Comments on: The Cost of Not Understanding Probability Theory

By: Eka

Eka — Tue, 29 Dec 2009 03:00:37 +0000

Dear enthusiastic math lover..after seeing comments above, i think i want to contribute a little here.
@alain: yes, real world problems are complex. That is why theories are developed: to simplify problems n get the solution of it. As stated above,we ASSUME tht the dice is fair. Then,from such assumption,we have the theory of measuring the probability to support our decision. If you consider that the assumption is not valid in real world problem you face,e.g. the dice is not fair,you may consider not to employing the theory. But then what happened? We may end up dealing with bigger complexity.. So,we may as well put things in the proper corridor in order to get the best theory we can employ to support our decision.
@rick: i believe you assume that the sequence of the outcome is not necessary. Taking such assumption into account,you should have known that the outcome of the coins are:
>over-over
>under-under
>over-under.
Just 3 instead of 4 since you consider that {over-under}={under-over} without realizing it. By knowing that the outcome of one coin is ‘over’, Therefore we have eliminated the outcome {under-under} and thus the options we have left are {over-under} n {over-over} which yields the probability 0.5 of having {over-over} as the outcome.. I hope this is comprehensible. *sigh,typing this comment via cellphone is surely tiring..
Be sure to check math.web.id to see the various colour of math undergraduate in blogging.merci.

By: Edward

Edward — Wed, 16 Dec 2009 20:53:08 +0000

This all sounds extremely similar to some of the rhetorical fallacies you learn in english class. Though, admittedly, there’s more data behind this.

By: Rick

Rick — Mon, 30 Nov 2009 09:37:33 +0000

So let me put this another way. A betting shop has determined a total for points scored in a football game and is taking bets on if you think the actual total will be under or over their number. Over the last 1000 games the bet shop’s numbers are below the actual total 50% of the time so of course they are also over the actual 50% of the time. It’s a 50/50 bet.

I will take a coin and designate heads as “over’ and tails as under, then flip it and and compare it to the outcome of the game. (the hidden coin)

There are four possible outcomes:
Over-Over
Over-Under
Under-Over
Under-Under
I flip the coin and its heads or “over”
so by your rule I have eliminated one outcome, the “under-under” and three outcomes remain. So again by your rule I have only a 33% chance the game will be “over”.

What a great idea! Before I bet I’ll just flip a coin and choose the opposite of the result and I’ll have a 66.67% chance of winning a 50/50 bet.

I’ll send you a postcard from my Villa in France.

By: Rick

Rick — Mon, 30 Nov 2009 07:27:14 +0000

“If I flip two coins simultaneously, then show you that one is heads… what is the probability that the other is heads?

Most people will say “50/50″, since they’ve heard that each coin flip is independent, or “1 in 4″, since that’s the chances that 2 heads were flipped originally. That’s not really the case here. There are 4 possible outcomes of the 2 coins flipped, but I showed you one was heads, which eliminates the tails-tails outcome. Of the 3 remaining possibilities, only one is heads/heads, so the probability that the 2nd coin is heads GIVEN that the first coin was heads is 33%.”

This is incorrect.

Ther are four possible outcomes:
1.Heads-Heads
2.Heads-Tails
3.Tails-Tails
4.Tails-Heads

By showing me the first coin is heads, you’ve eliminated TWO outcomes.(#3 and #4)

So the probability the other coin is heads is still 50-50.

By: Calcatraz

Calcatraz — Tue, 27 Oct 2009 07:27:54 +0000

Thanks for an interesting read. I knew about most of these, but not the specifics of the Italian Lotto. Another source of misunderstanding probability occurs when humans try to do mental calculations involving very large numbers. A million is about the same as a billion, right?

By: Antonio Cangiano

Antonio Cangiano — Sun, 18 Oct 2009 22:27:28 +0000

> Say that 5 bets are made and then the 6th one is a win.

#1: spends 1, wins 0
#2: spends 2, wins 0
#3: spends 4, wins 0
#4: spends 8, wins 0
#5: spends 16, wins 0
#6: spends 32, wins 32*11.232

So the net is:

won = 32*11.232 = 359.392
spent = 63 (1 + 2 + 4 + …)

net = won – spent = last_wager * 11.232 – (2*last_wager -1) = last_wager * 9.232 + 1

Sudeep’s formula is the correct one. I updated my post.

By: David Friedman

David Friedman — Sun, 18 Oct 2009 21:38:14 +0000

Sandeep wrote:

>
>For your calculation of the additional
>payout for martingale, i think the correct
>formula should have been
>(last wager * 9.232 ) + (initial wager)
>
>

Antonio wrote:

>Sudeep, the 11.232x payout includes your >last bet, and the initial wager is assumed >to be 1 Euro here. The general formula is: >(last wager * 8.232) + initial wager.

Say that 5 bets are made and then the 6th one is a win.

First bet: bets 1 loses 1
Second bet: bets 2 loses 2
Third bet: bets 4 loses 4
Fourth bet: bets 8 loses 8
Fifth bet: bets 16 loses 16

he loses a total of 31 before he wins.

When he wins, he wins 32*11.232 .

So the net is:

won = 32*11.232
lost = 31

net = won – lost
net = 32*11.232 – 31

net = last_wager*11.232 – (last_wager-1)
= last_wager*10.232 + 1

This seems very confusing to me since it isn’t either of the two other answers.

I originally thought I might have the same answer as Sudeep. My understanding is that by 11.232 what is meant is that if the bettor has only 5 euros in his pocket and he bets all 5 euros then if he wins he can put 11.232*5 = 51.16 euros in his pocket.

I see though that your answer to Sudeep and your formula in the blog post is consistent:

answer to Sudeep:

(last wager * 8.232) + initial wager

blog post:

(last wager x 8.232 + 1)

given that initial wager=1

By: EM

EM — Mon, 05 Oct 2009 18:20:36 +0000

@The Count:

Your example is nice, but it is not a fallacy. The problem by flipping 2 coins is defining the probability space. Different definitions yield different probabilities for the event that the 2nd coin is heads.

@Alain:

That’s why goodness-of-fit tests exist. If the deviation from the uniform distribution is significant, then we can assume that something is wrong with the coin.

By: t kanno

t kanno — Sat, 03 Oct 2009 12:22:23 +0000

hi there antonio,

this might not pertain to the article, but i saw that you probably have an italian background.

i’m trying to decide wether to enroll in the facolta’ di matematica or in chimica in the university of Pavia?

did you study in italy? what can you tell me about mathematics?

thanks

By: Alexandros

Alexandros — Tue, 22 Sep 2009 08:19:18 +0000

@The Count:

@Alain:

That’s why goodness-of-fit tests exist. If the deviation from the uniform distribution is significant, then we can assume that something is wrong with the coin.