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I'm trying to do the maths on a coin toss game after 100 games but think I am failing.

The rules are as follows.

  1. we start with 1000 coins
  2. we always bet that heads come up
  3. minimum bet is 10 coins
  4. maximum bet is 80 coins
  5. if tails comes up, we lose our bet
  6. if heads comes up, we win twice as much as we bet
  7. the coin is fair and so the probability of heads is 50%
  8. we start with a bet of 10
  9. if we lose we double our bet
  10. if we lose the maximum bet of 80 we start at a bet of 10 again
  11. if we win we start at a bet of 10 again

So in this set up, 50% of the time we would get our 10 coins back plus 10 additional coins.

When we lose we need to lose four times in a row which I've figured out will be 6.25% which will cost 150 coins.

50% (10 coins) * 50% (20 coins) * 50% (40 coins) * (80 coins) 50% = 6.25% (150 coins total)

Would I be right in thinking that this means that the chance of breaking even is 43.75%?

So after 100 games we would have the win and loss as below and end up with the balance after the wins and losses are applied.

Win: (50% win chance * 100 games) * 10 coins won = 500 coins won Loss: (6.25% loss chance * 100 games) * 150 coins = 937.5 coins lost Total: 1000 start + 500 win - 937.5 loss = 562.5 end balance

I am not confident that my maths is correct. I'd appreciate it if someone eyed it over and told me if I'm going horribly wrong.

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2 Answers 2

up vote 2 down vote accepted

No matter what your betting strategy is, your expected number of coins will always remain 1000 coins. This includes the faint possibility that you cannot even make your 100 games because you lost your 1000 coins already after 28 rounds.

Given this fact, what is the probability to have more or less than 1000 coins after very many (say, $N$) games? If $p_n$ denotes the probability to have $n$ coins after $N$ games, then the expected value is $$E(X)=1000=\sum_{n=0}^\infty n p_n$$ And the probability of having less or more than 1000 coins is $$p_{<1000}=\sum_{n=0}^{999} p_n$$ $$p_{>1000}=\sum_{n=1001}^{\infty} p_n.$$ With just a few games, we have a high probability of having slightly more than 1000 coins and a low probability of having much less than 1000 coins.

However, for large $N$, when it is possible to have gained much, much more than the original 1000 coins ans also not too unlikely that we may have gone bancrupt with $0$ coins (and canot afford to bet more than $0$ coins) the situation changes: $p_{<1000}$ is largely dominated by $p_0$, which makes no contribution at all to $E(X)$, whereas $p_n$ for large $n$ are non-zero. This is like the oppsoite situation of what the betting strategy intended: There is a high probability of a moderate loss (of 1000 coins) and a small probability of substancial gain. Especially, for sufficiently large $N$, we will have $p_{<1000}>p_{>1000}$. It would be interesting to know, at which number $N$ of rounds the switch occurs. I estimate that it will take more than your suggested 100 games, though.

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The mean of this game is 1000 coins, but I think Bushstar is asking for the probability that a given game achieves or exceeds the mean. –  Austin Mohr Nov 16 '12 at 20:41

Let's call a set of flips terminating in a reset of the bet size (to $10$ coins) a series. As you calculated, the only way to lose a series is to flip four tails in a row, which has probability $1/16$; in this case the series lasts four flips and has outcome $-150$. In every other case (probability $15/16$), the series has outcome $+10$, although its length can be anywhere from one to four flips. So the expected outcome of a series is $(1/16)(-150) + (15/16)(10) = 0$, as it must be (since each individual flip is fair). The odds of breaking even on a single series are much better than even. But when a large number of series are strung together, since the rare losses are larger than the frequent wins, the eventual break-even probability is going to approach fifty-fifty.

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