture

  The average across playersThe average along one path

• A fund that reports its average annual return over twenty years

  The average across playersThe average along one path

Check

A strategy has a positive expected value and a one-in-fifty chance per year of losing everything. Someone argues it is worth doing because the expectation is clearly favourable. What is the strongest objection?

Choose one answer

Over a thirty-year horizon the chance of the ruin event landing at least once is large, and after it lands the expectation stops applying to youThe expected value calculation is probably wrongOne in fifty is too small a probability for anyone to estimate reliablyPositive expected value strategies are generally too good to be true

Check answer

›Why the error runs in one directionOptional

Nobody wildly overestimates their chance of ruin. The mistake is one-sided, and there are two reasons for it.

The first is arithmetic. Losses need larger gains to undo them — down 50 requires up 100 to get back — and that asymmetry is invisible when you are adding percentages in your head.

The second is that everyone whose sizing was too aggressive and who survived anyway is available to tell you about it, and everyone whose sizing was too aggressive and who did not is not. The advice you hear about position size has been filtered by the exact process it should be warning you about.

Worth remembering

• Expected value averages across parallel sequences; you only ever live one sequence, in order.
• Zero is an absorbing state — every other balance is somewhere you pass through, and zero is somewhere you stop.
• A sequence of results behaves like a product rather than a sum, and one zero in a product sets the whole thing to zero.
• Assume a large edge and a stake worth half the bankroll: eventual ruin is still roughly two chances in three.
• Position size does not decide whether a bet is good; it decides whether you are present for the outcome.