andrewducker | Monty Hall

Why does the lack of MM make the MF possibilities less likely?

It doesn't. There is a 50% chance of FF or MF/FM. The location of Male or Female doesn't change the fact that each option is 50% as a whole. There is a 50-50 that the Male is in the 1st or 2nd slot so that is half of the 50% (25%) for each option.

Your method is not a Monty Hall problem. That is when you are given NEW information AFTER a decision has been made. This example gives informations BEFORE a decision has been made.

Nope. you want to know if you have two females.

Your scientist chooses one and checks it. You have information. You now want to know the odds of the other being female. That's a re-evaluation of the probability after you've been given information.

That doesn't seperate out the system.

Let me apply this to dice. I roll 2 dice. I reveal that one of them is a 6 but don't reveal which dice. The options are 1-6 2-6 3-6 4-6 5-6 6-1 6-2 6-3 6-4 6-5 6-6. You will notice that 6-6 only appears once in the list. Do you honestly believe that I have less chance of rolling a 12 than rolling a 7?

If you don't say which die is the 6 then you have to let both dice vary completely in your options. It should be:

1-6 2-6 3-6 4-6 5-6 6-6
6-1 6-2 6-3 6-4 6-5 6-6

not

1-6 2-6 3-6 4-6 5-6
6-1 6-2 6-3 6-4 6-5 6-6

6-6 has to appear twice, not once, in the possible outcomes because you're not telling me which dice you've looked at. Without position (whether it's dice or raptors) the odds are different than with position.

Edited 2011-08-17 18:48 (UTC)

Then you will understand that the revised explanation below I have illustrated marking the female are F* holds true.

You've actually got the answer right there:

Of the 36 equal possibilities of two dice, you eliminate the 25 that involve no sixes, leaving you with the 11 still-equally-likely where at least one die is a six.

Of those 11, one is 6-6, and the rest are not.

So your odds of the second die also being a 6 are 1/11.

Another one, since we're pulling out the dice:

Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6, the game is over. Sue rolls first. If she doesn’t roll a 6, Bob rolls the die; if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn?

Hints: The answer is not 1/6. It is not 5/36. It is not 125/1296.

The answer is 5/6 x 5/6 x 5/6 x 1/6

Nope! That's the odds of Bob winning on the second turn of any game.

What you want is, given that Bob won, what is the probability that he did so on turn 2?

Badly phrased, but given this then the answer is 1 x 5/6 x 1 x 1/6

Which is ofcourse 5/36 Which you have already thought was wrong. I'm afraid you clearly don't understand statistics.

You seem to be taking this personally, and I don't know why.

But no - "5/36" is still not the right answer. That's the odds of Bob winning in turn 2 if he's playing alone.

Er no. You just said to assume he wins. That makes Sue the loser. I don't have to take any of her rolls into account at all.

What you are saying is "Sue's FIRST roll is not a 6". Not the same thing at all.

You could calculate it that way, but it's harder, and now you have to filter out the games Bob doesn't win.

The fact that Sue can win in turn 2 changes what percentage of Bob's wins will come in turn 2.

Imagine Bob is playing alone. 1/6 of his wins will happens in turn 2, 5/36 of his wins will happen in turn 2, etc. Right? That's what you just calculated.

Now remember that, 1/6 of the time when you hit turn 2, Bob loses outright and you start over, and Bob loses that game, even if, had he been playing alone, he would have won it.

Sue's presence affects *when* Bob wins, even if we only consider the games where Bob wins. Sue is a spoiler.

Like I said in the other comment: Imagine that Bob is the 1000th player. For him to win at all, he needs to beat the odds massively, but we can run the trial for long enough that he gets a few wins.

In order for him to win on the second turn, though, he has to beat the odds *again*, just to get a chance. Most of the time, even IF Bob got a shot in turn 1, he'll never get a shot in turn 2 - so Bob's wins that come in turn 2 are going to be much rarer than his wins in turn 1, because of all the things that can go wrong with a game before Bob has a chance to win, right?

I already have filtered out all the other possibilities. That is why I multiplied the odds of rolling a 6 on the second time (1/6) by the odds of NOT rolling a 6 on the first time (5/6).

You clearly just don't understand this at all.

I already have filtered out all the other possibilities.

And in doing so, you have calculated the odds of Bob's win coming in turn 2 if nobody else is playing.

And that is why this is *the wrong answer*.

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chuma.livejournal.com - 2011-08-17 21:32 (UTC) - Expand

I'm afraid that's not correct. The odds of Bob winning ("BW") satisfy the equation

x = 1/6 + (5/6)(1-x)

so it's 6/11.

The odds of Bob winning on the second turn ("BW2") are as you calculated 5/6 x 5/6 x 5/6 x 1/6 = 125/1296.

So P(BW2|BW) = P(BW2 & BW)/P(BW) = P(BW2)/P(BW) = (125/1296)/(6/11) = 1375/7776.

This is wrong; 6/11 is SUE's chance of winning. Bob's is 5/11. So that's 275/1296. Sorry.

I realise this thread is all over the shop, so I will paste a bit from later on here.

"Also with regards to Sue and Bob, I think a better way to phrase it is "Of the times that Bob wins, what proportion will be on his 2nd turn?", which would rule out any ambiguity."

The way I read it was that there was a 100% certainty that Sue would lose rather than working out how often Bob would win on his 2nd turn of the times that he wins. One rules out Sue's dice rolls and the other is dependant upon them but only when the outcome is in favour of Bob.

The thread got derailed because it was about English rather than Maths; a different kind of expression :)

Wrong again, because that fails to account for the possibility that, in any given game, Sue can win in turn 2.

You're correct that Sue's first turn is actually irrelevant: In the set of all games that Bob wins, Sue *will always* lose on the first turn. However, the fact that Sue can end the game in turn 2 will change the percentage of Bob's wins that will come in turn 2.

A more intuitive way of looking at Sue's influence on Bob's results is to imagine a thousand players, with Bob going last. In order for Bob to win *at all*, the first 999 players must fail, so we can discard those on the first turn - but on the second turn, they all must lose *again* in order for Bob to have a chance to win on turn 2.

Can you see how that will affect the percentage of Bob's wins that come on turn 2, versus turn 1 or turn 3+?

"Bob rolls a 6 before Sue."

This is definitive statement. If you want me to assume this then I never need to worry about what Sue rolls.

Nope. Still wrong, for the same reasons. The fact that Sue can win changes *when* Bob wins, even if we only care about games where Bob wins.

It's an unintuitive problem. Simulate it if you don't believe me - which will, by the way, also converge on the correct result of 275/1296.

It it only unintuative to you. I suggest if you want me to take you seriously from this point on you start linking me to some sort of reference online which actually backs up your way of thinking. Something that isn;t just a friend of yours on LJ but an actual statistical reference. Mainly as at this stage I think you are probably trolling rather than just being stupid.

If you don't believe me, simulate it. Or take a look in the comments of xkcd, where I originally got it.

One of the correct answers is here.

And the short version is Bayes' Theorem and the sum of an infinite series.

(rather than sum the infinite series, a more elegant solution: Consider the two die rolls as simultaneous. If the first is a 6 (6/36 of the time), Bob loses. If the first is not a 6 and the second is a 6 (5/36) of the time, Bob wins. If neither is a 6, the test repeats.

Therefore, you can view this as an infinite loop with 11 possible exits, 6 of which are Sue wins and 5 are Bob wins. Given that SOMEONE wins, we see that Bob wins 5/11 of all games.)

If you don't like that, then yank out your geometric progression rules and sum the infinite series the hard way. I'll wait, you'll get 5/11.

Now: You've already shown that, for any game, the odds of Bob winning in turn 2 are 5/6*5/6*5/6*1/6: not-6, not-6, not-6, 6. That's 125/1296.

So, the odds of him winning in ANY game in turn 2, divided by the percentage of games he wins, gives you (125/1296) / (5/11).

Which is 275/1296.

275/1296 of Bob's wins will come in turn 2, which means that, for any given game where Bob wins, the odds of his win happening in turn 2 are about 21%.

Mainly as at this stage I think you are probably trolling rather than just being stupid.

You're demonstrably, aggressively wrong about the dino babies, the pub coins, and Bob and Sue, and you're calling *me* stupid because of it. You have no leg to stand on with accusations of "trolling"

Edited 2011-08-17 20:40 (UTC)

See other comment save me repeating myself.