Can someone tell me if this is true.

Someone told me that well basically if you have a 60% chance of success and your opponent has a 40% chance of success and you have one instance of probability, your odds of winning are 60%.

Now if you have 60% chance of success, your opponent has 40% chance of success, and the goal is to be the first to win 20 instances, then the odds of success are actually higher.

Is this true? He was using this to explain why more hp in a game makes it more tactical where as less xp makes it more chance based.

 

true
The odds take the probability of an event occurring and divide it by the probability of the event not occurring.
in a game
In general, the less likely the event is to occur, the higher the payoff when it happens.
If you win a bet on an event that has a high probability, you’re unlikely to win much money.
People are tempted to make bets where the return is high, even though the chances of them winning is negligible.

 

Ok so how do I calculate how the probability changes as I add or remove instances?

 

The odds of being the first to win N games is a good bit more complicated.

Let P = the probability of winning any 1 game.

The odds of being the first to win 1 game is just P.

The odds of being the first to win 2 games requires the odds of all of the ways that can happen. There are ways: WW, WLW, and LWW. So the probability is P^2 + 2xP^2x(1-P) = 65%

There are 10 ways to be the first to win 3 games: WWW, LWWW, WLWW, WWLW, LLWWW, LWLWW, LWWLW, WLLWW, WLWLW, & WWLLW. So the odds are: P^3 + 3*P^3*(1-P) + 6*P^3*(1-P)^2.

Here it is in a little spreadsheet:



For higher wins, you will need a series of combinations. It gets messy. But, as you can see, the odds are increasing with more wins.

 

Use the “not” trick. If probability of X is 0.35, then the probability of “not X” is 0.65 (35% and 65% respectively).

“Not X N times” is X ^ N. “At least one X in N trials is 1 - X ^ N.

The chance of X in 1 trial is 35%, in 2 trials is 1 - (0.65 ^ 2), in 3 trials is 1 - (0.65 ^ 3), etc.

if you calculate probability in Excel:

For data sets that include total instances that you have not yet converted into probability, include an additional column between the other two labeled "Instances." Step five details how to convert these instances into probabilities. For data sets where your category uses words as labels instead of numbers, add a column to the left of the category row labeled "Sort Order."

If your data set includes probabilities for each label, enter each probability into the "Probability" column. If you have not yet converted your data into probabilities, instead enter the instances into the "Instances" column. Find the total number of instances recorded by adding them. For large data sets, Excel provides a formula to quickly calculate a sum of a set of numbers. Enter the following formula into the cell underneath the last cell containing instance data in your chart:

=SUM( [top cell containing instance data] : [bottom cell containing instance data] )

The SUM function returns the total value of all cells in between the two cells separated by a colon in the formula and updates automatically if you change the value of a cell within the range.

Convert the instance data of the top row into a probability by entering the following formula in the top cell underneath the "Probability" label:

=[cell containing instance data] / [cell containing SUM function]

Repeat this for all cells in the "Probability" column to convert them

 

The only way I know of to do this is with a sum of N terms, which becomes unwieldy for N>5. My solution is to write an Excel UDF (user-defined function). Here's the code:

Code:

Function Odds(P As Double, N As Integer) As Double

Dim i As Integer
Dim Q As Double
Q = 1 - P

Odds = 0
For i = 0 To N - 1
  Odds = Odds + Application.WorksheetFunction.Combin(i + N - 1, i) * Q ^ i
Next i

Odds = P ^ N * Odds

End Function

And here are the results for P=.6 (60%):



And these are the results for P=.51 (51%):



Cheers