Help calculating expected earn

[whipsaw]

[ QUOTE ]
With all due respect, I don't believe the 23k figure is correct. Since you're bet size is changing with bankroll, I belive this will change you're EV. However, this is grad school level stats. You can try simulating this in MATLAB or Excel to make it a bit easier.

[/ QUOTE ]

Well golly gee, mister, thanks for coming off your throne to explain this to us idiots here. We surely a-per-see-ate it. Why don't you post the math and give us a final number rather than just trolling for threads relating to math and critiquing them.

Not trying to offend anybody here. I don't know how to do this using a formula. If anybody does, please let me know. Using MATLAB, i got $19,900 as the expected return. The only assumptions i used were that each bet paid off even money and there was a 56% chance of a good outcome.

Here's the code:
>> home
>> clear
>> TotalBets=500;

p=.56; %chance of winning
PercentBet=.05;

for trial=1:10000

Bankroll=10000;
for i =1:TotalBets;
betsize=Bankroll*PercentBet;
temp=rand(1);
if temp < .56
Bankroll=Bankroll+betsize;
else
Bankroll=Bankroll-betsize;
end
end
finalBank(trial)=Bankroll;
mean(finalBank)
end

[whipsaw]

[ QUOTE ]
Not trying to offend anybody here. I don't know how to do this using a formula. If anybody does, please let me know. Using MATLAB, i got $19,900 as the expected return. The only assumptions i used were that each bet paid off even money and there was a 56% chance of a good outcome.

Here's the code:
>> home
>> clear
>> TotalBets=500;

p=.56; %chance of winning
PercentBet=.05;

for trial=1:10000

Bankroll=10000;
for i =1:TotalBets;
betsize=Bankroll*PercentBet;
temp=rand(1);
if temp < .56
Bankroll=Bankroll+betsize;
else
Bankroll=Bankroll-betsize;
end
end
finalBank(trial)=Bankroll;
mean(finalBank)
end

[/ QUOTE ]

That looks more like high school level computer programming rather than "grad school economics" to me, but whatever. Raccoon explained the math pretty well in an earlier post, but I'll try and explain it again so that your elite brain can handle it.

First of all, you made two errors. First, it looks like you ran your recursive function 500 times, not 250 as DOTTT originally posted. Second, and more importantly, you assumed that bets pay out even money--they definitely don't.

In a usual football spread bet, the bettor is laying $110 to win $100, meaning if he wins he gets $210 back, which is approximately 1.909 (210/110) times the original bet. Therefore, for his original $500 bet, if he wins he will get back $954.50.

If his expected win percentage is 56%, then on his first individual bet his expectation is to receive $534.52 (.56*$954.50), or a profit of $34.52, which is 6.904% of the original bet.

Recursively adding the expected return of 6.904% and then adding it to the bankroll for the next bet (which gets resized to stay at 5% of the bankroll) over 250 bets yields a total bankroll of $23667.41.

Of course, as raccoon noted, there are downfalls to this calculation, because with each bet you really don't get your expected return, but instead win or lose.

First of all, I said the formula was grad school stats. Simulating this type of thing isn't so hard, but it is very effective. Second, the 2 things you just mentioned should only lower the EV. Third, getting the EV on each game is probably not a solid assumption. The changes to the simulation will be made soon and results posted.

[jmillerdls]

Well, you seem confident that someone with a 56% rate will not go broke betting 5%. I really hope this guy keeps us updated because I think there an extremely large chance he will go broke betting this much. 56% is not such a large advantage that you will not have pretty large swings in both directions. Hope I am wrong, but this seems like suicide.

[bradha]

Actually, I think the math is fairly simple - it reminds me of a math problem I had back in college. Assuming that you always bet 5% of your total bankroll and that you are betting at the standard bet 110 to win 100 rake, then: after each win new BR = BR * (1.04545) and
after each loss new BR = BR * (0.95).
After a series consisting of M wins and N losses, you have
new BR = BR * (1.04545)^M * (0.95)^N . This is true regardless of the order of the wins and losses. If over 250 trials you average exactly 56% winners, then you have M = 140 and N = 110. This is about 1.788 times your original bankroll, making $7,888 in the first year.

If you get a better rate, e.g. betting $107 to win $100, then you get new BR = BR * (1.04672895)^M * (0.95)^N, which is about 2.12 BR for M = 140 and N = 110, making $11,202 in first year.

Note: I have a lot more confidence in these numbers than I have that your system will actually yield 56% winners...

Brad