Fun Probabilty Problem.......

[ToraToraTora]

You have a bag of 10 keys, one of which opens a lock. You pick a key at random and try to open the lock. If the key you pick does not open the lock you must put it back in the bag shake them up and start again. On which try are you most likely to get the correct key?

[ClareTGold]

Sure, it's intuitively easy to argue as you say. Still, one of the ways you can prove it is that formula, which arises because this problem is a geometric distribution:

https://en.wikipedia.org/wiki/Geometric_distribution

[New Judge]

//So that's longer odds than the chance of getting it on the first attempt and it gets worse each attempt.//

Of course it is. Because once a trial succeeds, subsequent trials are no longer undertaken. So the likelihood of the second trial being successful is not solely a function of the chances of drawing the right key; it is also a function of the trial being undertaken in the first place. The chances of the one hundredth trial being the successful one are (if my arithmetic is correct) something a little over 2 in a million, which sounds preposterous. But for the 100th trial to be undertaken at all, the previous 99 must all have failed.

[Prudie]

No he said which is most likely and mathematically that means the probability. I get what you're all discussing but the fact is, taking the coins as example, if you throw 50 heads in a row our brains will tell us that the 51st is more likely to be another head and you can use your maths above to suggest why but the bottom line is the probability remains 50/50.

[ClareTGold]

I agree with Prudie that it's a little ambiguously worded. Comparing any given try to each other they must be the same as defined. These sorts of problem often have to be worded very carefully to avoid these sorts of trap.

[bobbinwales]

Of course your coin examples right prudie but it's different problem as your carrying on rather than stopping when you got a certain number of heads

[New Judge]

If you repeat Tora's problem with a coin, Prudie, there is less than one chance in a thousand of the tenth toss being (say) heads. But for the tenth toss to be heads, the previous nine must have been tails, otherwise the tenth toss would not have been made.

[ClareTGold]

You could also argue that if the first, say, hundred tosses of a "fair" coin all came up heads, then the probability that the coin is actually fair is much smaller than you thought...

[Arksided]

Still one in ten, then the toss is fifty fifty.

[Prudie]

I understand you are all saying the first try is the most likely to have the outcome you want because if you get what you want you give up. If that's simplified to a coin and say on which toss are you most likely to get a head then this argument says the first. So if you've tossed it once and got tails you are just a little bit less likely to get it on the 2nd etc however we all know the chance of getting a head on the 2nd remains 50/50. BUT we can use befuddling maths to disprove that in this case. That was my point really.

[dustypuss]

Call a locksmith.

[Bazile]

//On which try are you most likely to get the correct key?//

From the second one onwards

[ToraToraTora]

baz, did you read this thread?

[New Judge]

//So if you've tossed it once and got tails you are just a little bit less likely to get it on the 2nd//

No Prudie, that's not what this is saying. You are not less likely to get a tails on the second than you are on the first - the chances are 50:50 for both individual toss. But if you look at the two spins as a whole, the second is less likely to be a tails than the first because if the first results in tails then the second toss isn't made.

Look at it this way: toss a coin twice and there are four possible outcomes:

1: H - H
2: H - T
3: T - T
4: T - H

If you see Tails as a "success" (and you stop once tails is shown), options 3 and 4 will not see you toss again. Option 2 will see you "win" on the second toss whilst Option 1 will not see you win at all. So of those four possible outcomes, two have a chance of winning on the first toss (meaning you won't toss again) and one has a chance of winning on the second. So, **of the two tosses - if the first loses**, the first is more likely to be the winner than the second.

[Canary42]

The probability is identical for each try (i.e. 1 out of 10). The second try is unaffected by the first, and so on.

[ToraToraTora]

The BA is correct. People seem to be forgetting we are not talking about individual events we are talking about a desired outcome from the outset.

[ClareTGold]

Ambiguity, ambiguity everywhere...

There are, as I see it, three questions here.

1. "Assuming each key is equally likely to be chosen, and that the bag is indeed shuffled after every trial, what is the probability of picking the correct key on the Nth try, assuming that you have failed N-1 times?"

Answer: 1/10, although this is by definition, since this is describing a Bernoulli trial. https://en.wikipedia.org/wiki/Bernoulli_trial

2. "Prior to starting this experiment, what is the probability that it takes N trials to collect the key? What is the greatest value of this for a given N?"

Answer: p=(9/10)^(N-1)*(1/10), maximal at N=1. Again, this is more or less by definition, since we have set up a geometric distribution. https://en.wikipedia.org/wiki/Geometric_distribution

3. "What is the expected number of tries in order to collect the key, given that we are allowed as many tries as possible?"

Answer: 10 (or more generally 1/p for probability p of success). This can be intuitively seen as bobbinwales described, or proven using the trick I hinted at.

TTT presumably meant to ask (2), but it's clear that (1) is also a possible interpretation.