Easy Question, then the Pain Begins..

Anyone that can answer these two problems for me I will gladly name my first born after you.

The first question is relatively easy:

For a bet with a 33% chance of success (2/1) such as choosing the correct number to appear on a 3 sided die, if I was given odds of 21/10 for this bet what size bankroll would I need to ensure a 95% degree of certainty that I would not go broke.

The second question:

If the EV for a given bet is known (like in the above 2/1 vs. 21/10 example)what other variables need to be known to determine the bankroll to ensure a 95% degree of certainty that you would not go broke.

How does this formula to determine the bankroll change when there is a range of bets each with differnt EV's?


Hopefully this is enough information, if you need any clarication or questions answered, please post or PM me.

Thankyou in advance.

[AaronBrown]

You can have a 100% chance of not going broke with a one penny bankroll. All you have to do is not bet. Or you can have a 100% chance of going broke with any size bankroll, any chance of winning (less than 1) and any odds payoff; just bet the entire bankroll all and let it ride until you lose.

Whether or not you go broke depends on your betting strategy, not the payout or chance of winning.

If you assume you bet a constant amount each time and keep playing for a fixed number of bets, the calculation is only easy for a fair game (2 to 1 payout on 2 to 1 odds). Otherwise you can solve it numerically (that is you write a computer program to add up a bunch of calculated numbers) or by approximation.

If you give more details about why you are asking, we could help more.

Thankyou for the reply.

The reason for me asking this question is because I have discovered a bet where I can easily identify when it is a +EV situation and I can also calculate exactly what the EV is. From poker I know that no matter how +EV a game is, if you do not have the proper bankroll it is not profitable due to the standard deviation of the bet.

So what I am trying to work out is: when the exact EV of a situation is known how do you determine the bankroll required to invest in the bet profitably.

For clarification:

Suppose we are walking down the street and we come across a man who is playing our favourite game: the coin toss.

For a $1 bet he offers to pay us $2 if it comes up heads, he keeps our money if it comes up tails. Obviously a coin flip is a 1/1 situation but we are getting 2/1 so this is a +EV situation of $0.50 per flip.

Because this is a +EV situation and the man offering the game has a substantial amount of money, I want to make a living playing this game. As I have heard the words variance and standard deviation I know that in a +EV situation for it to be profitable in the long term I need a bankroll to withstand the downswings.

How large does my bankroll need to be?

My understanding (from Mason Malmuth's Gambling Theory And Other Topics p. 47-48 if you have it) is that the determination of the bankroll for poker is a function of the win rate and standard deviation. Perhaps the bankroll for my above bet can be calculated in the same way, but I do not have enough of a grip on the calculus or standard deviation as a concept to apply it to this example.

Are you devising a new bar bet? Or a street shill game?

/images/graemlins/confused.gif

[BruceZ]

[ QUOTE ]
Suppose we are walking down the street and we come across a man who is playing our favourite game: the coin toss.

For a $1 bet he offers to pay us $2 if it comes up heads, he keeps our money if it comes up tails. Obviously a coin flip is a 1/1 situation but we are getting 2/1 so this is a +EV situation of $0.50 per flip....How large does my bankroll need to be?


[/ QUOTE ]

Assuming that your opponent cannot go broke because he is very weathy, and you make this bet forever or until you are broke starting with a $1 bankroll, then your risk of ruin ror would satisfy:

ror = 1/2 + (1/2)*ror^3

This says that you can go broke on the first bet with probability 1/2, or you can win the first bet with probability 1/2, and then go broke sometime after that. The probability of going broke after winning the first bet is ror^3 since your bankroll will have grown to $3 after winning the first bet, so going broke at that point means losing a $1 bankroll 3 times.

I originally used Excel's goal seek function to solve this, but this particular cubic can be factored as (ror-1)*(ror^2 + ror - 1) = 0, and by the quadratic formula, the desired root of the quadratic is the golden ratio:

ror = -1/2 + sqrt(5)/2

If a bankroll B gives a risk of ruin of 5%, then B would satisfy:

[-1/2 + sqrt(5)/2]^B = 5%

B = ln(0.05)/ln[-1/2 + sqrt(5)/2] =~ 6.2, so you need $7, and that would give a risk of ruin of about 3.4%.

You might also be interested in the classic gambler's ruin formula that I derived in this post (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=0&Number=2197150&page=&vc=1). In that case, your advantage comes from the fact that the coin is rigged so that it comes up something other than 50/50, while your payoff is 1/1. Also, this derivation allows for a finite opponent bankroll, so that he can also go broke. For an infinitely wealthy opponent, let T -> infinity in the final formula to get a ror of (q/p)^B when the coin is weighted in our favor (q < p).


[ QUOTE ]
Because this is a +EV situation and the man offering the game has a substantial amount of money, I want to make a living playing this game. As I have heard the words variance and standard deviation I know that in a +EV situation for it to be profitable in the long term I need a bankroll to withstand the downswings.

[/ QUOTE ]

Since you know the odds and payoff for each bet, you do not need to use the EV and standard deviation in the above problem. If you only knew the EV and standard deviation, then you can compute the bankroll requirement and risk of ruin using the formulas I have given in this thread (http://archiveserver.twoplustwo.com/showflat.php?Cat=&Number=207100&page=&view=&sb=5&o =&fpart=all&vc=1) which are derived in this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=683150&page=0&view=ex panded&sb=5&o=14&fpart=2#Post682045683150), provided that the central limit theorem applies as discussed in the post with the derivation. Note that in the above example, the EV is $0.50, and the variance is 0.5*(-1)^2 + 0.5*(2)^2 - 0.5^2 = 2.25, so for a 5% risk of ruin the bankroll given by this method is

B = -2.25/(2*0.5)*ln(.05) = 6.7, so again you would need a $7 bankroll.

If the central limit theorem does not apply, then you must solve a more complicated equation for ror which you would set up similarly to the equation that I set up for the first question.