[Ganchrow]

Originally Posted by ohdear

1. Blindly betting 1,000 -110 spreads
    
2. Blindly betting 1,000 +140 ML's on a (+140/-160 line)
    
3. Blindly betting 1,000 -160 ML's on a (+140/-160 line)
    
4. Blindly betting 1,000 +750 ML's on a (+750/-1000 line)
    
5. Blindly betting 1,000 -1000 ML's on a (+750/-1000 line)

If I start with a $10,000 bankroll in all independent cases, and bet $10 on every game, what is my expected loss on all of these? Obviously I will expect to lose money in ALL cases, but betting which one would kill me the most?

Let's make this question a bit more interesting.

Let's say that the player were instead to risk $500 in the each of the 5 scenarios above (A, B, C, D, & E). What would be his (approximate) expected loss in each of the 3 cases? (For the sake of simplicity, further assume the vig to be spread evenly across the two pairs of related options, i.e. B/C & D/E.)

[Matt Rain]

A. 22,727
B. 20,000
C. 12,500

Final answer. I think. I risked $500 in every case and used -150 as the no-vig line for B. & C.

[Matt Rain]

nm

[Pancho sanza]

A -22727
b -15528
c -15528

[ohdear]

I think that fool would go broke in all cases

[Matt Rain]

A. (500*100*0.909)-(500*100) = -4,550

B. (400*100*1.4)-(600*100) = -4,000

C. (600*100*0.625)-(400*100) = -2,500

[Data]

Originally Posted by Matt Rain

401/599 is more like it.

[Ganchrow]

Being strapped for time I took the easy way out and just ran a few Monte Carlos.

We assume the player risks $500 per bet over 1,000 wagers or until he's exhausted his $10,000.

For simplicity's sake, I further assumed that a player with less than a full bet remaining in his positive bankroll may still place a full wager (going in to debt were he to lose). OTOH, were we to instead have assumed that a player with less than a full bet remaining were to stop playing, then this would obviously (slightly) lower expected losses.

After 10,000,000 trials for each scenario:

Scenario	Odds For	Odds Against	E(Loss)	StdDev(E(Loss))	E(Bankroll)
A	-110	-110	$9,075.74	$1.36	$924.26
B	+140	-160	$6,890.13	$2.83	$3,109.87
C	-160	+140	$8,078.61	$1.78	$1,921.39
D	+750	-1,000	$3,657.58	$6.46	$6,342.42
E	-1,000	+750	$9,198.09	$0.74	$801.91

So what we see is that due to the increased probability of going broke betting at longer odds, one on the average will expect to wager less (all elese being equal) and hence his expected loss by doing so will be lower.

[Wheell]

I've been drinking tonight, but as far as I can tell if he is RISKING $500 per bet his negative EV per bet is:

A. -22.7273
B. -20
C. -12.5
D. -64.1026
E. -6.4103

For 1000 bets you can simply multiply the EV per bet times 1000.

Now, there is another way of defining splitting the vig. In the case of B & C the middle is not in fact -150 (.6), but -149.6 (.599359).

If you want to use that more annoying but also more accurate method of splitting the vig then the EV's per bet are:

A. -22.7273
B. -19.2308
C. -13.0208
D. -56.8182
E. -7.3529

The simple conclusion is that if you hold RISK amount constant you'll go broke slower betting large favorites on the moneyline. Of course, if we were not using risk amount, but rather using what is commonly referred to as base amount we'd get different answers. In that case the new answers per bet are:

A. -25
C. -20.8333
E. -73.5294

This methodology allows for your EV amount to be larger than your bet amount. An example would be a case where you bet 1000 to win 100 dollars on a huge favorite who in reality will only win 60 times out of 100. In that case your EV would be -340 on a $100 bet.

[Ganchrow]

For those interested in the simple Perl code I used yp generate the aboev results:

#!/usr/bin/perl

# monte_broke.pl
# by ganchrow AT SBForum DOT com
# output to STDERR is trial#<TAB>avg bankroll<TAB>std dev of that average


use strict;
use warnings;

use constant	TRIALS		=>	10_000_000;
use constant	NUM_BETS	=>	1_000;
use constant	BET_SIZE	=>	0.05;	# as fraction of starting bankroll
use constant	ODDS_FOR	=>	-160;
use constant	ODDS_AGAINST	=>	+140;
use constant 	VERBOSE		=>	0;	# setting to TRUE prints all finishing bankrolls to STDOUT

my ($sum_bankroll, $sum_bankroll_sq,) = (0,0);

my $dec_odds = &us2dec(ODDS_FOR);

my $win_q = BET_SIZE*($dec_odds - 1);
my $loss_q = -1*BET_SIZE;

my $win_prob = &us2prob([ODDS_FOR,ODDS_AGAINST]);

for ( my $trial_num = 1 ; $trial_num <= TRIALS; $trial_num++) {
	my $bankroll = 1;
	for (my $bet_num = 1; ( ($bankroll > 0 ) && ( $bet_num <= NUM_BETS ) ); $bet_num++) {
		$bankroll += (rand() < $win_prob ? $win_q : $loss_q);
	}
	print STDOUT "$bankroll\n" if VERBOSE;
	$sum_bankroll += $bankroll;
	$sum_bankroll_sq += $bankroll*$bankroll;
	unless ($trial_num % 10_000) {
		select((select(STDOUT), $| = 1)[0]) if VERBOSE;    # flush standard output buffer
		my $avg = $sum_bankroll/$trial_num;
		my $stddev = ( sqrt( $sum_bankroll_sq  + $bankroll*$bankroll ) / $trial_num );
		print STDERR "$trial_num\t$avg\t$stddev\n";
		select((select(STDERR), $| = 1)[0]);    # flush standard error buffer
	}
	
}

sub us2dec {
	my $us = shift;
	return (
		$us >= 0 ? 1+$us/100 : 1-100/$us
	);
}

sub us2prob {
	# IN: takes pointer to array of us odds
	# OUT: implied zero vig prob of 1st element
	my $ar_p = shift;
	my $first = &us2dec( shift @$ar_p );
	my $sum = $first;
	$sum += &us2dec( shift @$ar_p ) while @$ar_p;
	return(1 - $first / ($sum || 1));
}

[Data]

Originally Posted by Ganchrow

So what we see is that due to the increased probability of going broke in less than 1000 bets betting at longer odds, one on the average will expect to wager less due to making a lesser number of bets (all else being equal) and hence his expected loss by doing so will be lower.

I think these clarifications are important.

Originally Posted by Wheell

This methodology allows for your EV amount to be larger than your bet amount. An example would be a case where you bet 1000 to win 100 dollars on a huge favorite who in reality will only win 60 times out of 100. In that case your EV would be -340 on a $100 bet.

When it comes to calculating EV that is actually a $1000 bet.

[Ganchrow]

Originally Posted by Data

I think these clarifications are important.

FWIW, it was in fact the OP who initially specified the player's $10,000 total bankroll.

[Data]

Originally Posted by Ganchrow

FWIW, it was in fact the OP who initially specified the player's $10,000 total bankroll.

I am afraid I did not get the point you are making. However, as specified by the OP, the player could not have gone broke in less than 1,000 bets.

[Ganchrow]

Originally Posted by Data

as specified by the OP, the player could not have gone broke in less than 1,000 bets.

Which had been my impetus for having suggested, "Let's make this question a bit more interesting."