Not more eggs....

dowie

dowie

dowie

Caporegime
Joined
29 Jan 2008
Posts
59,094
In light of the previous two threads how about this one:

You're playing roulette in a casino, ignoring the green(zero), the numbers are coloured red and black. You've decided to keep playing until one of two combinations appear. Which is the more likely combination to come up first in the course of playing:

Red, Red, Black

Or

Red, Black, Black
 
Question isn't spin three times... question is which is more likely to appear first, you keep playing/spinning until one of the two combinations. For example you could have red, black, red, black... then one of the combinations...

Is it 50/50 as some are claiming or do some combinations have a higher chance than others of appearing first?
 
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What if I told you that the fact the spins are independent can still allow for some combinations to be more likely to occur first than the other.
 
I'm not saying spin three times what is the probability if this combination. I'm saying which of those two combinations is likely to appear first if you keep playing until one of them appears.

The probability of the two possibilities must total 1 so 1/8 for each is clearly wrong. If you think they're equally likely then it is 1/2, but that is also wrong, one is more likely than the other.
 
Skeeter said:
I have, twice. As have most other people in here.
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No you haven't, all you've shown is that if I spin three times then there is a 1/8 chance of either two occurring right away. You've not considered the other possibilities at all... remember we stop when one of the two combinations appears.

The total probability is 1, there are two outcomes... telling me the probability of one combination occurring in the first three spins doesn't answer the question.
 
Skeeter said:
Its 50/50 if you only look at those two events, but there are 6 other possible outcomes. However that is still 1/8 for each outcome.

So regardless of how you look at it, the chances of each happening first are the same.

Well, at least they are to everyone other than the OP.
Click to expand...

There are only two outcomes, you're still only looking at the first three spins.. there are infinite possible combinations that could lead to one of the two outcomes so talking about 6 'other' outcomes is irrelevant. It might take 3 spins, it might take 4, 5, 6 spins... you might get a run of 20 blacks to begin with.
 
Skeeter said:
I think you need to word the question better, as it looks like what you think you have written and what you have actually written are very different.

For example, what if the first spin is black?
Click to expand...

The question is fine, you keep playing until you see one of the two combinations, a black comes up first... well spin again... keep paying until one of two combinations appears.
 
Judgeneo said:
bp3Xton.png

Funny that.
Click to expand...

There are only two combinations, your spreadsheet has some unfinished games... For whatever reason you've chosen to stop after 6 spins.

For example any row of those unfinished games ending in RR is guaranteed to end the game with the first combination.
 
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Judgeneo said:
6 is all you need, as that lets you show all 3 spin combinations with previous spins.
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No

For example

every unfinished game from that spreadsheet ending in RR is guaranteed to end with the first combination. Also every unfinished game ending in BB benefits neither combination

You get RR, the next spin is black, first combination appears, game ends... you get another red.. well spin again, as soon as you get a black the first combination appears...
 
joeyjojo said:
Think about the likelihood of getting one R compared to one B? Evens. What about RR compared to RB compared to BR compared to BB? Evens. It's the same for any length.
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Also think about that last bit... combinations of two colours

simpler problem... which is more likely to appear first out of BR and RR?

Answer BR

Roll a B first and you're guaranteed to get 'BR' ahead of 'RR'

Roll an R first and you then have a 1/2 chance of getting another R...
Or you get a B and 'BR' is then guaranteed again.

So 3/4 chance overall of getting BR ahead of RR. 1/4 chance of getting RR ahead of BR.
 
Skeeter said:
Once you have thrown RR the chance of the game ending RBR is zero, but the chance of it ending RRB remains 1/2. You have assumed eliminating RBR makes RRB 100% certain, but it doeant. You could continually spin R every go forever and never finish the game.
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Chance of infinite Rs is 0

once you have RR the chance of the game ending RRB is 1
 
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Skeeter said:
The chance of the game ending RRB is 1, but your missing that the chance of the game ending at all is still only 1/2. The game cannot end RBR, but what your missing is the probability the game never ends, which takes the place of the RBR result (and its 1/2 probability) as soon as RR is spun.

The chance of throwing infinate Rs is not zero, and is actually irrelevant. What your after is the chance of the NEXT spin being R.
Click to expand...

I think you've got yourself in a complete muddle... the game will end, in most cases fairly rapidly... you can simulate it yourself if you don't believe me...

The probability that the game never ends is not 1/2

As for 'RBR' well you've got one 'R' towards either combination... though RRB is the more likely one...
 
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