PDA

View Full Version : odds of losing half your bankroll

NoTalent
11-23-2004, 02:19 PM
Say you are playing a game with a house edge of .5%.

What is the probability of losing half your bankroll in x number of bets you make of y value?

Say you are playing blackjack and have a bankroll of \$500 and you make 50 bets of \$10. What is the probability you will have \$250 after those bets. What is the probability you will have \$750 after those bets? Is this the same thing as 'Risk of ruin'?

Regards,
NoTalent

fnord_too
11-23-2004, 03:01 PM
Probability of losing half your bankroll is the same as losing 2/3 of the bets. Lets call it losing 33 and winning 17 for simplicity's sake. The probability of this happening is:

C(50,33) * .495^33 * .505^17

For losing AT LEAST half, you need to sum from i = 33 to 50 for
C(50,i) * .495^i * .505^(50-1)

It seems related to RoR, but I really haven't looked at those equations.

For winning 2/3, just swap 33 and 17 in the first equation. ( C(50,33) = C(50,17), so you don't really need to swap that one.)

mannika
11-23-2004, 05:57 PM
[ QUOTE ]
Probability of losing half your bankroll is the same as losing 2/3 of the bets.

[/ QUOTE ]

If the bankroll is defined as the number of bets multiplied times the bet value, then shouldn't the probability of losing half your bankroll be the same as losing 3/4 of the bets?

If you're only losing 2/3 of the bets, then you have won 1/3, so you're essentially breaking even on all but 1/3 of your bets, which you lose.

However, if you're losing 3/4 of the bets, only 1/4 of that is covered by the ones you win, so that is the correct proportion for solving this problem.