AboutScotto Expertise Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.).
I also have answered some questions in
Physics (mass, momentum, falling bodies),
Chemistry (charge, reactions, symbols, molecules), and
Biology.
Experience Experience in the area: I have tutored students in all areas of mathematics for over 20 years.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
Publications Maybe not a publication, but I have respond to well oveer 3000 questions on the PC.
That's around 2,000 in basic math and 1,000 in advanced math.
Education/Credentials I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few junior level courses my sophomore year.
Awards and Honors I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
Past/Present Clients My past clients have been students at OSU, students at the college in South Seattle,
referals from a company, friends and aquantenances, people from my church, and people like you.
Question There are four independant systems that generate either a Win or Loss output.The Win % on each system per month is 28%,35%,45% & 62%. What is my overall winning % each month?TH? .
Answer THe win on the total of four systems, assuming they are all independent of each other and played separately, is found by adding up the probabilities. That will give you the number of wins you get each month at four games of gambling.
That would be 0.28 + 0.35 + 0.45 + 0.62 = 1.70.
That means out of of four games, you could almost expect to win two.
If this were done for ten months, the average says that 17 of the games would have been won out of the 40 played.
I don't know what TH? means.
Is that for these odds being Tail/Head
on four different biased coins?
The chance of winning all of the probabilitities would be
the product of 0.28, 0.35, 0.45, and 0.62.
That would be 0.027342.
The chance of winning all but one would be the sum of four sets of four probabilities where three were one and the other lossed. The chance of losing each one is one minus the chance of winning.
These would be 1 - 0.28= 0.72; 1 - 0.35 = 0.65;
1 - 0.45= 0.55; and 1 - 0.62 = 0.38.
The chane of getting at only three wins would be the sum
0.016758 + 0.033418 + 0.050778 + 0.070308 = 0.171262.
The chance of getting at only 2 wins would be the sum of 6 products.
You could win 1&2, 1&3, 1&4, 2&3, 2&4, or 3&4. The final answer is
0.373262.
I put the chances in a spreadsheet and calculated the odds of winning 0, 1, 2, 3 or 4 games.
For 0 games, the chance is 0.097812.
For 1 game, the chance is 0.330322.
For 2 games, the chance is 0.373262.
For 3 games, the chance is 0.171262.
For 4 games, the chance is 0.027342.
So even though two games is the most likely, the arerage is less than 2 since the chance of winning less than two is 0.428134 and the chance of winning more than two is only 0.198604.
