Winning Streak Anomaly (I think!)

evolving_doors

Basically for the 'Dare to Share' part of the game Marty says you have a 1 in 3 chance of getting the top prize each time. Something doesn't sit right with me for some reason I can;t put my finger on. here''s the gist of the game..

1. Player selects colour of balls depending on the level of risk.
2. Marty positions the 3 corresponding balls for that colour to drop down into the ring.In each ball it could be 50k,1k and 1k
3. Marty releases the 3 balls.. But (here's the niggly bit)..
4. Only two of the balls ..i.e. the outside ones can come around and drop down the hole to be revealed by marty.

My quandry here is; as the inside ball always comes after the other two outside balls, it never has a chance of dropping down.
So when stokes Kennedy Crowley (or whoever) mixes the 3 balls at the start one of the balls can never be chosen. In effect it is removed form the game.

Another analogy would be if I have 50eur 1eur and 1 eur in boxes. Mix them all up...take one box out and throw it away.. then offer you 2 boxes and tell you that you have a 1 in 3 chance of getting the 50eur. (When I could have well chucked the 50eur away so you've no chance of getting it really!).

Is it true to say you have a 1 in 3 chance when the game commences. Or is it like you have a one in 3 chance before the game starts but when marty flicks the switch your odds have changed.

Apologies I can't find a short video of this online.

fergalr

Armelodie wrote: »

Basically for the 'Dare to Share' part of the game Marty says you have a 1 in 3 chance of getting the top prize each time. Something doesn't sit right with me for some reason I can;t put my finger on. here''s the gist of the game..

1. Player selects colour of balls depending on the level of risk.
2. Marty positions the 3 corresponding balls for that colour to drop down into the ring.In each ball it could be 50k,1k and 1k
3. Marty releases the 3 balls.. But (here's the niggly bit)..
4. Only two of the balls ..i.e. the outside ones can come around and drop down the hole to be revealed by marty.

My quandry here is; as the inside ball always comes after the other two outside balls, it never has a chance of dropping down.
So when stokes Kennedy Crowley (or whoever) mixes the 3 balls at the start one of the balls can never be chosen. In effect it is removed form the game.

Another analogy would be if I have 50eur 1eur and 1 eur in boxes. Mix them all up...take one box out and throw it away.. then offer you 2 boxes and tell you that you have a 1 in 3 chance of getting the 50eur. (When I could have well chucked the 50eur away so you've no chance of getting it really!).

Is it true to say you have a 1 in 3 chance when the game commences. Or is it like you have a one in 3 chance before the game starts but when marty flicks the switch your odds have changed.

Apologies I can't find a short video of this online.

So, out of curiousity, I put it into google and found RTE player here:
http://www.rte.ie/player/ie/show/10207642/

(winning streak, not my favourite program).
At 18:18 they start dare to share.

Looks fine to me.

They say the balls are loaded randomly.

The balls go through a complex mechanism of rolling down the device, but that doesn't matter. As long as the positions the balls are loaded in is random, and as long as the balls are effectively identical in terms of their physical properties, then it doesn't matter how the balls get from the top of the device to the bottom.

Yes, as you say one of the balls never goes anywhere. But as long as the balls were loaded randomly, then there's an equal chance that any ball is in any position. So the chance that any of the three balls comes out the end is 1 in 3.

This depends on you trusting that the balls are indeed 'loaded randomly' as the host says.

Its not as transparent as putting the balls in a drum, which we can see rotating, using a selecting machine that has been vetted by a third party.

It'd be even more transparent if they were to just put the three balls out on a table, and say 'which ball do you want, A,B or C' and then show you all three.


But, as long as you believe them when they say the balls are loaded randomly, and as long as the balls are physically identical, (such that the show can't, e.g. make the more valuable balls heavier, and make the machine prefer lighter balls), then the ball you get is random, from the perspective of the contestant and the host.


Obviously, the 'universe' 'knows' which ball is which: if its the case that the ball you want is the one left behind, well, then your chance of winning that ball is zero, even before one of the other two balls reach the bottom.


But, believing the hosts that the balls are loaded randomly, then from your perspective, no one knows whether the ball you wanted has been left at the top, or is falling down through the device.

As no one knows that, at the exact point that the host opens up the ball that comes out of the machine, revealing what ball it was, the chance of it being the ball you wanted, from your perspective (and from the hosts, assuming honest game setup) is 1/3. That's what matters to you.




Incidentally, I also calculated the averages of each of the sets of balls.

The (50,0,0) with the most risk has the highest expected value, of 16.6.
The (12,11,10) with the least risk has the lowest expected value, of 11
Its not a smooth decrease either.

Which combination you'd pick would depend on your utility function.

If you were super rich, you'd probably always pick (50,0,0).
If you were poor, you'd probably always pick (12,11,10).

Assuming the guest are smart, I guess you could say they are revealing how much they need money by their selection in this game - whether they can afford to gamble getting nothing, in return for a higher average prize.

evolving_doors

Ya cheers, I spse the thing is all the money is pooled and divided evenly at the end anyhow...so even if I was poor i could go for the biggie 50k safe in the knowledge that other winnings would compensate for my crazy gamblings! I think the contestants discuss with each other what colours they will select beforehand.

But getting back to the one in three odds....if we extend the process a little bit would the same principals apply...
Say there are one hundred balls before the start .98 are discarded and you are asked to pick 1 out of the remaining 2. just before you pick are your odds still 1 in 100 of picking a particular ball? Considering you have zero chances of getting that ball if it's been removed beforehand.

Mellor

Armelodie wrote: »

My quandry here is; as the inside ball always comes after the other two outside balls, it never has a chance of dropping down.
So when stokes Kennedy Crowley (or whoever) mixes the 3 balls at the start one of the balls can never be chosen. In effect it is removed form the game.

That doesn't change anything.

Another analogy would be if I have 50eur 1eur and 1 eur in boxes. Mix them all up...take one box out and throw it away.. then offer you 2 boxes and tell you that you have a 1 in 3 chance of getting the 50eur. (When I could have well chucked the 50eur away so you've no chance of getting it really!).

you could have thrown it away, meaning there's no chance or you could have throw away 1 euro, which increases the chance of getting 50 euro.
The average of these chances is 1 in 3.

Is it true to say you have a 1 in 3 chance when the game commences. Or is it like you have a one in 3 chance before the game starts but when marty flicks the switch your odds have changed.

Your odds never change, they are 1 in 3 right up until the prize is revealed.

But getting back to the one in three odds....if we extend the process a little bit would the same principals apply...
Say there are one hundred balls before the start .98 are discarded and you are asked to pick 1 out of the remaining 2. just before you pick are your odds still 1 in 100 of picking a particular ball? Considering you have zero chances of getting that ball if it's been removed beforehand.

If the balls are removed at random, then your odds don't change.
Lets say there's one ball with a prize inside, the rest are empty.

98% of the time the prize is discarded and you get nothing.
1% of the time, the ball remains, but you pick the wrong one and get nothing.
1% of the time, the ball remains, and you pick the right one and get the prize.

You get the prize 1 time out of 100. Just like if you picked a ball from the start.

Mellor

There's a famous game theory problem sometimes called "The game show problem" where you make a 1 in 3 choice are asked if you want to switch your choice after the host discards one of the others.
In that case you should switch because the odds do change, - because the host is aware which prize is which, an important aspect that doesn't apply on winning streak.

Yakuza

^^^
It's best known as the Monty Hall problem: http://en.wikipedia.org/wiki/Monty_Hall_problem

Mellor

Yakuza wrote: »

^^^
It's best known as the Monty Hall problem: http://en.wikipedia.org/wiki/Monty_Hall_problem

I'm aware of that. I did say "sometimes".
I felt "Game Show" was more fitting in relation to the OP

fergalr

Mellor wrote: »

There's a famous game theory problem sometimes called "The game show problem" where you make a 1 in 3 choice are asked if you want to switch your choice after the host discards one of the others.
In that case you should switch because the odds do change, - because the host is aware which prize is which, an important aspect that doesn't apply on winning streak.

Are you sure its a good idea to introduce to monty hall problem, to a discussion where we are trying to explain to the OP basic ideas about subjective probability?

The monty hall problem seriously confuses people with a lot more experience dealing with basic probabilities.

Mellor

fergalr wrote: »

Are you sure its a good idea to introduce to monty hall problem, to a discussion where we are trying to explain to the OP basic ideas about subjective probability?

The monty hall problem seriously confuses people with a lot more experience dealing with basic probabilities.

I didn't think it was a good idea tbh, but I felt it was going to come up eventually, so I felt it was better to bring it up and point out the key difference early, rather than somebody else just dropping the link.


FWIW, i think the MH problem is pretty simple when its explained properly.


OP, ignore the monty hall problem until you figure out winning streak. Do one at a time.

fergalr

Armelodie, is this making any more sense to you than it did at the start?

Let me consider your most recent comment:

Armelodie wrote: »

But getting back to the one in three odds....if we extend the process a little bit would the same principals apply...
Say there are one hundred balls before the start .98 are discarded and you are asked to pick 1 out of the remaining 2. just before you pick are your odds still 1 in 100 of picking a particular ball? Considering you have zero chances of getting that ball if it's been removed beforehand.

That's a reasonable question to ask.

The answer is well understood, though, and if you persevere, it'll make sense in the end - but its not totally obvious at the start.



It might help think about a very simple game.

Lets say:

1) We have two cards, a King and a Queen.

2) In our game, the King is 'the winning card' because of its higher value.

3) We are sitting at a table.

4) The two cards are shuffled, and put face down on the table in a row, so that neither of us knows which card is which.


Lets consider the following scenarios:


Scenario 1:

You pick one of the two cards, by choosing one of them.

You are about to turn the card you picked face up, and find out whether you got the King.

What is the chance that you have the winning card?



Scenario 2:

You flip a fair coin, planning that you will pick the left hand card if the coin lands heads, and the right hand card if the coin lands tails.

It lands tails so you pick the right hand card.

You are about to turn the card you picked face up, and find out whether you got the King.

What is the chance that you have the winning card?


Scenario 3:

With intent to make you lose the game, I push one card forward towards you that I have chosen. (Remember, I don't know which card is which, either).

I throw the other card away.

You are about to turn the card I picked face up, and find out whether you got the King.

What is the chance that you have the winning card?



Scenario 4:

You always pick the left hand card.

You are about to turn the card you picked face up, and find out whether you got the King.

What is the chance that you have the winning card?



Scenario 5:

The rules of the game are that the right hand card is always thrown away before the game starts.
You are left with the left hand card.

You are about to turn the card you got face up, and find out whether you got the King.

What is the chance that you have the winning card?




I hope that if you think about it, you realise that the chances that you have the winning card, in all of the different scenarios, is the same. Its 0.5 or 50% or 50/50.


The reason for this, is because the cards were randomly shuffled at the start.

As a result of this random shuffling, neither you nor I in this game have any basis with which to distinguish which card is the King and which card is the Queen.


Given the information that we have, there is no reason to suppose that the card you are about to turn over is any more or less likely to be the King than it is to be the Queen. This is what we mean, when we say that the probability is 0.5


Now, it is true, that once you have picked a card, you have already won or lost the game. By the time that card is in your hand, the outcome of the game has been decided. An observer, watching through a camera underneath the table (like in TV poker) would be sure whether you were going to win or lose.


Equally, an observer who had perhaps marked the top of the cards, with a subtle mark, that neither you nor I could perceive, consciously or unconsciously (maybe you need special glasses to see it) could know whether you were about to win or lose the game.


But, from your perspective as the player, and my perspective as the player, the probability that you are about to win or lose, remains 0.5 until you turn over the card.


If each of us had to bet on the outcome of the game, it would be rational for us to bet if we got better than even money odds on the outcome.


Does this make sense?

This is using the definition of probability as the subjective degree of belief that its rational to have in a statement such as 'I have the winning card in my hand.'
I think thats a useful way to think about probability.

There are other ways to think about probability, but they lead to the same conclusion here: if you were to play the game a great many times, and always say, when you have the card in your hand, but before you turn it over, in each of those scenarios 'I am going to win' you would be right approximately half the time (0.5).



I hope that considering each of those scenarios in a simple game is helpful.

I hope you can see how the different scenarios correspond to scenarios between what you saw on winning streak, and what you intuitively understand about probability, such that you can see there isn't any one of those scenarios where you think the probability suddenly changes.

If you really want to be convinced, you could get some cards and a friend, and play each of those games, take records, and see how it works out in the long run - you'll find that in every scenario, the proportion of times you win approaches 0.5 as the number of games you play gets large.

dh0011

OP there is a well known problem in statistics called the monty hall problem. Google it and it might make things clearer for you.

evolving_doors

fergalr wrote: »

Are you sure its a good idea to introduce to monty hall problem, to a discussion where we are trying to explain to the OP basic ideas about subjective probability?

The monty hall problem seriously confuses people with a lot more experience dealing with basic probabilities.

Ok Folks, thanks a million for yissr replies... funnily enough the Monty Hall Problem was in the back of my head when I initially posted as I was viewing it as 2 consecutive scenarios (I accept that on W.Streak though that a 'choice' is not given to the player for the 3(or 2!) balls, in that the result is achieved by the machine 'chosing' ) i.e....
Scenario 1. load the 3 Balls (A goat,A goat, A Car)
Scenario 2 Middle Ball is removed ..Have the odds changed?

I accept now that the odds are still one in 3. I suppose i feel sorry for that middle ball that can never get chosen! Am I the only one here who is nagged by the 'removal' of that ball during the game?

Speaking of the Monty Hall.. is the Deal or no Deal scenario at the end where the banker offers the player to switch essentially the same as the Monty Hall... in that you should always switch?

fergalr

Armelodie wrote: »

Ok Folks, thanks a million for yissr replies... funnily enough the Monty Hall Problem was in the back of my head when I initially posted as I was viewing it as 2 consecutive scenarios (I accept that on W.Streak though that a 'choice' is not given to the player for the 3(or 2!) balls, in that the result is achieved by the machine 'chosing' ) i.e....
Scenario 1. load the 3 Balls (A goat,A goat, A Car)
Scenario 2 Middle Ball is removed ..Have the odds changed?

In W. Streak, even if the player was allowed choose which ball to remove, it would make no difference. The reason it would make no difference is the player has no information about which ball is which. Therefore, any choice they make is in effect random. So you might as well always remove the middle ball, as its random which ball is which. If the player chose a random ball to remove, it'd be the same.

What changes things with the Monty Hall problem is that it is not the case that a *random* door is removed. Rather, it is guaranteed that a *losing* door is removed.

The host, who chooses which of the remaining 2 doors to remove, has *extra information* about which prize is where, which he uses.
If, in the monty hall problem, the contestant, who did not have any information about which prize was where, was asked to choose one of the remaining doors to remove, then it would make no difference whether the contestant subsequently choose to stick or switch.

The reason is because the contestant has no information about which prize is where.

The monty hall problem can be hard to get your head around until you get practice thinking about probability.

If in doubt, perhaps write out each of the scenarios, and each of the paths through the scenario, to get a good sense of what is going on.

Armelodie wrote: »

I accept now that the odds are still one in 3. I suppose i feel sorry for that middle ball that can never get chosen! Am I the only one here who is nagged by the 'removal' of that ball during the game?

Assuming the balls are all positioned randomly at the start, then there is nothing to be worried about.

*However* if the balls are *not* positioned randomly - if the game show hosts are deciding to cheat - then it would be dodgy. They could cheat by placing the winning ball on the place where it will get left out, say, 9 times out of 10. This would mean that the chance of the person winning the big prize (in the 50,0,0 scenario on winning streak) would be greatly reduced.

So, setting up the gameshow in such a way that they choose which ball to leave out, and so that you aren't shown the balls being randomly put into place, might nag you, because it makes the game look less transparent.

Clearly, if you could see the balls being put into place by a big rotating drum, or some other mechanism that looked obviously random, then if they wanted to cheat, it'd be harder for them to. (although still pretty easy, in fairness).


However, as long as you trust them when they say that the balls are all positioned randomly, then it shouldn't bother you.

A lot of probability stuff is unintuitive, so don't feel bad if you are bothered by something you shouldn't be. I'm still trying to build my intuition in this area, and have spent a good bit of time reading about it.

Armelodie wrote: »

Speaking of the Monty Hall.. is the Deal or no Deal scenario at the end where the banker offers the player to switch essentially the same as the Monty Hall... in that you should always switch?

Don't know - what happens in deal or no deal?

CJC86

Armelodie wrote: »

Speaking of the Monty Hall.. is the Deal or no Deal scenario at the end where the banker offers the player to switch essentially the same as the Monty Hall... in that you should always switch?

No, the Monty Hall problem is quite different from what happens at the end of Deal or No Deal.

I think there are 30 boxes with different values in them at the start of the show, so the box that you have has 1/30 chance of being the top prize. If you then work out the probability that you don't select the top prize with your 28 guesses it works out as (28/29)*(27/28)*(27/26)*...*(1/2)=1/29, but that is assuming that you didn't have the box, which happens 29/30 times, so the odds that either box has the top prize in it is exactly 1/30 (at the start of the game), hence the top prize is equally likely to be in either box.

Mellor

Armelodie wrote: »

Speaking of the Monty Hall.. is the Deal or no Deal scenario at the end where the banker offers the player to switch essentially the same as the Monty Hall... in that you should always switch?

The player didn't know where the top prize was when he was picking, so it was just as likely to be in any single eliminated box. As it happened, it wasn't eliminated this time, and the remaining boxes have an equal chance of being top prize.