Dean Mullen Posted February 8, 2011 Share Posted February 8, 2011 (edited) If you have 3 possibilities in a random number generator or lets say as an analogy, 3 balls in a bag, the odds of any particular ball coming out is 3/1, yet the odds of 2 coming out 5 times in a row is 3x3x3x3x3 = 243, so the odds of that occurring is 243/1, so lets say we did our random picking, and 2 came out 4 times in a row, then the odds of it continuing would be extremely unlikely, at 243/1 for it to come out a 5th time, so it shows that 2 coming out a fifth time is unlikely, yet at the same time there is only 3 options so then you conclusively speaking can realize that the odds of 2 coming out is both 243/1 and 3/1 at the same time? how can this be? Paradox? PS: apologies for spelling error, I meant of course probability paradox. Edited February 8, 2011 by Dean Mullen Link to comment Share on other sites More sharing options...

imatfaal Posted February 9, 2011 Share Posted February 9, 2011 If you have 3 possibilities in a random number generator or lets say as an analogy, 3 balls in a bag, the odds of any particular ball coming out is 3/1, yet the odds of 2 coming out 5 times in a row is 3x3x3x3x3 = 243, so the odds of that occurring is 243/1, so lets say we did our random picking, and 2 came out 4 times in a row, then the odds of it continuing would be extremely unlikely, at 243/1 for it to come out a 5th time, so it shows that 2 coming out a fifth time is unlikely, yet at the same time there is only 3 options so then you conclusively speaking can realize that the odds of 2 coming out is both 243/1 and 3/1 at the same time? how can this be? Paradox? PS: apologies for spelling error, I meant of course probability paradox. Dean Normally we would say probability of any one ball is 1/3 or 1:3 not 3/1. Ie with a fair dice the probability of throwing a 5 is 1 in 6 - 1/6. your second proposition is strangely phrased - " the odds of 2 coming out 5 times in a row". Is that two different results coming five times on the trot or the result #2 coming solely five times... from your calcs I guess it is the second. The next part of your post is the mistake - once you have got four "2"s on the trot the random generator DOES NOT remember this fact and still produces the same odds for each option. You have already got the four "2"s - and the next throw is as random as any other; the dice, the cards, the ball etc do not remember what has got before!! This is vital in gambling - unless you are working on a restricted deck then there is no memory of the previous event. to be explicit - if you have just rolled 2,2,2,2 on a fair die, you are just as likely to roll another 2 than any other number. the high chance of 1 in 243 is for 5 times a certain number - but that is only applicable before the first ball is chosen. If the first ball is "1" then the chances of five "2"s is zero, ditto if a "3" is chosen . If you know the first ball is a "2" then the chances of 5 "2"s on the trot is simply the probability of the next four balls being a "2". In essence, basis a fair game (and if it is not a fair game then walk away) probabilities only work forwards and remember the dice/cards have no memory Link to comment Share on other sites More sharing options...

Dean Mullen Posted February 9, 2011 Author Share Posted February 9, 2011 Dean Normally we would say probability of any one ball is 1/3 or 1:3 not 3/1. Ie with a fair dice the probability of throwing a 5 is 1 in 6 - 1/6. your second proposition is strangely phrased - " the odds of 2 coming out 5 times in a row". Is that two different results coming five times on the trot or the result #2 coming solely five times... from your calcs I guess it is the second. The next part of your post is the mistake - once you have got four "2"s on the trot the random generator DOES NOT remember this fact and still produces the same odds for each option. You have already got the four "2"s - and the next throw is as random as any other; the dice, the cards, the ball etc do not remember what has got before!! This is vital in gambling - unless you are working on a restricted deck then there is no memory of the previous event. to be explicit - if you have just rolled 2,2,2,2 on a fair die, you are just as likely to roll another 2 than any other number. the high chance of 1 in 243 is for 5 times a certain number - but that is only applicable before the first ball is chosen. If the first ball is "1" then the chances of five "2"s is zero, ditto if a "3" is chosen . If you know the first ball is a "2" then the chances of 5 "2"s on the trot is simply the probability of the next four balls being a "2". In essence, basis a fair game (and if it is not a fair game then walk away) probabilities only work forwards and remember the dice/cards have no memory Of course I know that but the odds of it re-occurring five times in a row is 2x2x2x2x2 = 243 isn't it? or is that not true. Link to comment Share on other sites More sharing options...

timo Posted February 9, 2011 Share Posted February 9, 2011 (edited) - You accidentally wrote 2x2x2x2x2 instead of 3x3x3x3x3 - It is not common to say "the odds are five". One says the odds "are one in five" (though probably less common except for "one in a million"), "one fifth", or "one to four". - Realizing that the odds of an event happening can change when the conditions have changed is cool. But you'll hardly impress anyone with this finding. Obviously, the odds of someone knocking on the door of my flat in the next five minutes are lower in the middle of the night than just after someone rang the bell downstairs. In a similar (but even more trivial) way the odds of getting five 2s in a row are higher when you already threw four. EDIT FOR CLARIFICATION: by "the odds getting five 2s when you already threw four" I mean the probability to throw 2 in the fifth trial after four 2s have already been already-thrown, which would then make it five 2s in total (i.e. the presumably paradoxical scenario that this thread is about). Edited February 10, 2011 by timo Link to comment Share on other sites More sharing options...

Bignose Posted February 9, 2011 Share Posted February 9, 2011 (edited) In a similar (but even more trivial) way the odds of getting five 2s in a row are higher when you already threw four. Not on a fair random number generator (like a fair die) it isn't. The results of past rolls or trials have zero influence on future rolls or trials. On a fair die, the chances of rolling five 2's in a row are the same whether you've previously just rolled one 2, two 2's, seven hundred seventy-seven 2's. This isn't a "paradox" as much as it is you have to be very careful with how you word the question. What are the chances of rolling five 2's in a row on a fair six sided die? [math]\frac{1}{6^5}[/math] I've just rolled four 2's , what are the chances of rolling another five 2's in a row? Again, [math]\frac{1}{6^5}[/math] I've just rolled four 2's, what are the chances of rolling another one, resulting in five in a row? [math]\frac{1}{6}[/math] Because the fact that you have already rolled the die, has no influence on future events. The above is why you have to be very careful with the way statements or words are phrased. You get very different results with very similar-sounding statements. Edited February 9, 2011 by Bignose Link to comment Share on other sites More sharing options...

timo Posted February 9, 2011 Share Posted February 9, 2011 Bignose: I said "getting five 2s in a row when you already threw four", not "getting five additional 2s in a row". Hence the adjective "trivial". Link to comment Share on other sites More sharing options...

Bignose Posted February 10, 2011 Share Posted February 10, 2011 Yeah, that just emphasizes how tricky this is. I still look at what you wrote there, and my brain still thinks "another five 2's...." even though I have read your clarification. Must be very, very clear talking about these things. Link to comment Share on other sites More sharing options...

timo Posted February 10, 2011 Share Posted February 10, 2011 Guess that's hard to argue with. Dean Mullen: Perhaps it's easier to just hint you into the right direction rather than trying to explain something in own words. The keywords you should probably look for is "Conditional probability". I recommend reading the Wikipedia article, it seems to talk about something very similar in its introduction. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now