Not more eggs....

[dowie]

I have, twice. As have most other people in here.

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]

I'd also say 50/50. Both combinations have an equal chance of occurring as each individual event is also 50/50.

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.

 

[Andreas.96]

Ah yes, compared to each other they're 50/50 but compared to other events they're not. That makes sense in my head anyway

 

[dowie]

Can you explain why you are correct?

Yes I can, I'll post it later if needed

 

[joeyjojo]

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..

This is the tricky bit - you keep spinning until a given combination comes up. But the question was "Which is the more likely combination to come up first" and the answer is, both combinations are as likely, because they're the same length (3) and red is as likely as black.

Now if you'd asked "what is the expected number of spins before getting one of these combinations?", that's a much harder question.

 

[Skeeter]

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.

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?

 

[kaiowas]

For RBB to come up first the preceding spin must have been black (if it was red you'd get RRB first). On that basis the second combination is less likely to come up first.

 

[dowie]

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.

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.

 

[Judgeneo]

Funny that.

 

[dowie]

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?

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.

 

[dowie]

For RBB to come up first the preceding spin must have been black (if it was red you'd get RRB first). On that basis the second combination is less likely to come up first.

Yup

 

[dowie]

Funny that.

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.

 

[Judgeneo]

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

6 is all you need, as that lets you show all 3 spin combinations with previous spins.

 

[dowie]

6 is all you need, as that lets you show all 3 spin combinations with previous spins.

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...

 

[Skeeter]

So what your saying is what is the probability of spinning a black after spinning 2 reds?

 

[joeyjojo]

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...

But "ending" with RR is just as likely as RB, BR, and BB.

Of course if you start with RR then getting RRB is very likely to happen soon, but there are so many possibilities that don't have an RR that it all averages out.

The point is spins are independent, so once you start going, any specific run of outcomes is as likely as any other (RRB, RBB, BBB, RRR, ...).

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.

 

[dowie]

Yes spins are independent

But once you've got RR... you're guaranteed to get RRB before RBB

 

[estebanrey]

Red, Red, Black

Or

Red, Black, Black

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

I think the OP maybe trying to put forward Penney's Game....

http://en.wikipedia.org/wiki/Penney's_game

Penney's game, named after its inventor Walter Penney, is a binary (head/tail) sequence generating game between two players. At the start of the game, the two players agree on the length of the sequences to be generated. This length is usually taken to be three, but can be any larger number. Player A then selects a sequence of heads and tails of the required length, and shows this sequence to player B. Player B then selects another sequence of heads and tails of the same length. Subsequently, a fair coin is tossed until either player A's or player B's sequence appears as a consecutive subsequence of the coin toss outcomes. The player whose sequence appears first wins.
Provided sequences of at least length three are used, the second player (B) has an edge over the starting player (A). This is because the game is nontransitive such that for any given sequence of length three or longer one can find another sequence that has higher probability of occurring first.

So if you swap 'tails' for 'red' and 'heads' for 'black'....

 

[Jono8]

Yes spins are independent

But once you've got RR... you're guaranteed to get RRB before RBB

So what?

 

[Skeeter]

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.

Once you have reached RR all you are asking is one simple thing, what is the probability of spinning B. Which is 1/2.