Two Plus Two Older Archives

Two Plus Two Older Archives (http://archives2.twoplustwo.com/index.php)
-   Probability (http://archives2.twoplustwo.com/forumdisplay.php?f=23)
-   -   Two Questions (http://archives2.twoplustwo.com/showthread.php?t=546)

08-05-2002 05:06 PM

Re: Two Questions
 


My Guesses:


1) Depending on how you read the question: 4 or

Never.


2) 1.


(The number of plays needed to have a greater than 50% to hit would be 27726 .)




08-05-2002 05:34 PM

Re: Two Questions
 


1. The probability of a 7 is 1/6 since out of 36 combinations the 6 combinations that give 7 are 1-6, 2-5, 3-4, 4-3, 5-2, 6-1. The probability of not getting a 7 on a roll is 5/6. The probability of not getting 7 for n rolls is (5/6)^n. This becomes less than .5 for n = 4 rolls.


2. This is intended to be a trick. The probability of being right if you predict it will hit on the nth play is the probability that it will not hit on the first n-1 plays and then hit on the nth play. This is

(1/40,000)(39999/40000)^n. This is maximum for n=1, so you should predict it will happen on the first play for the prediction to have the highest probability of being right.


This is not the same thing as asking how long it will take to have a greater than 50% chance of hitting as we did in the first problem, nor is it the same as asking the average time it will take. It also doesn't matter that the machine just hit as long as the machine has no memory.

08-05-2002 10:23 PM

Funny ...
 


1: Never


2: 1

08-05-2002 11:22 PM

Re: Two Questions
 


many years ago before almost any poker player could figure out these things i won large winnings from two other players who were thought to be more knowlegedable than me with each of these props. funny thing is on one of them we even used david as the deciding person to determine who was right. do you remember david the 25,000 bet i had with mickey.

08-06-2002 02:29 AM

Re: Two Questions
 


Minor correction:


The probability of being right if you predict it will hit on the nth play is the probability that it will not hit on the first n-1 plays and then hit on the nth play. This is

(1/40,000)(39999/40000)^n



Make that (1/40,000)(39999/40000)^(n-1) with the same conclusion. The idea is that the first play has the highest probability (1/40000) because all the other plays require n-1 consecutive failures immediately prior to the first success.

08-06-2002 02:38 AM

answers please????
 


Are you going to give us the answers?

appears like you keep giving questions, which has been great, but never the answers.

08-06-2002 04:04 AM

Re: answers please????
 


4 1

08-06-2002 04:25 AM

Re: Two Questions
 


1) After four rolls you are more likely to have rolled at least one seven.


2) Because each trial is equally likely to have a royal flush, and we will play until we get a royal flush, the most likely time to get the Royal flush is on the first trial, because the chance that the royal occurs then is 1 out of 40,000. The chance that it occurs on the second trial is slightly less, 1 minus the chance that it occured on the first trial times 1 out of 40,000. So if I have to get within 100 of the correct answer, I will choose the 100th trial.


Bob T.

08-06-2002 05:26 AM

Learn to think!
 


I suspect David's answer would have been longer if we had all fallen for his trap. Perhaps something like this:


I told you to think and not plug numbers mindlessly into your formulas and calculators. I asked for the prediction most likely to be correct, not the time it would take to be more likely than not to win. I only required you to be within 100 and many of you were off by thousands. The correct prediction is the first play since its probability of 1/40000 is maximum. If you don't know how I got that you should stop reading this forum at once. You don't know enough about probability, and you deserve to be taken out back and shot.


How'd I do?

08-06-2002 09:40 AM

I disagree with a)
 


David,

This is the same fallacy that allows all of those gambling system-hawkers to tout their crap.


Each roll ALWAYS gives you 6/36 chances to roll a 7. If you were going to phrase the question as "Over how many rolls, all considered together, would it take until rolling a 7..." then you might be able to sell the 5/6 *5/6 *5/6 *5/6 = 48.22% chance of NOT rolling a 7 anytime during those 4 rolls....


But since I know you know that already, I'm wondering what your point was... unless it was to set up the next question (good trick, by the way)


All times are GMT -4. The time now is 05:18 PM.

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, vBulletin Solutions Inc.