Simple Probability?

[Tretiak]

I haven’t read too closely the coin flip example. But to properly mimic the problem you need to flip three coins. The first two to determine EW (HT) and the third to determine who speaks let’s say H=1, and T=2.

So then you want to remove all those with WW* (* = H or T). As with two coins you will get 1/3 EE*, 1/3 EW*, and 1/3 WE* (approximately). But now here’s the kicker. If we know that E is spoken by one, but with equal proability it could be 1 or 2 then we have to eliminate all the EW2 (HTT) and all the WE1 (THH), however we can keep all the EE*. On average we will get rid of half the EW* and half the WE*, thus leaving us twice as many EE*.

Therefore, 1/2 is the correct answer. I suggest someonee try three coins with their coin flipping program.

[DrDoom]

*Originally posted by Tretiak *
I haven’t read too closely the coin flip example. But to properly mimic the problem you need to flip three coins

Sorry – but I don’t understand? Only two passengers got into the cab, not three. Why flip three coins?

[Tretiak]

The third coin is to see who answers the cabbie’s question.

[MinkMan]

Tretiak said:

I suggest someonee try three coins with their coin
flipping program.

If you look at the perl program I posted on page two, it does just that, here’s the relevant portion:

$person_a_wants = $direction[coin_toss];
$person_b_wants = $direction[coin_toss];
$first_speaker_is = $people[coin_toss];

The results of the program confirm the 50/50 result. The program flips the coins, and disregards outcomes where the first speaker picks west, regardles of what the second speaker picks, which I think is the correct interpretation of the OP.

[DrDoom]

*Originally posted by MinkMan *
**Tretiak said:

I suggest someonee try three coins with their coin
flipping program.

If you look at the perl program I posted on page two
<snip>

The results of the program confirm the 50/50 result. The program flips the coins, and disregards outcomes where the first speaker picks west, regardles of what the second speaker picks, which I think is the correct interpretation of the OP.**

Using the third coin flip to pick “a” particular passenger to speak first, especially on a 50%/50% basis, seems to be completely contrary to the OP. The OP only says that one of the passengers says “East”, no more. By flipping a coin to determine which distinct passenger, A or B, speaks first, clearly violates the OP.

[Kat]

DrDoom, you might want to think twice next time before starting threads in IMHO and GD to advertise threads in GQ.

Thanks bunches.

[Tretiak]

*Originally posted by DrDoom *
**
Using the third coin flip to pick “a” particular passenger to speak first, especially on a 50%/50% basis, seems to be completely contrary to the OP. The OP only says that one of the passengers says “East”, no more. By flipping a coin to determine which distinct passenger, A or B, speaks first, clearly violates the OP. **

No, the coin flip perfectly emulates the OP. Since one of them does speak (i.e. heads or tails does occur), but the cabbie does not know which one, he has to assume they are both equally likely.

[DrDoom]

*Originally posted by Tretiak *
**No, the coin flip perfectly emulates the OP. Since one of them does speak (i.e. heads or tails does occur), but the cabbie does not know which one, he has to assume they are both equally likely. **

The cabbie doesn’t assume anything … he just hears “East” … he doesn’t know if it is 50%/50% as to who said it and doesn’t know which one said it …

[ZenBeam]

The cabbie doesn’t assume anything … he just hears “East” … he doesn’t know if it is 50%/50% as to who said it and doesn’t know which one said it …

The cabbie knows that one of them said it, and he knows that the other one didn’t say it. So he knows with 100% certainty (within the bounds of the question) that the one he heard is going east. He knows nothing about which way the other person is going. It doesn’t matter that he doesn’t which of the two passengers in the back seat is the one who spoke.

It also doesn’t matter whether each of the two passengers is equally likely to speak. If one of them is more likely to speak (e.g., the one on the left or the one with the bigger shoe size), as long as there is no correlation between who is most likely to speak and which direction that person is going, it’s 50/50 that the two passengers are going the same direction.

[Tretiak]

There is another way to think about the problem. It is not the identical problem but it has really the same rationale. Basically, the fact that one of them says East is more likely to happen if they are both going East.

Assume that the probability of one of them answering the question is 50/50. Let us assume that EE, EW, WE all happen with 1/3 probability.

What is the probability that one of them will say East, given that they are both going East? 100%

What is the probability that one of them will say East given that they are EW? 50%

What is the probability that one of them will say East given that they are WE? 50%

So, EE must be twice as likely as either EW or WE separately. Or equally likely as the two of them together. Ergo, 50/50.

[DrDoom]

The answer to this OP was, in fact, .3333 (1/3). Flip two coins, “reflipping” a WW (TT) result … CIAO

[bicyclops]

4 possibilities:

EE: Speaker says east, non-speaker goes east
EW Speaker says east, non-speaker goes west
WE: Speaker says west, non-speaker goes east
WW: Speaker says west, non-speaker goes west

Since the speaker said east, WE & WW are eliminated. The probability is 50%