odiousgambit
Joined: Nov 9, 2009
• Posts: 8111
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
Joined: Nov 15, 2009
• Posts: 1258
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.
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Wizard
Joined: Oct 14, 2009
• Posts: 20402
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.
It's not whether you win or lose; it's whether or not you had a good bet.
nyuhoosier
Joined: Feb 16, 2010
• Posts: 248
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
Joined: Jan 14, 2010
• Posts: 7895
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
Joined: Jan 14, 2010
• Posts: 7895
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
Joined: Mar 28, 2010
• Posts: 2001
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
Joined: Mar 28, 2010
• Posts: 2001
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

Joined: Nov 14, 2009
• Posts: 5441
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
Joined: Oct 14, 2009
• Posts: 20402
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.
It's not whether you win or lose; it's whether or not you had a good bet.