Defeating Derren Brown

In one Derren Brown show, “Inside Your Mind“, he plays his “favorite game” with 3 Psychology students. He puts money in 1 of 2 envelopes, and the 3 Psychology students have to guess which envelope contains the money. If they guess right, they get to keep the money.

2 envelopes

Derren first plays with 5 pounds and shuffles while the 3 Psychology students face the other way. When they look back, he says some things and makes certain gestures in an attempt to influence their choice of envelope. They pick and, of course, lose.

The second round is for 10 pounds, and Derren shuffles under the table. He again says some things and makes certain gestures in an attempt to influence their choice of envelope. Again, they pick and, of course, lose.

The third round is for 20 pounds, and Derren shuffles under the table. When he puts them on the table, he pushes one envelope towards them. He again says some things and makes certain gestures in an attempt to influence their choice of envelope. Again, they pick and, of course, lose.

The fourth round is for 50 pounds. This time the Psychology students hide the 50 pounds in 1 of 3 envelopes, and Derren has to guess which envelope contains the money. They, of course, lose.

The reason why they lost every time is because their strategy is to pick whatever they feel like picking, which is wrong since what they feel like picking is influenced by a mentalist.

Their strategy has an expected value of  \Sigma _{-\infty}^{\infty} xp(x) = (5)(\frac{0}{2}) + (10)(\frac{0}{2}) + (20)(\frac{0}{2}) + (50)(\frac{0}{3}) = 0.

Here is an alternative strategy based on the first arc of the second series of Yu-Gi-Oh!:

In Yu-Gi-Oh!, the main characters, Yugi and Atem, share a body and play a children’s card game hoping to save Yugi’s grandfather from the arc villain Maximillion Pegasus, who has the ability to read minds. In the first part of the game, Atem plays. Pegasus, being a mind reader, obviously counters all of Atem’s strategies. Eventually, they come up with the idea of Yugi playing instead of Atem and then switch to Atem whenever Pegasus tries to read their mind. Thus, they are able to successfully hide their ace card, Dark Magician, in Magical Hats and defeat Pegasus’ Toon Monsters.

The application: Get Derren to be unable to read your mind/influence by choosing an envelope randomly.

This new strategy has an expected value of \Sigma _{-\infty}^{\infty} yp(y) = (5)(\frac{1}{2}) + (10)(\frac{1}{2}) + (20)(\frac{1}{2}) + (50)(\frac{2}{3}) > 0 .

Comments on Ren Arceño’s Epic February 23 10:13pm Speech in teh group

here‘s teh epic speech. What my Facebook note didn’t include were the comments in teh group. So, here (+1 means I like it, and I don’t know why I didn’t before):

p1 p2 p3

Fender Avril Lavigne Giveaway

I entered the “Fender Avril Lavigne Giveaway.”

It says

“Want to double your chances of winning?

Share this sweepstakes using the share icons below.
When one of your friends enters…
You get an extra entry*!
* Maximum of one extra entry person

Image

Double?

Let x, y > 0. Let P(me winning|I do not share)  = \frac{1}{x} , and let P(me winning|I share)  = \frac{2}{x+y} where  x = \left | A \right | and  y = \left | B \right | where A = {people who would have known about this even if I did not share} and B = {people who know about this specifically because I shared} = {people who know about this} \ A.

 \underline{Assumption \ 1} : The number of people who would have known about this had I not shared is greater than the number of people who know about this specifically because I shared.

 \underline{Proposition \ 2} : The probability of me winning if I share is greater.

 \underline{Pf:} **
 y < x (by assumption 1)
 x+y < 2x
 \frac{1}{x} < \frac{2}{x+y} .

My increased probability is  f(x,y) = \frac{2}{x+y} - \frac{1}{x} = \frac{x-y}{(x)(x+y)} .


 \underline{Corollary \ 3:} My chances of winning is not doubled if I share it.

 \underline{Pf:}

y≠0 => 2P(me winning|I do not share)≠P(me winning|I share)

 \underline{Exercise:} Explain the meaning of  f(x,y) as  y \to x , if any.

Note: P(me winning) = P(I do not share) P(me winning|I do not share) + P(me winning|I share) P(I share) by Bayes’ Theorem