Odds or evens

I think I understand this game correctly, we both pick a number between 0-5 and then add the two numbers and are attempting to pick if the resulting number is odd or even? Your friend is correct in that even has an edge but I think the reasoning is not entirely correct. Player A and B has 6 options 0,1,2,3,4,5 so 3 Even options (0 is considered Even) and 3 odd options. If Player A and B was to pick at random then the numbers they pick 50% of the time the number is even and 50% odd. The reason that Even has the edge is because Even +Even=Even and Odd+Odd=Even. So its only in the Case where one player picks odd and the other picks even will you have an odd outcome. 3 Possible Cases and 2/3 times its going to be Even. I never realized this game was rigged until I sat down and thought about it LOL

i think even more important than there being more number of evens vs odds, is the psychological bias to pick certain numbers. I don't think the probability of picking 2 is the same as 3. There might be cultural and human biases that make us tend to prefer certain number over others.

That is definitely true based on various cognitive psychology experiments I've read. Numbers like 3 and 7 are much more likely to be picked than for example 13 or 4 ( well 13 doesn't really matter in the current examples in odd or even ). Sadly I can't really remember the case of "4" although its general meaning is positive, but people usually related it to something bad.

On August 29 2016 13:40 FMLuser wrote: I think I understand this game correctly, we both pick a number between 0-5 and then add the two numbers and are attempting to pick if the resulting number is odd or even? Your friend is correct in that even has an edge but I think the reasoning is not entirely correct. Player A and B has 6 options 0,1,2,3,4,5 so 3 Even options (0 is considered Even) and 3 odd options. If Player A and B was to pick at random then the numbers they pick 50% of the time the number is even and 50% odd. The reason that Even has the edge is because Even +Even=Even and Odd+Odd=Even. So its only in the Case where one player picks odd and the other picks even will you have an odd outcome. 3 Possible Cases and 2/3 times its going to be Even. I never realized this game was rigged until I sat down and thought about it LOL

I think you have fallen into a trap there. Though you are right that even + even = even and odd + odd = even, that only accounts for 50% of the scenarios. If you look at each individual scenario (like counting the number of combos of a given hand):

even-even = even
even-odd = odd
odd-even = odd
odd-odd = even

so if both players have a 50% chance of picking each, half the scenarios give an even sum, and the other half get an odd sum. Game isn't rigged

Raidern, assuming I understand the game correctly, your intuition is correct. If you are forced to choose an integer from 0-10 at random where each integer has an equally likely chance of being picked, and your friend gets to pick odd or even before hand, he wins 6 out of 11 times, or more than 50%. However as stated there is no random choosing, so the amount of numbers that are odd or even is not really relevant; in the rules of the game it only matters if the number is odd or even, so essentially he should be choosing 'odd' or 'even', then pick any number that fits that criteria.

For an extreme example, let's say that he could either choose 2, or any odd number between 1 and 1,000,000. There is only 1 even choice and a many odd numbers, but that doesn't need to have any impact on the frequency with which an odd or even number is picked. The only thing that matters (and should effect decision making) is whether the number is even or odd, and not what the number itself actually is.

On August 29 2016 20:21 HungarianGOD wrote: Show nested quote + On August 29 2016 13:40 FMLuser wrote: I think I understand this game correctly, we both pick a number between 0-5 and then add the two numbers and are attempting to pick if the resulting number is odd or even? Your friend is correct in that even has an edge but I think the reasoning is not entirely correct. Player A and B has 6 options 0,1,2,3,4,5 so 3 Even options (0 is considered Even) and 3 odd options. If Player A and B was to pick at random then the numbers they pick 50% of the time the number is even and 50% odd. The reason that Even has the edge is because Even +Even=Even and Odd+Odd=Even. So its only in the Case where one player picks odd and the other picks even will you have an odd outcome. 3 Possible Cases and 2/3 times its going to be Even. I never realized this game was rigged until I sat down and thought about it LOL I think you have fallen into a trap there. Though you are right that even + even = even and odd + odd = even, that only accounts for 50% of the scenarios. If you look at each individual scenario (like counting the number of combos of a given hand): even-even = even even-odd = odd odd-even = odd odd-odd = even so if both players have a 50% chance of picking each, half the scenarios give an even sum, and the other half get an odd sum. Game isn't rigged

The order doesn't in which you pick the numbers has no effect on the results so counting even+odd and odd+even is double counting it. Like 1+2 is the same as 2+1

On August 29 2016 20:18 Spitfiree wrote: That is definitely true based on various cognitive psychology experiments I've read. Numbers like 3 and 7 are much more likely to be picked than for example 13 or 4 ( well 13 doesn't really matter in the current examples in odd or even ). Sadly I can't really remember the case of "4" although its general meaning is positive, but people usually related it to something bad.

4 sounds similar to "death" in Chinese (and many asian languages influenced by them), and so has some negative connotations.

Daut?

On August 30 2016 07:29 FMLuser wrote: Show nested quote + On August 29 2016 20:21 HungarianGOD wrote:   On August 29 2016 13:40 FMLuser wrote: I think I understand this game correctly, we both pick a number between 0-5 and then add the two numbers and are attempting to pick if the resulting number is odd or even? Your friend is correct in that even has an edge but I think the reasoning is not entirely correct. Player A and B has 6 options 0,1,2,3,4,5 so 3 Even options (0 is considered Even) and 3 odd options. If Player A and B was to pick at random then the numbers they pick 50% of the time the number is even and 50% odd. The reason that Even has the edge is because Even +Even=Even and Odd+Odd=Even. So its only in the Case where one player picks odd and the other picks even will you have an odd outcome. 3 Possible Cases and 2/3 times its going to be Even. I never realized this game was rigged until I sat down and thought about it LOL I think you have fallen into a trap there. Though you are right that even + even = even and odd + odd = even, that only accounts for 50% of the scenarios. If you look at each individual scenario (like counting the number of combos of a given hand): even-even = even even-odd = odd odd-even = odd odd-odd = even so if both players have a 50% chance of picking each, half the scenarios give an even sum, and the other half get an odd sum. Game isn't rigged The order doesn't in which you pick the numbers has no effect on the results so counting even+odd and odd+even is double counting it. Like 1+2 is the same as 2+1

you're not double counting it... Those are two entirely different things and you have to count both of them... You can't just say they both equal 3 so you can get away with only counting a single 3.

Raidern, think of it like this - There may be more even than odd results but the result of 0 ONLY comes about if BOTH players throw 0, and the result of 10 ONLY comes above if BOTH players throw 5. Now consider the result of 5, you can get it with 0-5 or 1-4 or 2-3, or 3-2, or 4-1, or 5-0, and suddenly odds aren't looking so bad, eh?

It's not a very complex problem. You and your opponent each pick a number between 0 and 5. That's 6 numbers each. So there are 36 possible outcomes.
possible ways to get each number:

0 - 1
1 - 2
2 - 3
3 - 4
4 - 5
5 - 6
6 - 5
7 - 4
8 - 3
9 - 2
10 - 1

Add those up and you will see there are 18 ways to get an even number and 18 ways to get an odd number. Or 50/50.

I think they were talking about 2 different games: in Raidern's game there aren't 2 numbers chosen, but rather just one. One person picks another and the other says odd or even. Or maybe I'm confused.

2 people each choose a number between 0-5 those numbers are simultaneously revealed and then added to each other. The players are trying to pick if the result of the two numbers are odd or even

Sorry HungarianGod I wasn't clear in my op. The is exactly how FMLuser explained it. Also ty blackjacki I think that's what we were looking for and indeed it's very simple, we were trying to come to a conclusion based over wrong premises.

Blackjacki solved it, but HungarianGOD delivered the cleanest solution in his first post. And HungarianGOD didn't double count; FMLuser half counted.

It is dependent on how much poker you have played. A good player knows to select "higher odds" or "lower odds" based on the odds alone. If you tend toward picking "higher odds" than "lower odds" you are selecting in favor of independent events laws. If you don't believe in independent events laws (i.e. if you are not a mathematician) then you might for instance have a theory like the so-called mantis shrimp theory. The Mantis Shrimp theory in effect is a resolution-logic theory in which often times the scale of probability regarding an event happening favors different probability resolutions.

So in a quantum death mechanic "I am going to die on one of three days: Wednesday Thursday Friday"

In the exercise the result favors Friday although the traditional probability of dying Friday is only 1/100.

This resolution exercise is a condensation of various Gaussian distributions and the result of dying Friday is something like:

infinity / R

in other words it's a ratio between two infinities and for that reason it's the most likely result although traditionally depicted 1/100 : 99/100 in which we would tend to select 99/100 by law of independent events.

so it's just a variety of modes of describing probability

http://www.bing.com/images/search?q=m...f51211622eb8980c1070o0&ajaxhist=0

there are some related paradoxes in philosophy like the Raven Paradox