Beware: the following involves math and FargoRate, which some of you might like, and others not so much. If you don’t like math, you may choose to move on. For those inclined to do so, please check my work. It is always a good idea.
Because FargoRate is a function of player odds (instead of probabilities) , there is a fascinating and simple formula for calculating player probabilities of winning “M-ahead” sets. Of course, we must assume infinitely (infinitesimally?) precise and accurate FR, that you and your opponent’s probabilities of winning/losing are constant over games and unaffected by psychology and past history, etc. For the math below, it's important that you and your opponent have different FRs.
Scenario A: each situation below has the same probability of 0.20
A1. You beating someone 200 FR points ahead of you, in a single game.
A2. You beating someone 100 FR ahead of you, in a match to 2-ahead
A3. You beating someone 40 FR points ahead of you, in a match to 5-ahead
A4. You beating someone 20 FR points ahead of you, in a 10-ahead match.
A5: You beating someone 200/M FR points ahead of you, in a M-ahead match.
General rule: If you play someone with a different FR, the probability you win a “M-ahead” match depends only on the value you get by multiplying M and the FR difference between you and your opponent, nothing else.
Scenario B: Imagine your opponent is stronger and brings M dollars to a match to play you $1 a game until one of you busts. Suppose you have unlimited wealth and refuse to quit. General rule: If you're the weaker player and your opponent is has R rating points above you and brings $M, the probability you’ll eventually bust him is 2^(-M*R/100). If you’re the stronger one, the probability is 1.
If you're a FR 630 with infinite wealth and refusal to quit, and SVB (currently FR 830) brings $1, your probability of busting him at $1 per game, eventually, is 25%. If SVB brings $2, this likelihood drops to 6.25% If you're a FR 730, you respective probabilities for both cases are 50% and 25%.
More Examples:
B1. If your opponent is 200 FR ahead of you,
If he brings $1, you’ll eventually bust him with probability 1/4
If he brings $2, you’ll eventually bust him with probability 1/16
If he brings $3, you’ll eventually bust him with probability 1/64
If he brings $M you’ll eventually bust with with probability 2^(-2M)
B2. If your opponent is 50 FR ahead of you,
If he brings $1, you’ll eventually bust him with probability 1/sqrt(2)
If he brings $2, you’ll eventually bust him with probability 1/2
If he brings $3, you’ll eventually bust him with probability 1/(2sqrt(2))
If he brings $M, you’ll eventually bust him with probability 2^{-0.5M}
Regarding situation A: for both of you playing a set to M-ahead, your probability to be victorious is
(1) probability you win the M-ahead set = 1 / ( 1 + 2^(DM) )
where D = ( opponent FR - your FR ) / 100 and M is the number of games ahead required for both of you. This is derived from the general formula
(2) general formula : probability you win = ( 1 - 2^(DN) ) / ( 1 - 2^D(N+M) )
which represents the probability of you getting ahead of your opponent by M games before he’s ahead of you by N games. Don’t take my word for it. The formula is obtained by combining Christiaan Huygens’ (1657? ) solution to the gambler ruin problem (see section 3.2 of en.wikipedia.org/wiki/Gambler's_ruin ) along with the FR definition
2^D = your odds of losing a single game
= probability of losing a single game divided by probability of winning a single game
you can get (1) from (2) by letting M=N and simplifying. Also, if you bring N dollars to a match and your opponent brings M dollars, and you play each other $1 a game until bust, then the formula (2) give the probability you bust your opponent. So assuming you have infinite wealth, analyzing (2) for N astronomically large (N tending to infinity) shows
Probability your opponent busts = 1 if D < 0 .... (case where you have the higher FR)
Probability your opponent busts = 2^(-DM) if D > 0 .... (case where you have the lower FR)
The second probability explains the statements made in B. Neither formula (1) nor (2) can be used when you and your opponent have the same FR. (zero divided by zero issue in the general formula)
Because FargoRate is a function of player odds (instead of probabilities) , there is a fascinating and simple formula for calculating player probabilities of winning “M-ahead” sets. Of course, we must assume infinitely (infinitesimally?) precise and accurate FR, that you and your opponent’s probabilities of winning/losing are constant over games and unaffected by psychology and past history, etc. For the math below, it's important that you and your opponent have different FRs.
Scenario A: each situation below has the same probability of 0.20
A1. You beating someone 200 FR points ahead of you, in a single game.
A2. You beating someone 100 FR ahead of you, in a match to 2-ahead
A3. You beating someone 40 FR points ahead of you, in a match to 5-ahead
A4. You beating someone 20 FR points ahead of you, in a 10-ahead match.
A5: You beating someone 200/M FR points ahead of you, in a M-ahead match.
General rule: If you play someone with a different FR, the probability you win a “M-ahead” match depends only on the value you get by multiplying M and the FR difference between you and your opponent, nothing else.
Scenario B: Imagine your opponent is stronger and brings M dollars to a match to play you $1 a game until one of you busts. Suppose you have unlimited wealth and refuse to quit. General rule: If you're the weaker player and your opponent is has R rating points above you and brings $M, the probability you’ll eventually bust him is 2^(-M*R/100). If you’re the stronger one, the probability is 1.
If you're a FR 630 with infinite wealth and refusal to quit, and SVB (currently FR 830) brings $1, your probability of busting him at $1 per game, eventually, is 25%. If SVB brings $2, this likelihood drops to 6.25% If you're a FR 730, you respective probabilities for both cases are 50% and 25%.
More Examples:
B1. If your opponent is 200 FR ahead of you,
If he brings $1, you’ll eventually bust him with probability 1/4
If he brings $2, you’ll eventually bust him with probability 1/16
If he brings $3, you’ll eventually bust him with probability 1/64
If he brings $M you’ll eventually bust with with probability 2^(-2M)
B2. If your opponent is 50 FR ahead of you,
If he brings $1, you’ll eventually bust him with probability 1/sqrt(2)
If he brings $2, you’ll eventually bust him with probability 1/2
If he brings $3, you’ll eventually bust him with probability 1/(2sqrt(2))
If he brings $M, you’ll eventually bust him with probability 2^{-0.5M}
Regarding situation A: for both of you playing a set to M-ahead, your probability to be victorious is
(1) probability you win the M-ahead set = 1 / ( 1 + 2^(DM) )
where D = ( opponent FR - your FR ) / 100 and M is the number of games ahead required for both of you. This is derived from the general formula
(2) general formula : probability you win = ( 1 - 2^(DN) ) / ( 1 - 2^D(N+M) )
which represents the probability of you getting ahead of your opponent by M games before he’s ahead of you by N games. Don’t take my word for it. The formula is obtained by combining Christiaan Huygens’ (1657? ) solution to the gambler ruin problem (see section 3.2 of en.wikipedia.org/wiki/Gambler's_ruin ) along with the FR definition
2^D = your odds of losing a single game
= probability of losing a single game divided by probability of winning a single game
you can get (1) from (2) by letting M=N and simplifying. Also, if you bring N dollars to a match and your opponent brings M dollars, and you play each other $1 a game until bust, then the formula (2) give the probability you bust your opponent. So assuming you have infinite wealth, analyzing (2) for N astronomically large (N tending to infinity) shows
Probability your opponent busts = 1 if D < 0 .... (case where you have the higher FR)
Probability your opponent busts = 2^(-DM) if D > 0 .... (case where you have the lower FR)
The second probability explains the statements made in B. Neither formula (1) nor (2) can be used when you and your opponent have the same FR. (zero divided by zero issue in the general formula)
