Maths question.

[ferrariF50lover]

Ok, to keep this car related, I'm going to tell a story:

It's a raffle. There are five Ferraris to be won. Tickets are £5 each and there are 100,000 tickets per draw and no limit to the number of tickets any one person may buy but I can't win more than one car.

Clearly, the initial maths is simple. 1 ticket = 1/100,000 chance of winning. 2 tickets means a 1/50,000 chance and so on up to five tickets where the odds are 1/20,000.

My question is this. If I buy five tickets, am I better off having all five in one draw (at one set of 1/20,000) or do I have one ticket in all five draws (5 sets of 1/100,000).

My maths suggests that the odds of winning are identical no matter that I do, but I was wondering if this might not be the case. It's a bit like the Monty Hall problem, where the answer seems obvious, but might not necessarily be right. For those of you who don't know, go here >>> http://en.wikipedia.org/wiki/Monty_Hall_problem

Thanks for the help.

Simon.

Edited by ferrariF50lover on Thursday 20th March 23:10

[Sir Humphrey]

I'd say one in all five as there's a chance you could win more than one Ferrari. But I haven't got the energy to sit down and work out probabilities at 11pm, so that's assuming the odds are the same.

[Hackney]

I'm sure someone will prove me wrong but I'd think all in one draw would be better.
The odds may be the same but if you put them in the same draw there are fewer "other" tickets in play, making your 5 tickets 5 in 100,000 and ignore the rest, rather than 5 in 500,000

[ferrariF50lover]

Sir Humph, I've amended my OP to reflect the fact that I can't win more than one car. Thanks for pointing that out, I'd forgotten to make that clear.

Simon.

[dazwalsh]

Hmm i would say that having all the tickets in one draw would be better, and i would take a 1 in 20,000 shot over 5 lots of 1 in 100,000 any day of the week.

Edited by dazwalsh on Thursday 20th March 23:18

[marshalla]

If you buy one ticket in each of five draws, then your odds of winning are 1 in 100000 across all 5 draws and 1 in 100000 in any one draw.

If you buy 5 tickets in one draw, then your odds of winning are 1 in 100000 across all 5 draws, but 1 in 20000 in the one draw you've entered and 0 in the other 4. Ergo, your odds of winning are better if you have all tickets in one draw.

The problem is that each game is really played independently of the others, unless you win. Spreading tickets across multiple draws would only make sense if you could win in each one, since you can't - if you in any one of them your chance of winning again automatically becomes 0 - because of this, it makes more sense to improve your chances in one game and ignore the others.

I think it's more like the complete inverse of the Monty Hall problem.

Edited by marshalla on Thursday 20th March 23:24

[AJB]

I think all in one would be better, but only because of the limit of winning one car max. If not for that, each ticket would be a straight 1 in 100,000 regardless if which draw it's in, but the 1 car max changes it I think.

Consider if you were buying 100,000 tickets not 5. If you buy them all in 1 draw, you'll definitely win 1 car. If you spread then evenly, you might not (4/5 chance of losing in each draw, so (4/5)^5 chance of losing every time =33% chance of no car.

If not for the 1 car max, spreading would give you a reasonable chance of winning more than one car which cancels out the chance of winning none. But, given you automatically lose every draw after you've won 1 car, you get the disadvantage of spreading without the advantage.

Having said that, I really don't like probability/stats, so I might be very wrong!

[AJS-]

If I'm reading it right there are 5 draws of 20,000 and a Ferrari to be won in each.

I think you're far better having 5 tickets in one draw than 1 in each of the 5 as each draw is independent and 1/20,000 times 5 is not equal to 5/20,000 (or 1 in 4,000). But it is always a bit of a head scratcher.

[Snowboy]

Make the numbers smaller and it's easier to see.
Imagine it's just 10 tickets per draw.

If you buy 5 tickets in one draw you have a 50% chance of winning.
If you buy one in each draw you have a 10% chance of winning.

Or, imagine just 5 tickets per draw.
Buy 5 tickets in one you are 100 guaranteed to win.
Buy 1 ticket in each draw and you probably won't win.

[ferrariF50lover]

Thanks very much to all respondents. You're all far too kind.

Simon.

[RizzoTheRat]

Snowboy said:

Make the numbers smaller and it's easier to see.
Imagine it's just 10 tickets per draw.

If you buy 5 tickets in one draw you have a 50% chance of winning.
If you buy one in each draw you have a 10% chance of winning.

Or, imagine just 5 tickets per draw.
Buy 5 tickets in one you are 100 guaranteed to win.
Buy 1 ticket in each draw and you probably won't win.

Interesting alterative way of looking at it but it screws up my maths

In the OP's example:

5 tickets in one draw = 0.99995 probability of not winning

1 ticket in each draw = 0.99999 probability of not winning in each separate draw so his probability of not winning any Ferraris is 0.99999^5 = 0.99995, exactly the same as before. Although there is a finite probability of winning more than one so this should be the better option.

However in Snowboys version with only 10 tickets:

5 tickets in one draw = 0.5 probability of not winning

1 ticket in each draw = 0.9^5 = 0.59049, so less likely to win than all in one draw.

What am I doing wrong? The probability of wining more than one is going to be a lot higher in Snowboys version, but I'm only looking at the probability of not winning any

[ikarl]

I tried explaining something similar to a friend a few years ago that my odds of winning the Euromillions changed when I realised that someone in the UK had won

Prior to the draw I think the odds are 1 in 76 million

After the draw has taken place, if the winner is announced as being from the UK, I have a 1 in 6 million (the odds of winning the raffle, so is directly proportionate to how many tickets are sold) if I have a ticket

[marshalla]

RizzoTheRat said:

Interesting alterative way of looking at it but it screws up my maths

In the OP's example:

5 tickets in one draw = 0.99995 probability of not winning

1 ticket in each draw = 0.99999 probability of not winning in each separate draw so his probability of not winning any Ferraris is 0.99999^5 = 0.99995, exactly the same as before. Although there is a finite probability of winning more than one so this should be the better option.

However in Snowboys version with only 10 tickets:

5 tickets in one draw = 0.5 probability of not winning

1 ticket in each draw = 0.9^5 = 0.59049, so less likely to win than all in one draw.

What am I doing wrong? The probability of wining more than one is going to be a lot higher in Snowboys version, but I'm only looking at the probability of not winning any

You're treating the draws as completely independent, but there is a conditional dependency - if you've won one, then your chances of winning any subsequent draw automatically become 0.

Edited by marshalla on Friday 21st March 15:18

[RizzoTheRat]

But I'm treating all cases the same by looking at the probability of not winning aren't I?

[CYMR0]

RizzoTheRat said:

But I'm treating all cases the same by looking at the probability of not winning aren't I?

Without doing the maths, the probability of not winning, once you have won once, is 1 on all remaining tickets.

[marshalla]

RizzoTheRat said:

But I'm treating all cases the same by looking at the probability of not winning aren't I?

I'm not so sure.

Case 1 : 10 tickets per draw, 5 draws. 5 tickets in first draw.
P(win1)=0.5
P(win2-5)=0
P(lose2-5)=1

P(lose1)=0.5
P(lose1,win2)=0
P(lose1lose2,win3)=0
P(lose1lose2lose3,win4)=0
P(lose1lose2lose3losed4,win5)=0


So you have a 0.5 chance of winning in draw 1, 0.5 chance of losing in draw 1 and a 100% chance of losing in draws 2-5. So, while you have a 0.5 chance of winning a car, that's only in the first draw, If you lose that, you have nothing left in the other draws.

Case 2: as above but 1 ticket in each draw.
P(win1)=0.1 = 0.1 (stop)
P(lose1, win2)=0.9 x 0.1 = 0.09 (stop)
P(lose1lose2,win3)=0.9x0.9x0.1=0.081 (stop)
P(lose1lose2lose3,win4)=0.9x0.9x0.9x0.1=0.0792 (stop)
P(lose1lose2lose3lose4,win5)=0.9x0.9x0.9x0.9x0.1=0.06561

Each one counts as a win, but some possibilities only exist IF the previous events didn't generate a win, so the chances of those possibilities existing depend on the previous sequence. The chances of a ticket winning a car in its own draw stays at 0.1, but the chances of you being eligible to win that draw are reduced each time there's a previous draw, because you have a chance of winning one of the earlier draws. In effect, some of the price of the tickets for later draws is wasted because there is a chance your tickets will not be eligible (and hence other tickets have a greater probability of winning a car in later draws).

P(lose1)=0.9
P(lose1lose2)=0.81
P(lose1lose2lose3)=0.792
P(lose1lose2lose3lose4)=0.6561
P(lose1lose2lose3lose4lose5)=0.59409 (worst outcome).

Your "best" option might actually be to buy 5 tickets for the last draw as you will be competing against fewer eligible tickets by then.

[RizzoTheRat]

CYMR0 said:

Without doing the maths, the probability of not winning, once you have won once, is 1 on all remaining tickets.

Again true but is that relevant? I've assumed they're sequential draws and worked on the probability of not winning in any of them, ie you only go on to the second draw if you haven't won the first, therefore winning in any of the individual draws should just be 1-P(lose) shouldn't it?

I do some of this kind of stuff for a living so I'm quite interested to work out where I've gone wrong and why.

[marshalla]

RizzoTheRat said:

Again true but is that relevant? I've assumed they're sequential draws and worked on the probability of not winning in any of them, ie you only go on to the second draw if you haven't won the first, therefore winning in any of the individual draws should just be 1-P(lose) shouldn't it?

I do some of this kind of stuff for a living so I'm quite interested to work out where I've gone wrong and why.

But you have to be eligible to proceed to each draw. i.e. you must have a ticket for it and NOT have won a previous draw.

(and, thinking about it, I think my numbers are off because they don't fully account for other people having multiple tickets for multiple draws - the chances of winning any one draw vary depending on how many entries the winner of a previous draw had).

Edited by marshalla on Friday 21st March 15:53

[RizzoTheRat]

Aren't I already taking that in to account by calculating the probability of losing all 5 draws though?

My probability of losing the second draw is 0.9x0.9, ie you have to lose the first draw to be able to lose the second draw, and lose the second to be able to lose the third (0.9^3) etc

You're also coming up with 0.59049 probability of not winning anything, so 0.40951 of winning a car, which draw you win on is irrelevant in that calc. This is clearly different from 5 tickets in one draw, yet the same maths for the 100000 ticket version gave the same answer as 5 in one draw.

Edited by RizzoTheRat on Friday 21st March 16:03

[marshalla]

Right, let's try again.

There are 5 draws, 10 tickets in each and 10 people. Each person buys 1 ticket for each draw. Each time there's a winner his tickets are taken out of all subsequent draws.

So, for any one person :

P(win1)=0.1 (1/10)
P(win2)=0.11 (1/9)
P(win3)=0.125 (1/8)
P(win4)=0.142 (1/7)
P(win5)=0.1666 (1/6)

P(win1)=0.1
P(lose1win2)=0.9x0.11
P(lose1lose2win3)=0.9x0.89x0.125
P(lose1lose2lose3win4)=0.9x0.89x0.875x0.142
P(lose1lose2llose3lose4win5)=0.9x0.89x0.875x0.858x0.1666

and repeat for as many combinations of draws, entrants and tickets as you like The chances of losing the 2nd draw aren't a nice neat 0.9 x 0.9 because someone else may be guaranteed NOT to be a winner, so your odds improve. The simple multiplication only works if you are the only one playing. (i.e. the only one who can be eliminated from any draw because of an earlier win). As soon as other players come into it, the odds change in every draw after the first one.

Edited by marshalla on Friday 21st March 16:13