Gambler's ruin - 1

I have a friend who walked into a casino with one dollar and walked out with (the cash equivalent of) a BMW. Now, I can tell you think I'm making this up, but I am not. Think about it. There are a bunch of casinos/lotteries out there. Some people (very few, I know) win them. Those people have some friends (or at least they did before they won that jackpot, whether they do later or not is anyone's guess), and those friends are people who can truthfully say "I had a friend who won the lottery". I just happen to be one of them.

Still, most of us, if we gamble at casinos at all, do not get that lucky. In fact, one bottom line about gambling at casinos is that the house always wins. Really. In many jurisdictions, if the casino management perceive that a player is winning too much, they have the right to refuse to allow that player to play any more, and you can be sure they exercise that right.

For the sake of mathematical discussion, let us posit that somewhere out there there is a fair casino called "Martin Gale's Betting Room" (<snicker, snicker> see footnote if you don't get the reference). In other words when you go there, and you get to place fair bets. What does this mean? It means that when you pay $1 to place a bet, your expected return on the bet is $1, or your expected profit is $0. (Which is also your expected loss.)

There are, of course, any number of ways to offer a fair bet, but Martin decides to keep it simple, and only offers one kind of bet. The game is as follows:

  • You pay Martin a dollar.
  • Martin tosses a fair coin
  • If the coin comes up heads Martin pays you two dollars. If it comes up tails, you get nothing.

Is this a fair game? Yes it is. The fair coin has probability 1/2 of coming up heads, and 1/2 of coming up tails, so your expected payout is 2×1/2+0×1/2=12\times 1/2 +0 \times 1/2 = 1.

Now we get to the interesting bit. Of course you do not have to play just once. Martin never sleeps, so you can gamble all night, as long as you still have the money.

You walk into Martin Gale's Betting Room with $2.

You can play the following game any number of times (if you have what it costs)

  • You pay Martin a dollar.
  • Martin tosses a fair coin
  • If the coin comes up heads Martin pays you two dollars. If it comes up tails, you get nothing.

You decide that you will play for a while, but if you ever get up to $5 you will cash out and go home. And, naturally, if you are left with no money then also you will go home.

What is the probability that you will go home "ruined", i.e with nothing?

There are the number of ways to tackle this, but the approach I've taken in the solution I posted is to set up a system of linear equations. The same approach can be taken with the next problem, that tackles how long it takes until you are ruined.

You walk into Martin Gale's Betting Room with $2.

You can play the following game any number of times (if you have what it costs)

  • You pay Martin a dollar.
  • Martin tosses a fair coin
  • If the coin comes up heads Martin pays you two dollars. If it comes up tails, you get nothing.

You decide that you will play for a while, but if you ever get up to $5 you will cash out and go home. And, naturally, if you are left with no money then also you will go home.

What is the expected number of games you will play before you stop?

$22 $28 $25 $24 $23 $20 $26

You walk into Martin Gale's Betting Room with $20.

You can play the following game any number of times (if you have what it costs)

  • You pay Martin a dollar.
  • Martin tosses a fair coin
  • If the coin comes up heads Martin pays you two dollars. If it comes up tails, you get nothing.

You play for a while, and then you are pleasantly surprised to find that you now have $24. You glance at the clock and see it is almost midnight. You decide you will play exactly 4 more games and then go home. How much money do you expect to go home with?

To be continued...


Footnotes:

  1. A stochastic process X0,X1,,Xt,X_0, X_1, \dots, X_t, \dots for which all steps are unbiased, that is, E(XtXt1X0,X1,,Xt1)=0\mathbf{E}(X_t - X_{t-1} \mid X_0, X_1, \dots, X_{t-1}) = 0 , is called a martingale. Obviously, the owner of the casino has a very suitable name.

  2. In case you are wondering how Martin's business makes any profit, it sells fancy mixed drinks (and ad space on the cocktail napkins).

Note by Varsha Dani
7 months, 2 weeks ago

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