In roulette, if you begin by betting one dollar on red, then double your bet if you lose, and double again...

probability/martingale Added material

Problem

In roulette, if you begin by betting one dollar on red, then double
your bet if you lose, and double again if you lose again...but return
to your initial bet of one when you finally win, you will break the
bank one dollar at a time, and the odds of ever losing all your money,
if you have a significant stockpile, are small enough to be
inconsequential.  Is this true?

Show solution

Fallacy number one. Yes, the chances of losing on *one* *specific*
*bet* are small. But keep in mind that you're only winning one dollar
on each bet. The number of runs needed to break the bank (or even win a
significant sum) is very large. And "small" * "very large" yields a
number that is large enough for discomfort.  Better to keep your day job.

Fallacy number two.  Casinos have betting limits. Typically, the
difference between the maximum and minimum bets are enough for 6 or 7
doubling runs. This is why it doesn't work. A run of 7 consecutive
blacks happens once out of 64 runs.

Assume you have a stockpile of $1023 (enough for 10 straight losses)
and the wheel includes 0 & 00, so that your odds of winning are 18/38.

It's easy to figure the probability of losing, and that half of the
expected value equation:

p = (1-(18/38))^10 = .001631
ev(l) = (1+2+4+8+...+512)*.001631 = 1023*.001631 = -1.668551

But, if the probability of NOT losing is 1-p, and the value of the win
is $1, the other half of the expected value equation would be:

1*.998369 = .998369

Thus:

ev = .998369 - 1.668551 = -.6701

This is your dollar expectation every time you begin your doubling
series, ending with either your win of $1 or your ruin. And it is about
negative 67 cents.

On each bet, your rate of return is:

r = 18/38 - 20/38 = -0.05263 dollars

If you just bet on red 10 times for a dollar each time, you'd expect to
gain 10r ~= -0.5263 dollars (i.e. lose about 53 cents).

Now of course you won't always play the whole 10 games, because you
might win and cash out early.  In fact, your chance of playing game
number k is only (20/38)^(k-1), or for k > 1, a little better than
1/2^(k-1).  But if you *do* play the game, you'll wager 2^(k-1) dollars
on it, so its contribution to the overall expectation (amount won or
lost times the probability of that happening) is just as high as the
first game's, in fact it gets higher as k increases.

Since the later (possible) spins of the wheel actually weigh in against
you a little heavier than the first, it's no surprise that your
strategy is somewhat worse than 10 straight spins at $1 each, i.e., the
expected value is

  E ~= -.6701 ~= 12.7 r < 10r

as you computed.

The strategy of doubling the amount of losing bets, ad infinitum, until
one wins, is called the martingale betting strategy and was first
proposed by a Frenchman a long time ago.  Some claim that this lead to
the name "martingale process" for certain stocastic processes but I've
also heard other stories.  This strategy does guarantee gains if you
start with an infinite bankroll but then again, it's hard to invent a
losing strategy assuming you start with an infinite amount of money.
Also, if you already have an infinite amount of money, how can you win
more?

Additional material

Consider this: suppose the house had NO ADVANTAGE. The calculations would
become:

      p(l) = 1/1024
     ev(l) = -1023/1024

      p(w) = 1023/1024
     ev(w) = 1023/1024

        ev = 0

In other words, following the doubling strategy (often known as the
'Martingale') confers on you no advantage whatsoever.

-- Jonathan Haas <positron@quip.eecs.umich.edu>

Precisely, the strategy works like this.
Bet $1  if win make $1 start betting $1 again
    if lose down $1,  bet $2,   if win up $1 start betting $1 again
    if lose down $3,  bet $4,   if win up $1 start betting $1 again
    if lose down $7,  bet $8,   if win up $1 start betting $1 again
    if lose down $15, bet $16,  if win up $1 start betting $1 again
    if lose down $31, bet $32,  if win up $1 start betting $1 again
    if lose down $63, bet $64,  if win up $1 start betting $1 again
.
.
.

My puzzle is this.  Assume you are playing a game with even odds, 50%
chance of winning, 50% chance of losing and you can make even money
bets.  You start with 2^N - 1 dollars where N is a natural number.
You play using the martingale strategy making $1 initial bets until
you either double your money or lose N times in a row, which blows
your initial bankroll.  Realize that if this happens, you are left
with your prior winnings.  What is the probability of doubling your
money as N -> inf?

spoiler follows

Prob of losing N times in a row = 1 / 2^N
prob of winning $1 = 1 - Prob of losing N times in a row = 1 -  (1 / 2^N)
prob of winning $1, 2^N - 1 times in a row = (1 -  (1 / 2^N)) ^(2^N-1)

X = 2^N - 1 [Editor: was "2^N+1"]

prob of winning $1, 2^N - 1 times in a row = (1 -  1/(X+1) ) ^X

                                           =  (X / (X+1) ) ^X

                                           = 1 / ( (X+1) / X) ^ X

                                           = 1 / ( 1 + 1/X ) ^ X
lim N-> oo = lim X-> oo
                                           = 1 / e

                                           ~ 0.368

this can also be done by taking logs and using l'Hopital's rule

-- Philip Weissman <pweissma@lehman.com>

In article 1603962324170001@laurel24.onramp.net, tomt@onramp.net (Tom Turcich) writes:
> A semi popular system to beat the roulette table is called the
> cancellation system.
>
> Basically, to play this system, you write the numbers 1 through 10 on a
> piece of paper. Make an even money (18/38 expectation of success) bet in
> an amount equal to the first and last number on your list. Your first bet
> is 10 + 1, or 11 units. Your unit is $1 and the house limit (highest bet)
> is $500
>
> If successful (win), cross out the first and last number and bet again
> using your new first and last number as your bet size.
>
> If failure (loss), append the amount lost as your new last (highest) number.
>
> Continue playing until you have crossed out all the numbers and won $55.
>
> 3) I have been absolutely stumped by the math - I "know" that there will
> eventually be a big loss (caused by being unable to make the next bet
> because of the house limit) that wipes out all the wins and more but I
> can't seem to figure out how to calculate it.
>
> The puzzle I present is as follows:
>
> A) how many times do you succeed, on average, for each failure?
>
> B) how much will the average failure cost?

I agree that this is (in some sense of the word) equivalent to the
doubling strategy, but that doesn't answer Tom's question.  A
significant difference is that in doubling, you hit the house limit and
lose all your money at once.  In this one, you lose all your money
gradually.  To calculate the odds, just consider what happens if you
lose every time:

   Round 1: Lose $11, add 11 to the list
   Round 2: Lose $12, add 12 to the list
   Round 3: Lose $13, add 13 to the list
   Round 4: Lose $14, add 14 to the list

Continue until you lose all of your money or hit the house limit.
To hit the house limit, lose 11+12+13+14+...+500 which is

500*(500+1)/2 + 500*10 = 125250 + 5000 = $130250

[ Editor: 11+12+...+500 = 500*501/2 - (1+2+...+10) = 125250 - 55 =
$125195. ]

So if you walk into the casino with less than that (which I assume your
friend does), you lose all of your money without ever reaching the
limit.  That means that the answers to A) and B) depend on how much
money you start with; the more you start with (on the average), the
smaller your chance of losing, and the bigger your loss is when you
lose.

Without proof, I claim that the actual numeric answers for A) and B)
are pretty close to

   (chance of winning * amount won) / (chance of losing * amount lost) = 2*(18/38)

-- Bob Morris <morris@phx.sectel.geg.mot.com>