All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
Sins of the Gambler (Posted on 2004-08-23) Difficulty: 3 of 5
Can you think of a method of playing standard casino roulette that gives the player a better than 90% chance of winning each time he plays a game?

A game is considered to be at least 10 bets on consecutive spins of the roulette wheel. A game is considered won if the gambler stops with more money than he started. The player starts with $1100 and table limits are min: $1 max: $1000).

See The Solution Submitted by ThoughtProvoker    
Rating: 2.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: Classic martingale | Comment 3 of 6 |
(In reply to Classic martingale by Old Original Oskar!)

(19/37)^10 is 0.13%, not 1.3%.

Considering wheels with 00 in addition to 0, 20/38 is the appropriate chance of losing an individual bet, leading to 0.16% probability of ultimately failing in the martingale.

In that small chance of losing the martingale, the player will have lost $1023, for the chance of winning $1. He will be left with $77.

The martingale need have gone only for 4 bets: $1, $2, $4 and $8 (or at the other extreme 73, 146, 292 and 584, leaving the player $1095 in the hole if he failed to gain $73). It's 7.67% likely that he would lose overall, so his chance of winning $1 (or $73) is over 90%.


  Posted by Charlie on 2004-08-23 10:20:16
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information