Not more eggs....

Skeeter said:
Ah OK I see your thinking (and error) now.

In your infinite game that plays until one of the 2 combinations appear, as soon as you spin RR the probability of a game ending RBR is then zero as the B immediately ends it with RRB.

You've incorrectly assumed this means the RRB will happen first. What you've missed is that the probability of spinning that RR is the same as spinning RB, or BR, or BB. Every spin is independent.

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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He's not incorrect, see the Wiki page I posted above about 'Penney's Game'.

Red, Black, Black is more likely to appear before Black, Red, Red by 2 to 1
 
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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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estebanrey said:
He's not incorrect, see the Wiki page I posted above about 'Penney's Game'.

Red, Black, Black is more likely to appear before Black, Red, Red by 3 to 1
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Penny's Game places the advantage on the player, not the specific choice.

In the OPs question he's asking which is more likely, there is no information about the order the possible outcomes were chosen in or players. See my post earlier about the OP not asking what he thought he did. The OP is not Penny's Game.

There are a lot more combinations than that wiki pic shows, a lot of which will put player 1 at the advantage.
 
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dowie said:
Chance of infinite Rs is 0

once you have RR the chance of the game ending RRB is 1
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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.
 
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Oops, screwed up myself :), I've edited my posts above so it is Red, Black, Black the optimal choice...doh!

Skeeter said:
Penny's Game places the advantage on the player, not the specific choice.
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The two are the same.

Skeeter said:
In the OPs question he's asking which is more likely, there is no information about the order the possible outcomes were chosen in or players. See my post earlier about the OP not asking what he thought he did. The OP is not Penny's Game.
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He IS using Penney's game. Unless he has edited his OP he is asking which permutation is most likely to appear FIRST if you spin forever until it happens.

This is exactly what Penney's game is all about.

Your earlier post isn't taking into account that the game ends as soon as one of the permutations appears.
 
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Recently I started flipping a coin and it was heads 17 times in a row. A standard 2p. Was beginning to question my reality until a tails popped up.
 
Penny's game says Player 2 has the advantage. Not what their choice is. Player 2 has an advantage in being able to make an educated choice based on what player 1 has chosen. Its not the same as simply picking 2 of the 64 possible pairs and asking which is more likely.

Edit: actually its 56 isn't it (8x7), as both players can't pick the same?
 
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Skeeter said:
Penny's game says Player 2 has the advantage. Not what their choice is. Player 2 has an advantage in being able to make an educated choice based on what player 1 has chosen. Its not the same as simply picking 2 of the 64 possible pairs and asking which is more likely.
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Why does it matter if it is two players picking the options or two options being presented before you? The end result is the same.

If someone picks Black, Black, Red then Penney's game says Red, Red, Black is twice more likely to happen. It doesn't matter whether that second choice was chosen by someone based on intelligence, or given to them like in the OP it is still has 2 times better odds.
 
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Interesting link estebanrey, will mull on that. You've definitely answered the OP though (choose RBB over RRB!).

The core thing it seems to me is that it's like rock paper scissors - one string might (on average) beat one other string, but pick two strings at random and you're back to 50/50.
 
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dowie said:
Red red black is more likely to occur first than red black black
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You're right, I messed up and hold my hands up. I didn't read your two options properly doh! I was comparing Red, Red, Black to Black, Red, Red (I've had a long day at work :D).

Red, Red, Black is more likely than Red, Black, Black by 2 to 1
 
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