How poker machines vacuum up your money

Poker machines are unique in the gambling world. They are the only form of gambling that has been designed and crafted for the purpose of making money, and where there is absolutely no chance of influencing the outcome.

Tom Cummings, Poker machine maths, 27 May 2011

The article, linked above, on poker machines is an excellent insight into the mathematics of poker machine gaming and how, despite a guaranteed return-to-player percentage which is high, a punter that plays long enough ends up with nothing. The vital point that Tom Cummings makes is this:

It’s common knowledge that poker machines have a return-to-player percentage of anywhere between 85 per cent and 90 per cent, depending on where you live and what kind of establishment you’re playing in. For the sake of the story, let’s assume that Gladys’ poker machine was set to 90 per cent. That means that over a long period of time, the machine will return 90 per cent of money gambled to players, and keep 10 per cent as profit.

But wait, I hear you cry. Gladys didn’t lose 10 per cent; she lost it all! Well, no… not according to poker machine mathematics. The rule is that the poker machine has to return 90 per cent of money gambled… not money inserted. And there’s a huge difference.

The way it works, mathematically, is, I think, very interesting. Take the example used in the article of a 2-cent gaming machine, at $1 per play, with guaranteed return of 90%. When you load that dollar in the slot, sometimes you win, and sometimes you lose, but the long-term average dictates that at each “turn” the dollar is losing 10% of its value. $1 turns into 90c, turns into 81c, turns into 72.9c, … , turns into 2c at which point the game ends.

More generally, at “turn” n, an amount A1 initially fed into a slot machine with return R is worth:


or solving for n:

n=\frac{\displaystyle ln(A_{n}) - ln(A_{1})}{\displaystyle ln(R)} + 1

What does this mean? It means that it takes, on average

n = [ ln(0.02) – ln(1.00) / ln(0.90) ] + 1 = 38

iterations to turn $1.00 into 2 cents on a 90% return poker machine. More importantly, the total amount gambled isn’t $1.00, it’s actually (because you’re re-investing your winnings and following your losses until the game ends): $1.00 + 90c + 81c + 72.9c + … + 2c. Or more generally,

A_{1}\sum_{1}^{n} R^{i-1}

which you’ll remember converges to:

A_{1} \frac{\displaystyle 1-R^{n-1}}{\displaystyle 1-R}

So a single dollar coin will generate, on average (on a 2-cent, 90% return machine):

[ 1-0.9(38-1)] / [ 1-0.90 ] = $9.80 in total bets.

At $1.00 per game, at 10 games per minute, that’s about 1 minute of play for every dollar put in, or 5 hours to burn through Gladys’ $300.

Those who defend poker machines often point to the high rate of return as one of the reasons that pokies are just “good, clean fun” for most people. The reality is that every poker machine can meet this “rate of return” requirement while still leaving the gambler broke. That’s poker machine mathematics.


Further reading:

ABC Hungry Beast, The Beast File: Pokies


One Response

  1. Stanley, glad you liked the article… and quite pleased to see that my maths stands up to the more rigorous examination and explanation you’ve undertaken! 🙂

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