# Expected Value question / Roulette scenario

• Jul 20th 2009, 12:36 PM
prapoport
Expected Value question / Roulette scenario
Hello, I am trying to explain to someone mathematically why this roulette strategy is a poor idea.

It has been a while since I have calculated more than the simplest probability problems, but I was thinking that demonstrating that this has negative expected value is a way to show that it's a bad bet.

Here's the 'system': You bet \$100 on a spin in which there is 12/37 probability of winning and a 2:1 payout. If you lose, you double your initial \$100 bet and wager \$200 on the following spin... etc. Any time you win, you begin the series again with the initial \$100 unit. I'm assuming a max of 12 spins in the series since you are bound by table limits of some sort. Since it is a 2:1 payout, you can win more than your initial \$100 betting unit if you are further along in the series of spins.

I think the expected value of spin #1 is:
(12/37) * 200 + (25/37) * -100 = -2.703

I'm trying to figure out what the EV is for the whole game, but I'm having some difficulty. I did a forum search and there was some reference to situations like this, but nothing that fit the bill.

It's my first post, so I hope it's not poor form to start with a question, but any help is greatly appreciated. Thanks!
• Aug 12th 2009, 12:28 AM
garymarkhov
Quote:

Originally Posted by prapoport
Hello, I am trying to explain to someone mathematically why this roulette strategy is a poor idea.

It has been a while since I have calculated more than the simplest probability problems, but I was thinking that demonstrating that this has negative expected value is a way to show that it's a bad bet.

Here's the 'system': You bet \$100 on a spin in which there is 12/37 probability of winning and a 2:1 payout. If you lose, you double your initial \$100 bet and wager \$200 on the following spin... etc. Any time you win, you begin the series again with the initial \$100 unit. I'm assuming a max of 12 spins in the series since you are bound by table limits of some sort. Since it is a 2:1 payout, you can win more than your initial \$100 betting unit if you are further along in the series of spins.

I think the expected value of spin #1 is:
(12/37) * 200 + (25/37) * -100 = -2.703

I'm trying to figure out what the EV is for the whole game, but I'm having some difficulty. I did a forum search and there was some reference to situations like this, but nothing that fit the bill.

It's my first post, so I hope it's not poor form to start with a question, but any help is greatly appreciated. Thanks!

As far as I can tell, your expected value calculation is correct. The expected value is indeed negative, and playing more times and with higher stakes as you go along only makes the EV worse.

However, if practical considerations like time and money were of no concern, it's possible to cut down the casino's odds enough to make them feel uncomfortable. As long as you can continue to double down, you are extremely likely to eventually win (although it's possible that you could play once per second for 10,000 years and never win!). In practice, the problem is that you will either run into a max-bet limit set by the casino OR you will run out of money.

I once engaged in this strategy (using play money at a fun casino) by always betting on black, so that my chances of winning were 18/37 per spin. I started with enough money to continue doubling down even if I lost 5 times in a row, which had a probability of occurring of (19/37)^5 = 3.57%. I played all night and racked up quite a lot of cash, enough to get me to being able to sustain 6 losses in a row, which had a probability of occurring = 1.83%. Eventually, I lost it all because the improbable happened - I lost 7 times in a row, which had a probability = 0.94%!

By the way, if your base bet amount is b, the amount you would need to make one more double down bet after n consecutive losses is equal to 2^(n)*b. So if you want to be able to survive 1 loss and your base amount is \$100, you need to have at least 2^1*100 = \$200. if you want to be able to survive 7 losses in a row, you need to have a bankroll of 2^7*100 = \$12,800. If you want to survive 15 losses in a row, you need 2^15*100 = \$3,276,800.
• Aug 12th 2009, 04:00 AM
bruxism
oh dear, the dreaded martingale strategy.

Everything you need to know about it is right here.

http://en.wikipedia.org/wiki/Martingale_(betting_system)

The short answer is....don't do it. Betting on black/red, the odds of having a losing streak of 7 in 250 spins is

= 1 - (the odds of having no losing streak of 7 in 250 spins)

= 1- ((1 - (20/38)^7)^250)
= 94% chance of losing seven spins in a row in only 250 spins. Almost a certainty!

If you're initial bet is \$5(which is the minimum bet on red/black at my casino, at some point in 250 spins you have a 94% chance of being forced to make the following bets.

\$5, \$10, \$20, \$40, \$80, \$160, \$320,

After making these 7 bets and losing each, you must bet \$640 in order to recoup all your losses and to attempt to make a grand profit of \$5. How much would you be willing to put on the table? What if you lose 10 spins in a row? 15? 20?

The interesting thing, and the thing that trips up a lot of gamblers, is the gamblers fallacy. If red has come up 20 times in a row, what are the odds that on the 21st it comes up black?

Same as always of course, 18/38, or a little under 50%.
• Aug 12th 2009, 04:23 AM
garymarkhov
Yeah, I was being a bit tongue in cheek when I said the improbable happened when I lost. If you play long enough, as I did, a 0.94% chance of something happening can become very likely indeed!