Is Ronda's mum really a statistician?

That's assuming that the chance of losing remains a constant no matter what you do. Recklessly trying to finish a fight quickly is certainly going to raise the odds of getting beat.
 
Rhonda's mom's logic is sound.

It is impressive how you can obfuscate a what is otherwise a simple and straight forward logical concept though.
 
her ph.d is not in stats/math

google for the win.

but i think she has experience with it


either way from the looks of it she's a great judoka and teacher

lol@thread tho
 
casio298 said:
Hello, fellow CFA charters or actuaries. Ronda's mom's comment in UFC 190 Vlog Epi 3 really bugs me.

She said.

"Even if someone has a one in a million chance per second of beating you, if you let it go 60 seconds where you could have done it in ten, you give them six times the chance, that's just stupid"​

Now there are three possibility in each second, Ronda wins, Ronda loses or proceed to the next second. Therefore this is basically a Trinomial Option Pricing Model, right?

She implied that the probability of an opponent winning in a 60 second is six times greater than an opponent winning in 10 seconds, that must mean the probability is additive, which is erroneous.

A simpler example, say Bethe has 1 in 1000 chance of winning per second, if it's additive, than that means after 1000 seconds, the Bethe has a 100% chance of winning. Which is not true.

If you do a two step trinomial model starting from the 10th second of round 1, which we denote as t_0, and for the sake of easier of calculation, let Ronda chance of winning per second be 3/10, Bethe winning = 1/10 and proceeding to the next second be 6/10.

probability at t_1, or 1 second after t_0, is 3/10 for Ronda (R), 6/10 going into next second (N) and 1/10 for Bethe (B). Or mathematically, P(R)_1 = 0.3, P(N)_1 = 0.7 and P(B)_1 = 0.1

At t_2, which means they have to go through the previous second, which had a 0.6 probability, therefore P(R)_2 = 0.3 * 0.6 = 0.18, P(N)_2 = 0.6^2 = 0.36 and P(B)_2 = 0.1 * 0.6 = 0.06

etc

At each second, the probability of Ronda beating Bethe is always the same,

At the 60th second, or t_50, Bethe's chance is 0.1 * 0.6^49 = 1.34714 E-12, the cumulative chance from t_1 to t_50 and to the of the match is 0.25 via a KO, TKO or submission for Bethe. 0.25 is not 50 times bigger than 0.1.

I mean she's got a PhD and an expert in SAS apparently, how can she make up statistically incorrect statement?
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I think Ronda's mom sees the number of win-lose rounds as a binomial distribution. I think it is not completely correct as they are not engaging each second, but you can take (1-second in which she loses) vs (second in which she loses). Fairly simple calculation. Although you can better use that binomial distribution converges to a poisson distribution when time gets large and as p(second in which she loses) is very small, makes it much more easy to calculate.
 
TS, nice explanation. However it only works if you don't take into account the wave function factoring. but this goes of course outside the real of pure statistics.

in the heart of if, it appears like ronda's mom chose a static form of Ronda (one not determined to stop the fight as fast as possible) and an offensive form of her opponent (that does her best to end the fight as fast as possible) to prove her point. and this is where the wave appears, since static ronda at time A and offensive opponent at time A is not a repeatable sequence, in the real world.

other than that, good post.
 
Ronda's mum's thread gets more clicks than mighty mouse threads
 
I don't think you know what you're talking about.

P(! beat Ronda in 1s) = 0.999999 = 1 - 0.000001
P(! beat Ronda in 6s) = 0.999999^6

Since !win = lose,

P(Opponent beating Ronda in 6s) = 1 - P(! beat Ronda in 6s) = 1 - 0.999999^6 = 0.00000599998


Wayweary said:
On the actual problem, what Ronda's mom is talking about is Ronda changing her own probability for finishing the fight by holding back to make a point like she claims she's going to do (thus reducing what they consider a near certain finish to a non-finish) . If she does that, her own chances of winning shrink and Beche's go up for the longer she maintains that posture. The principle is obviously sound, but the way it was related was not technically correct and it's somewhat splitting hairs.

If a fighter has a 1/1 million chance to win a fight in 10 seconds, and that chance remains the same for as long as the fight progresses, with one fighter removing their own probability of winning, the math goes something like...


Chance to win in 1st period: 0.000001 = 1 in 1 million
2nd 10 seconds: 0.000000999999 = 1 in 1 million, times chance to not win in 1st period
3rd: 0.000000999998= 1 in a million, times chance to not win in 1st or 2nd periods
4th: 0.000000999997= 1 in a million, times chance to not win in 1st, 2nd or 3rd periods
5th: 0.000000999996=1 in a million, times chance to not win in 1st through 4th periods
6th: 0.000000999995 = 1 in a million, times chance to not win in the 1st through 5th periods
Summing all of the above probabilities: 0.000005999985, which is another way of saying 5.999985 out of a million (very close to 6 times the chance to win in the first 10 seconds).


Of course, those of us acquainted with reality know that probabilities are not applicable to the scenario in such a way. It is certainly the case that the chance to finish in the first ten seconds is astronomically low compared to the chance to win in the second ten, or the third through some later period of the fight. As the fight draws out, the probabilities to finish the fight fluctuate in a punctuated, not gradual manner, but go down as both fighters tire, where as if one tires more, they can go up. This could be the case with Ronda, which is why I think that the chances of Beche winning go up at a much more than steady rate (exponentially) as the fight progresses. We are talking about a dynamic system and such a simplistic application of statistics (or indeed, any) is not appropriate in modelling it. The only reason to bring up probabilities is to illustrate a point, really.
Click to expand...
 
s1fys said:
I don't think you know what you're talking about.

P(! beat Ronda in 1s) = 0.999999 = 1 - 0.000001
P(! beat Ronda in 6s) = 0.999999^6

Since !win = lose,

P(Opponent beating Ronda in 6s) = 1 - P(! beat Ronda in 6s) = 1 - 0.999999^6 = 0.00000599998
Click to expand...

From where I'm standing, it looks like you gave the same answer except that you truncated the value at the end, genius. Looks like neither of us knows what we are talking about.
 
I'm proud of all of you; I'm putting this thread on the refrigerator.
 
1235644960_50_cent_does_not_approve.gif


*Drives off in a supercar*
 
HockeyBjj said:
I went to college too ts

I'm not an overcompensating dick about it
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Great comment. I also went to college and even studied math and statistics; posting on sherdog about trinomial pricing is just not legit.
 
Hiding from Cyborg 100% of the time = 0% chance of losing to Cyborg

Great Mom indeed! :icon_chee
 
DrRodentia said:
I think I've see this or something similar said about every single degree at this point.

People seem to really love to put down other people's degrees.
Click to expand...

Med school? That's where the dopes who couldn't become astronauts go!
 
Real science : Bethe has no chance , she will get butchered

Source : i'm a statistician ;)
 
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