Thread: How do you think through this?

1. Originally Posted by Provoker
Infinity has a half point? Please specify.
I was talking about the world population at that point.

2. Originally Posted by JustHer
What is wrong with you people and sticking to the word "infinity", it was obviously not literal.
JustHer,

(1) Did it ever occur to you that one might draw more utility from the debates surrounding infinity than the post itself?

(2) In order to be able to back the currency of your comment up you must have intellectual reserves. Having the proper reserves entails that you must be aware of the difference between the literal as opposed to the figurative definition of infinity, which I propose you lay down. upon laying down this definition/conception, you will then need to demonstrate how the figurative conception can still be infinity and yet allow for fractions of it to exist--i.e. 1/4 of infinity. And so on.

3. Originally Posted by forzen
The same words that came out my mouth when I saw your first post .
I was not trying to divide infinity by infintiy in my first post. I was trying to divide infinity into the three seperate battles.

4. Very poorly worded question.

As for yeur chances, they suck, 1/4, 1/4, and 1/2 is beaten by nearly every possible incarnation he could come up with.

Two armies of 1/2 each have a 33% chance of beating yeu depending on where they were sent for the matchup, three of 1/3rd has 100% chance to win, so on and so forth, generally yeu're not in good shape using such a crappy choice, especially if it's blind luck.

Drop the word problem because this is absolute fail of a word problem. Just go with the raw facts and ask those.

I've only lightly touched on probability so am not sure the exact % chance.

5. I have one question that relates to context of this problem.

Let's say we have a game where a game master chooses a number randomly out of 0 to 100. This number is not shown to the other players and the other players also have to choose a number from 0 to 100. Whichever player has the number closest to the game master's number is the winner.

Now let's say we change the number of choices from '0 to 100' to '0 to 200'. And another game is played.

Concerning each game before the players have chosen their numbers, are we assuming that each player has a 50% of winning in each game? Thusly eliminating a finite amount in this case and establishing a safe correlation between the two?

6. Dude, everything is 50/50. You either win, or you don't. duh.

7. Originally Posted by hilo
Dude, everything is 50/50. You either win, or you don't. duh.
Well I have geometrical solution in mind, but it relies on the answer to my above question being 'yes'.

Essentially you choose a value, going from 0 to 100 and then create a square of your next possible choices for the next two battles for each of these game sequences. Then the opponent goes through each sequence from 0 to 100 for each of your 0 to 100 and you compare the squares and areas.

If the opponent loses the first battle, then they have to win the next two battles, so you only consider the area in its square where both battles have numbers above all possible numbers of the player's square.

If the opponent wins the first battle, then any choices that overlap with the player's square are considered 50/50 and ignored from factoring in winning conditions. The area where the opponent doesn't overlap is divided by 2 and you have the area of that scenario. Now do this for all possibilities and record the total scores, noting the smallest score that the first battle brings and you essentially have your maximum probability. So if one battle is a certain maximum probability then it doesn't matter what you put for the next two battles since they correlate directly and your probability stays the same as long as you choose that maximum probability number for any of the battles. It's hard to explain in text...I wish I had a tablet pc.

8. From my understanding of the OP, all 3 battles occur concurrently... It's not sequential. So you can't wait for the results of the first battle before setting the remainder of your army for the other 2

9. Assuming it is indeed a finite number, and assuming the forces are equal size and ability, the randomized one would (most likely) win as that would (most likely) mean an aproximate distribution of 1/3 on each. 1/3 beats 1/4 twice and 1/2 beats 1/3 once. It's quite simple.

10. I'll supply some answers without going into explanations as of yet. (1/2,1/2,0) has a 50% chance of winning. The (1/2,1/4/,1/4) has a 62.5% chance of winning.

Here's a question. What is the expected value of the largest of three randomly generated numbers between 0 and 100? It isn't the same as the expected value of any one.

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
Single Sign On provided by vBSSO