Please put me out of my misery

joeguevara

Ok, I am driving myself nuts and have come up with a logic that has convinced me I am right even though I am probably as wrong as wrong can be.

Please can someone put me out of my misery.

My contention that if there is a raffle with 100 tickets and I have 99 tickets then I have a 99% chance of winning. That goes without saying.

But the next is where I’m coming unstuck. Is my chance of losing still 50/50 in that I have the winning ticket or I don’t. Or am I just bat sh1t crazy, which is more than likely true.

On a separate but similar note, (and forgive me if asked before) does buying 2 lottery tickets double my chance of winning. It obviously does improve chances but I am unsure if it doubles it. But the first one is the one that is doing my head in.

Thanks in advance for any feedback.

Pherekydes

joeguevara wrote: »

My contention that if there is a raffle with 100 tickets and I have 99 tickets then I have a 99% chance of winning. That goes without saying.

Yes.

But the next is where I’m coming unstuck. Is my chance of losing still 50/50 in that I have the winning ticket or I don’t. Or am I just bat sh1t crazy, which is more than likely true.

Look at the previous part: if you have a 99% chance of winning then the holder of the other ticket has a 1% chance. What's the chance that they lose? How could it possibly be 50%?

On a separate but similar note, (and forgive me if asked before) does buying 2 lottery tickets double my chance of winning. It obviously does improve chances but I am unsure if it doubles it.

Yes. It's just the same as above. The person with one ticket has 1% chance, but if they bought another ticket then they would have a 2% chance (assuming there was still only 100 tickets in the draw).

Yakuza

You are confusing a priori and a posteriori probabilities.

In the first case you are estimating the probability based on statistical likelihoods prior to the event happening. (99/100). The other case, after (post) the event, you either won or you didn't (50/50).

To answer your last question, yes, buying a second lottery ticket will double your chances of winning, but you can't guarantee if you will win it outright (you might have to share).

fergalr

Yakuza wrote: »

You are confusing a priori and a posteriori probabilities.

In the first case you are estimating the probability based on statistical likelihoods prior to the event happening. (99/100). The other case, after (post) the event, you either won or you didn't (50/50).

That post doesn't make sense, sorry to say; to anyone reading, don't try and follow it, it'll just be confusing :-)

Yakuza, post the event there's no '50/50' probability; because as you point out, you've either won or not. (Either 100% or 0%).

And prior to the event there's no reason to think the probabilities are 50/50.

Yakuza

I may have phrased it better (not at my most cogent post midnight) but what I was trying to convey (in a tongue-in-cheek manner) was that there are two events (winning and not winning) and if exactly one of them can come true, then the odds for each could be 50:50. In the real world, this is obviously nonsense that any rational argument can prove.

What I meant by "post" was that the draw has taken place (but you're not aware of the results) so you're in the Schrödinger's cat type of situation of either having won or lost.

This is more in the territory of philosophy and I realise I'm on much shakier ground in when trying to talk about it.

Edit : 5000th post. Maybe I'll stop now...(naah!)

fergalr

Congratulations on your 5000th post :-)

Yakuza wrote: »

I may have phrased it better (not at my most cogent post midnight) but what I was trying to convey (in a tongue-in-cheek manner) was that there are two events (winning and not winning) and if exactly one of them can come true, then the odds for each could be 50:50. In the real world, this is obviously nonsense that any rational argument can prove.

What I meant by "post" was that the draw has taken place (but you're not aware of the results) so you're in the Schrödinger's cat type of situation of either having won or lost.

I'm not sure how serious you are being here, and that's fine!; but to be completely clear:

Given the OP's setup:

100 tickets and I have 99 tickets then

Before the draw, you have a 99% chance of winning.

After the draw, you have either won or not - that's true.

If, after the outcome of the draw, you know the outcome of the draw, then your chance of winning is 100% if you find out you have won, and 0% if you find out you have lost.

If, after the outcome of the draw, you don't know the outcome of the draw, or any other information resulting from the draw, then your chance of winning is still 99% until you hear some relevant information.

Yakuza wrote: »

This is more in the territory of philosophy and I realise I'm on much shakier ground in when trying to talk about it.

The above would be entirely consistent and typical usage of a subjective interpretation of probability, which is what most folk would be using here.

Bayesian stats would be super comfortable here, but I think you'd be hard pressed to find a statistician to disagree with the framing above, pretty uncontroversial in practical terms. (historically, some folk might have argued you can't really make probability statements about one-off events and stuff, but that's pretty rare for people to say in practical settings now.)

Yakuza

fergalr wrote: »

Congratulations on your 5000th post :-)

Thanks

fergalr wrote: »

I'm not sure how serious you are being here, and that's fine!; but to be completely clear:

I was (in a bit of a ham-fisted way) and with my tongue in my cheek attempting to show how some folks could perceive a probability in that fashion. It veers of into the territory of philosophy rather than maths, and I'm not really qualified to comment on that field.

fergalr wrote: »

Given the OP's setup:

Before the draw, you have a 99% chance of winning.

100% agree, for avoidance of doubt!

fergalr wrote: »

After the draw, you have either won or not - that's true.

If, after the outcome of the draw, you know the outcome of the draw, then your chance of winning is 100% if you find out you have won, and 0% if you find out you have lost.

If, after the outcome of the draw, you don't know the outcome of the draw, or any other information resulting from the draw, then your chance of winning is still 99% until you hear some relevant information.

The above would be entirely consistent and typical usage of a subjective interpretation of probability, which is what most folk would be using here.

Bayesian stats would be super comfortable here, but I think you'd be hard pressed to find a statistician to disagree with the framing above, pretty uncontroversial in practical terms. (historically, some folk might have argued you can't really make probability statements about one-off events and stuff, but that's pretty rare for people to say in practical settings now.)

Agreed. Subjective probability was the phrase I was looking for - thanks! Obviously (to me at any rate) the notion of the probability of winning or losing being 50:50 is bonkers in the real world.

However...I work in an actuarial field and some of my colleagues' roles involve trying to predict the likelihoods of future events (lapses, claims, interest rate movements etc) partly based on past data (and perhaps some macroenomic information) and you could give the same data to two people and they could come up with different (but equally valid and well-argued) sets of predictions. But in well-defined situations like the OPs, the only rational argument for predicting the chances of winning are (#tickets owned / #tickets in total).

Stanford

Have a look at BAYES THEOREM

filbert the fox

joeguevara wrote: »

Ok, I am driving myself nuts and have come up with a logic that has convinced me I am right even though I am probably as wrong as wrong can be.

Please can someone put me out of my misery.

My contention that if there is a raffle with 100 tickets and I have 99 tickets then I have a 99% chance of winning. That goes without saying.

But the next is where I’m coming unstuck. Is my chance of losing still 50/50 in that I have the winning ticket or I don’t. Or am I just bat sh1t crazy, which is more than likely true.

On a separate but similar note, (and forgive me if asked before) does buying 2 lottery tickets double my chance of winning. It obviously does improve chances but I am unsure if it doubles it. But the first one is the one that is doing my head in.

Thanks in advance for any feedback.

It's not much of a raffle - the prize would wanna be really great to bother buying 99 tickets out of 100. Shudda bought the other one as well just in case!

MathsManiac

Going all the way back to the original post, I think the following might be why you couldn't get your head around what you were doing wrong.

Your reasoning when you said your chances of winning were 99% was that you have 99 out of the 100 tickets. Since 99 of the 100 outcomes favour you, you have a 99/100 = 99% chance of winning.

When reasoning that you had a 50% chance of losing, you reasoned that there were two possible outcomes: winning or losing. Since one of the two outcomes is of interest here, the probability is 1/2 = 20%.

In both cases, you are using the idea that the probability is the number of favourable outcomes divided by the total number of outcomes. As you're using the same basic idea in each case, why is it going wrong?

The answer is that this basic rule of probability only applies to EQUALLY LIKELY outcomes. When all outcomes are equally likely, the probability of an event is the number of outcomes in that event divided by the total number of outcomes in the 'sample space' or 'outcome space' (the 'universe' of the scenario). This rule does NOT apply when the outcomes are not equally likely.

When the outcome space is considered to be {ticket 1 is picked, ticket 2 is picket, ... , ticket 100 is picked}, these 100 outcomes can be taken to be equally likely - as the description of the situation involved makes that the most reasonable assumption - so the probability of (my ticket wins) is the number of outcomes in the set of my tickets divided by the total number of outcomes.

But when the outcome space is considered to be {I win, I lose}, there is no reason to believe these are equally likely, so you cannot use the same principle and it is therefore incorrect to conclude that the probability is 1 out of 2.