odiousgambit
odiousgambit
  • Threads: 326
  • Posts: 9557
Joined: Nov 9, 2009
April 26th, 2010 at 11:12:18 AM permalink
I expect this will be right up your alley, some of you, for the rest of us you need to check out the OJ Simpson part at least.

http://opinionator.blogs.nytimes.com/2010/04/25/chances-are/
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
Croupier
Croupier
  • Threads: 58
  • Posts: 1258
Joined: Nov 15, 2009
April 26th, 2010 at 11:37:46 AM permalink
Thanks for posting that. It was easy to read and informative and has definately helped clear a few things up for me.
[This space is intentionally left blank]
Wizard
Administrator
Wizard
  • Threads: 1491
  • Posts: 26435
Joined: Oct 14, 2009
April 26th, 2010 at 4:13:46 PM permalink
On the topic of conditional probability, the most controversial problem by far on my math' rel='nofollow' target='_blank'>http://mathproblems.info/]math problems site is number 16. It goes like this:

Suppose you have a bag with two coins. One is a normal coin, with one heads and one tails. The other has two heads. You pull out a coin at random and observe one side, which is heads. What is the probability you chose the two-headed coin?

I've had people passionately argue that my answer is wrong. Of course none using any probability theory, but just their "common sense." I won't tell you the answer, by the way, in case you want to solve it on your own.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
nyuhoosier
nyuhoosier
  • Threads: 31
  • Posts: 248
Joined: Feb 16, 2010
April 26th, 2010 at 4:43:48 PM permalink
I'm not much of a math guy, but I'm wondering how this concept applies to gambling. I imagine that it has relevance in how the house edge runs up against bankroll and player habits.

For example, if the following are true:

A) I am playing a blackjack game with a .65 house edge at $10/hand.
and
B) 90 percent of the time I won't leave the table unless I'm up at least $60 (or have lost my $300 session backroll)

Then the likelihood I'll leave the table a winner is nowhere near 99 percent. It would be quite difficult to calculate (which is part of the NYT writer's frustration).

For an explanation of conditional probability, see the link above. I won't venture an explanation. However, I'm interested in the thoughts of the math guys here.

*** MOVED from separate post
pacomartin
pacomartin
  • Threads: 649
  • Posts: 7895
Joined: Jan 14, 2010
April 26th, 2010 at 4:47:53 PM permalink
Quote: Wizard

Suppose you have a bag with two coins. One is a normal coin, with one heads and one tails. The other has two heads. You pull out a coin at random and observe one side, which is heads. What is the probability you chose the two-headed coin?



Warning: partial spoiler

I guess one way to look at it is if you pulled the coin out of the bag and held it clutched in your hand without looking at (i.e. you have no information) the probability that the coin in your hand is the two headed coin is 50/50 .

By looking at the coin you now have some information. Additional information must change the probability.
* If you saw that it was a tail, then you would use that information to conclude the probability that you had picked the two headed coin was zero.
* If you see it is a head, there also must be some change in the probability. It can't remain the same.

Just like the blackjack expected values. All information changes the EV's. It may not change them enough to effect a change in the decision matrix, but it changes the EV's. A cut card, additional decks, the number of cards played, etc.
You can think of the Wizard's abbreviated strategy as the best decision you could make if you did not look at your cards but somebody told you if you were stiff or pat. It is a vital piece of information, but by looking at the actual cards you can refine your strategy.
pacomartin
pacomartin
  • Threads: 649
  • Posts: 7895
Joined: Jan 14, 2010
April 26th, 2010 at 5:09:33 PM permalink
Quote: nyuhoosier

I'm not much of a math guy, but I'm wondering how this concept applies to gambling. I imagine that it has relevance in how the house edge runs up against bankroll and player habits.

For example, if the following are true:

A) I am playing a blackjack game with a .65 house edge at $10/hand.
and
B) 90 percent of the time I won't leave the table unless I'm up at least $60 (or have lost my $300 session backroll)

Then the likelihood I'll leave the table a winner is nowhere near 99 percent. It would be quite difficult to calculate (which is part of the NYT writer's frustration).

For an explanation of conditional probability, see the link above. I won't venture an explanation. However, I'm interested in the thoughts of the math guys here.

*** MOVED from separate post



The article was lost when you moved this post.
I also assume you dropped the percent symbol on .65

If you are playing an even money game with a .65% house edge and you have the modest goal of earning $60, you will reach that goal 79.92% of the time without going bust. If your objective is to double your $300 the probability is 40.37% that you will achieve your goal. If you are playing baccarat the probability is only 30%.

Actual blackjack is a little harder to calculate since the blackjacks pay a bonus.

===================

As to the article, he is urging people to try to run through the problem with integers instead of just trying to plug percentages into a formula. If you teach math you do this all the time with students.

For instance if you get 10% simple interest on your principal every year, how long will it take you to double your money? Answer is 10 years.

If you get 10% compound interest on your principal every year, how long will it take you to double your money? Answer had better be less than 10 years or you've screwed up. Before calculators people used to use a logarithmic approximation to this question called the rule of 70. The approximate answer is 7.2 years. Similarly if you want to know how long it takes to double your money at 9% a year, the answer is 8 years (where 9*8 is about 70).
For the more mathematical you are using the first term approximation of the McClauren series for ln(1+x) which is x and the fact the the natural log of 2 is 0.693 or roughly 0.7
rudeboyoi
rudeboyoi
  • Threads: 27
  • Posts: 2001
Joined: Mar 28, 2010
April 26th, 2010 at 5:19:04 PM permalink
spoiler text in white

2:1

theres 2 heads on one coin and 1 heads on the other coin.

since it was revealed it wasnt tails, the tails is irrelevant

rudeboyoi
rudeboyoi
  • Threads: 27
  • Posts: 2001
Joined: Mar 28, 2010
April 26th, 2010 at 5:50:57 PM permalink
this math problem site you have is pretty fun.

i always enjoyed problems like 14 and 20.
teddys
teddys
  • Threads: 150
  • Posts: 5527
Joined: Nov 14, 2009
April 26th, 2010 at 5:51:08 PM permalink
Edit: Never mind, thought you said 1:2.
"Dice, verily, are armed with goads and driving-hooks, deceiving and tormenting, causing grievous woe." -Rig Veda 10.34.4
Wizard
Administrator
Wizard
  • Threads: 1491
  • Posts: 26435
Joined: Oct 14, 2009
April 26th, 2010 at 5:55:59 PM permalink
Quote: rudeboyoi

spoiler text in white

2:1

theres 2 heads on one coin and 1 heads on the other coin.

since it was revealed it wasnt tails, the tails is irrelevant



I'm going to just show what you wrote in black, so others can comment. Without revealing the answer, I'll just say that most people would probably agree with you.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
RaleighCraps
RaleighCraps
  • Threads: 79
  • Posts: 2501
Joined: Feb 20, 2010
April 26th, 2010 at 6:10:13 PM permalink
Well, I'm sure this is probably not the right answer either, but since I leave to play craps on Thursday night, it's time for me to get used to being wrong.......

Here goes. You have a 25% chance of pulling out a tail. Since the coin was not a tail, you now have 2 out of 3 chances the coin is the 2 headed coin, therefore your chance was 66.67%.
Always borrow money from a pessimist; They don't expect to get paid back ! Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, don't matter!
DJGenius
DJGenius
  • Threads: 8
  • Posts: 45
Joined: Mar 5, 2010
April 26th, 2010 at 7:58:07 PM permalink
I agree. Actually, I had already seen this on the Wizard's math site and solved it for myself in excel without too much trouble. I can see why there's so much controversy though, because it's not the most intuitive answer. If I'm not mistaken there's a whole story on the site about a school that was divided over this problem. The principal had even announced to the whole school that the wrong answer was right!
"The Quest stands upon the edge of a knife. Stray but a little, and it will fail, to the ruin of all." - Elf Queen Galadriel, teaching Frodo about the importance of blackjack basic strategy.
Wizard
Administrator
Wizard
  • Threads: 1491
  • Posts: 26435
Joined: Oct 14, 2009
April 26th, 2010 at 8:02:24 PM permalink
Quote: DJGenius

If I'm not mistaken there's a whole story on the site about a school that was divided over this problem. The principal had even announced to the whole school that the wrong answer was right!



You're right. That whole story is in the problem 16 solution.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
pacomartin
pacomartin
  • Threads: 649
  • Posts: 7895
Joined: Jan 14, 2010
April 26th, 2010 at 10:35:35 PM permalink
This problem has a probability space of only four options. It is a simple matter to write out the probabilities and see which ones are eliminated.

If I ask you what is the probability of getting a streak of 3 or more heads or tails in a row out of 5 coin tosses, the probability space is 32 possibilities. You could just write them out and count the ones that comply with the question.

If I ask you what is the probability of getting a streak of 6 or more heads or tails in a row out of 10 coin tosses, the probability space is 1024 possibilities. It would be difficult to write out the probability space by hand and count them. You could write a computer program, but you probably would be better off if you could develop the formula, or the Markov Transition matrix or the recursion algorithm.

In public relations the biggest problem you have is often the saying you can't prove a negative. The public often wants you to prove a negative. Lawyers complicate things with the precautionary principle. The precautionary principle says that if the hazard to the public or the environment is severe enough you must choose not to do it because of the consequences. Since you can't prove a negative, and there is always some possibility, the precautionary principle always kicks in. I used to manage a low frequency sonar program, and their was always a risk to marine life.
  • Jump to: