Suppose a game where a fair coin is flipped. Tails pays out $0.5$x the bet, and heads pays out $1.75$x. With a bet of \$1, the expected value of a single game is $0.5\cdot 0.5 + 0.5\cdot 1.75 = 1.125$. So if $1000$ games of this is played, \$125 is expected to be gained.

But what if instead of betting \$1, the entire bankroll is bet? So starting with \$100, after a single round there is a $50$% chance of \$50, and a $50$% chance of \$175. Regardless of outcome, the entire amount is bet again, so if it was heads, the next round \$175 would be bet and so on. What is the expected amount after $1000$ rounds?

My intuition is that you'd end up with \$0, since it would take more than one win to make up for a single loss as $0.5\cdot 1.75 = 0.875$, and when I run a few trials programmatically it would come out to $0$ (as in, $<\epsilon$) each time, but I don't know how to prove this

  • $\begingroup$ How can you ever reach zero if you always receive some positive payout? $\endgroup$ – Math1000 Feb 26 at 18:54
  • $\begingroup$ well, not zero but a negligible amount, once you start splitting pennies $\endgroup$ – Ally Feb 26 at 19:59

Computer simulations suggest that the bankroll is expected to get vanishingly small after the 1000 rounds. This is false, as will be shown below.

Without loss of generality let the initial bankroll be 1. If the number of heads obtained is $k$, the bankroll after playing 1000 rounds is $$(1.75)^k(0.5)^{1000-k}$$ Thus the expected value of the bankroll after 1000 rounds is $$\sum_{k=0}^{1000}\frac{\binom{1000}k}{2^{1000}}(1.75)^k(0.5)^{1000-k}$$ $$=\sum_{k=0}^{1000}\binom{1000}k(0.875)^k(0.25)^{1000-k}$$ $$=(1.125)^{1000}\sum_{k=0}^{1000}\binom{1000}k(0.875/1.125)^k(0.25/1.125)^{1000-k}$$ The sum in the last expression is over the probabilities of every possible result in a binomial $n=1000,p=0.875/1.125$ random variable, so is 1. The whole thing reduces to $1.125^{1000}=1.421\cdot10^{51}$.

This is counterintuitive, and the reason for this is that the distribution of ending values has a very long tail (positive-skewed) which is not very apparent from the simulation.

  • $\begingroup$ what is this method of using the base-2 logarithm called? I was trying to find material on this to read about it but not sure what to search for $\endgroup$ – Ally Feb 26 at 20:04
  • $\begingroup$ @TannerSwett Does this please you now? $\endgroup$ – Parcly Taxel Feb 27 at 4:21
  • $\begingroup$ Looks good now, thanks for making the edits! $\endgroup$ – Tanner Swett Feb 27 at 4:23

Let $\ X_n\ = 1\ $ if the $\ n^\mathrm{th}\ $ toss is heads, $\ 0\ $ otherwise, $\ Y_n = 0.5 + 1.25\,X_n\ $, and $\ B_{n-1}\ $ the player's bankroll just before the $\ n^\mathrm{th}\ $ toss. His bankroll $\ B_n\ $ after that toss will then be $\ B_n = Y_n\,B_{n-1}\ $. After $\ 1000\ $ tosses, the bankroll will therefore be $\ B_{1000}=Y_{1000}Y_{999}\dots Y_1B_0\ $. Since $\ Y_1, Y_2, \dots, Y_{1000}\ \ $ are independent, all with mean $\ \frac{1}{2}\times 0.5 + \frac{1}{2}\times 1.75 = 1.125\ $, then $\ E\left(B_{1000}\right) = 1.125^{1000}\,B_0\approx 1.4\times 10^{51}\,B_0\ $.

Thus, far from becoming vanishingly small, the player's expected bankroll will instead grow enormously large. The key here is that the "average" amount by which the bankroll gets multiplied on each toss is the arithmetic mean of $\ 0.5\ $ and $\ 1.75\ $—which is greater than $\ 1\ $—, rather than their geometric mean of $\ \sqrt{0.5\times1.75}\approx 0.94 \ $.

Additional observation: Despite the player's huge expected bankroll after $\ 1000\ $ tosses, he or she only has a very small chance of coming out ahead. If there are $\ n\ $ heads and $\ 1000-n\ $ tails in the the $\ 1000\ $ tosses, the player's bankroll at the end will be $\ 0.5^{1000-n}\,1.75^n\,B_0\ $, and this is greater than $\ B_0\ $ if and only if $\ n\ge 554\ $. The probability that the player will come ot ahead is thus $\ \sum_{n=554}^{1000} {1000 \choose n}2^{-1000}\approx\ 3.5\times 10^{-4}\ $, or about $\ 1\ $ in $\ 2800\ $.


After 1000 rounds, we cannot say, what the expected emount is, as that is down to chance.

However, taking the worst possible outcome, every time (that is, losing), you will be left, with an extremely small number, which, however, is unequal to zero.

The amount will only me equal to zero, is if someone loses the game, an infinite amount of times

  • 2
    $\begingroup$ "we cannot say as that is down to chance"... You absolutely can figure out what the expected amount will be. Expected Value is quite rigorously defined and is a rather important and useful thing which is commonly studied for questions related to random process and random variables. $\endgroup$ – JMoravitz Feb 26 at 20:05

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