ThePortForum.com

DRT wrote: ↑22:27 Tue 28 Feb 2017If the average rate of taint is 1 in 20 and you have 20 corks that are precisely average quality then with each re-corking from the same batch the chances increase. First corking is 1/20, second 1/19, third 1/18, etc. (unless one of those is the tainted one)

Incorrect. You're describing the wrong experiment. You're not using average corks, you're using a set of 20 corks in which one is guaranteed to be flawed. Because in your experiment, on the 20th cork there's a 1/1 chance of infecting the bottle (assuming you make it that far), while in reality there's about a 36% chance that a set of 20 bottles will be pristine even if each cork as a 1/20 chance of being tainted.

The correct way to run the experiment is:

(19/20) * (19/20) * (19/20) * ... etc., for as many times as you want to re-cork the bottle.

19/20 is the chance of a cork being clean. And every single (re-)corking must use a clean cork or your bottle will become tainted, so it's (19/20)^X where X is the number of times you want to (re-)cork the bottle. (Don't forget to include the original cork.)

(19/20)^20 is ~36% chance that the bottle is still clean after 19 recorkings (20 total counting the original cork).

(19/20)^2 is 90.25% chance that the bottle is still clean after 1 recorking (2 total corks). Or as Julian put it, a 9.75% chance that the wine as become damaged.

Put another way:

The chance of any given cork causing the wine to become tainted is independent, but the chance of the wine becoming tainted is cumulative because it doesn't matter which cork causes the problem... corked is corked.

Does anyone know where Phil's thread about port corks being branded on the ends is? This bottle of Senhora do Convento had a cork branded on the end with the date. Until the dog ate it.

Sorry to interrupt. Found it. Why can't you delete posts anymore?

Everything Glenn just said, I knew, but just couldn't explain. Honest.

Glenn E. wrote: ↑23:18 Tue 28 Feb 2017(19/20)^20 is ~36% chance that the bottle is still clean after 19 recorkings (20 total counting the original cork)

So how does that equate to the original assertion that re-corking a bottle once doubles the chances of it being corked?

I was taught that the best way to get your head round questions of probability was not to dwell on the chances of something happening, but on the chances of it not happening.

If you play Russian roulette with a six chamber revolver once, you have a 5/6 chance of survival. Play it twice and you have a 5/6 x 5/6 chance, or 25/36 chance or survival - which is slightly greater than the 2/3 chance (or 24/36) one instinctively assumes.

Similarly if you play it six times you are not assuredly dead - the chance of survival is 15625/46656 - which is very close to 1 in 3

DRT wrote: ↑22:55 Wed 01 Mar 2017

Glenn E. wrote: ↑23:18 Tue 28 Feb 2017(19/20)^20 is ~36% chance that the bottle is still clean after 19 recorkings (20 total counting the original cork)

So how does that equate to the original assertion that re-corking a bottle once doubles the chances of it being corked?

Round 1. 20 good bottles + 19 good corks + 1 tainted cork gives 19 good bottles + 1 corked bottle

Round 2. 19 good bottles + 1 corked bottle + 19 good corks + 1 tainted cork only gives 19 good bottles and 1 corked bottle if you are lucky enough to put the tainted cork in the corked bottle, else you get 18 good bottles, 1 newly corked bottle, and 1 old corked bottle with a new good cork in it.

DRT wrote:

Glenn E. wrote: ↑23:18 Tue 28 Feb 2017(19/20)^20 is ~36% chance that the bottle is still clean after 19 recorkings (20 total counting the original cork)

So how does that equate to the original assertion that re-corking a bottle once doubles the chances of it being corked?

(19/20)^2 gives a roughly 10% chance that the bottle is corked (90% chance that it is still clean), which is double the chance that it was corked with its original stopper.

Ergo, re-corking a bottle roughly doubles the chance of it being corked.

Note that "double the chance" is only roughly accurate because the chance of any given bottle being corked is so low to start with. If the chance were 50%, like flipping a coin, then re-corking a bottle would only increase its chances of being corked by 50%. (From 50% to 75%.)

DaveRL wrote: ↑14:12 Thu 02 Mar 2017Round 1. 20 good bottles + 19 good corks + 1 tainted cork gives 19 good bottles + 1 corked bottle

That is precisely the problem with this debate. The hypothesis being put forward relies on a captive batch of corks with a 1/20 taint rate.

My point is that because the first corking and the second corking are separated by many years or decades there is no connection between the corks and they will have vastly different taint rates. Each re-corking is an entirely independent action which has a singular chance of resulting in a corked bottle that is at the rate that is relevant to the batch of corks being used for that event.

Put another way - If you tossed a Farthing in 1945 and it landed on tails, what is the probability of a Trump Dollar landing on heads in 2017?

DRT wrote:Put another way - If you tossed a Farthing in 1945 and it landed on tails, what is the probability of a Trump Dollar landing on heads in 2017?

Except that isn't the question.

First, though, you are correct about DaveRL's scenario. His assumes a fixed set of non-independent corks which is not correct. The math I have used (as has jdaw1) assumes independent corks.

The correct time-spanning question is no different than what I asked before: what is the chance that you will get at least one head if you flip two coins?

The answer is 75%. Not 50%. 50% assumes you already know that the first coin was tails, but with a bottle of wine you do not know that.

Thus the formula I gave you.

Put another way: we will split the cost of a case of Port. To choose who gets each bottle, we will flip 2 coins. If either coin comes up heads, I get the bottle. If both are tails, you get the bottle.

Math say I will get 9 and you will get 3, on average. Mine are the corked bottles, though, so you shouldn't care.

If: (a) the first cork used has a 1 in N chance of causing the bottle to be corked, and (b) the second cork used has a 1 in M chance of causing the bottle to be corked, then the chance of the bottle being corked is 1-(((N-1)/N )*((M-1)/M)) with additional brackets used for the hard-of-priority. This can be continued as needed. N and M need not, but might be the same. Per earliest posts, with N=M=20, the probability of being corked is 9.75%.

flash_uk wrote: ↑09:14 Fri 03 Mar 2017 Who's on first?

Very funny.

Ha ha. I just knew I shouldn't get involved in a maths question.

Old Bridge wrote: ↑16:39 Fri 03 Mar 2017

flash_uk wrote: ↑09:14 Fri 03 Mar 2017 Who's on first?

Very funny.

+1

flash_uk wrote: ↑09:14 Fri 03 Mar 2017 Who's on first?

I'm going to concede and withdraw on this basis that now is not a good time in history to be compared to a mildly racist New Yorker

Ok - I get it. I have been thinking of the corks when I should have been focusing on the bottle. Each cork has the same chance, the bottle is being subjected to multiple chances.

DRT wrote: ↑20:24 Fri 03 Mar 2017Ok - I get it. I have been thinking of the corks when I should have been focusing on the bottle. Each cork has the same chance, the bottle is being subjected to multiple chances.

Exactly!

Now go have some Port. I'm drinking 1995 Souza at the moment.