Scott Steiner's WWE promo

Question

In Scott Steiner's WWE promo video, Scott mentions that in a three way match, everyone has a 33⅓ chance of winning, which is fair if you consider all fighters to be equal. But then, he proceeds to say that Kurt Angle cannot beat him so his chances of winning increase to 66⅔ %. How does that even work?

For example, in a three way situation, if I know in advance that one of the guys for sure will beat me, in a wrestling match how does the probability of my winning change?

I'm basically asking this, how does the probability of an event change when it makes it transition from a fair outcome to a biased/unfair outcome.

Example question that I framed: One coin among three simultaneously tossed coins will always show heads. What is the probability of getting two heads?

Answer 1

Let's model this "wrestling match" situation (I find the question quite amusing so I will try to give it justice; besides I think OP has an interest in math so might as well encourage it!)

Let's say we have wrestlers A, B, and C, and the match goes as this:

-You pick two wrestlers uniformly at random -Then with some probability one wins over the other -Then the remaining ones fight

Let's say P(A wins over B) = 0, P(B wins over A) = 1, and all other win probabilities are 1/2.

So, what is the probability that A (i.e. the guy that is for sure going to lose to B) wins? Well, for that to happen, it better be the case that B loses the first round, and then A wins the next one. So, the first round needs to be (B - C), which will be w.p. 1/3, then C wins (w.p. 1/2), and then A wins the next round (w.p. 1/2), and since these events are independent, the final probability is 1/12.

Let's also quickly disprove the claim that Bs chances are 2/3:

If the first match is A-B, then B wins w.p. 1, and then B wins w.p. 1/2 If the first match is A-C, then A wins w.p. 1/2, and then B wins w.p. 1 If the first match is A-C, then C wins w.p. 1/2, and then B wins w.p. 1/2 If the first match is B-C, then B wins w.p. 1/2, and then B wins w.p. 1

Giving a total probability of 1/3(1/2 + 1/2 + 1/4 + 1/2) = 7/12, which is not that far away from 2/3 (8/12)

You can model many things around you using simple models like this; they are fun toy problems!

Answer 2

I've been a wrestling fan since 1996, and I have to say I LOVE that promo. Always makes me laugh.

Even though this isn't the question you're asking here, I did attempt to correct the calculations in Steiner's promo to something more feasible and less ridiculous, just for fun.

Now I don't have a stellar history with probability, but here's how I'd initially approach trying to figure out the probabilities of each contestant winning the match with the constraints Steiner gave in the promo.

First, I'd use abbreviations where I'd abbreviate the contestants. A = Angle, S = Steiner, and J = Joe.

And the way I'd compute pretty much any probability would be to look at number of possibilities of the DESIRED event happening and divide that number by the number of TOTAL events that can possibly happen.

In a normal triple threat match, here are all possibilities:

A beats S

A beats J

S beats A

S beats J

J beats S

J beats A

But Steiner claims he's a genetic freak and is not normal. (He did graduate from the University of Michigan by the way.) He claims that the chances of him beating anybody one-on-one are 75% and not the 50% you'd see in a normal fair contest.

I'm not completely sure how to represent that probability-wise, but I'd say that can be interpreted as for each possibility that exists of an opponent beating Steiner, there are THREE possibilities that exist of him beating his opponent.

If that thought is true, then I guess for every J beating S there are THREE ways of S beating J, and for every A beating S, there are THREE ways of S beating A.

So that adds these possibilities to the ones above:

S beats A

S beats A

S beats J

S beats J

But Steiner made it clear that A beating S is not a possibility since Angle knows he can't beat Steiner and won't even try.

So, strike the A beats S possibility.

Now we have 9 possible outcomes, and out of those 9, there is one where A wins, 2 where J wins, and 6 where S wins.

So, Steiner actually had a 6/9 = 66.6% chance of winning, Angle a 1/9 = 11.1% chance, and Joe a 2/9 = 22.2% chance by all of these assumptions.

But my understanding of probability may be a bit naive, and unlike some other mathematics I may do, I'm very open to correction here.

It's my understanding that Angle didn't end up competing in that event at all at TNA Sacrifice 2008. Instead, it ended up being Kaz vs. Steiner vs. Joe. And Joe ended up winning the thing anyway. Yep, after all of that 141% chance of winning stuff, Steiner actually lost.