Can anybody demonstrate that options (a) and (b) don't have the same risk?

If you set up the game and liked deep fried testicles in ketchup and wanted to keep $1,000,000, how many red balls would you put in box B ?

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Can anybody demonstrate that options (a) and (b) don't have the same risk?

If you set up the game and liked deep fried testicles in ketchup and wanted to keep $1,000,000, how many red balls would you put in box B ?

When the balls were put in the box by the adjudicator and watched by Pat - there was 97 black balls and only 3 red balls

Pat feels safe that most will go with the A option - ie 50/50 - and so his money is safe

However Goldman Sachs get their mate to say here's $50k - can I have first pick - and I will go with option B

Pat feels really great - he gets $50k and is sure this guys is going to have to be really lucky to find one red ball - in amongst 97 black ones

What Pat did not realise that the special balls owned by GS convert from black to red within 1 minute of going into a dark closed box

The Punter wins - Pat's been ripped off - and GS want their balls back

A bit like trading FX really

Can't be answered!.....as Box B is unknown.Can anybody demonstrate that options (a) and (b) don't have the same risk?

Choice is between certainty of risk percentage V unknown risk percentage.

So 50/50.....or.....take a chance!

Can't be answered!.....as Box B is unknown.

Choice is between certainty of risk percentage V unknown risk percentage.

So 50/50.....or.....take a chance!

I don't think that is right. The risk is 50% ((0+100)/2)/100 for (b) which is the same as (a).

It does have an unknown distribution though.

I think it is 50/50 with a known distribution or 50/50 with an unknown distribution.

If that is the case, what is the difference and why the selection bias to (a) ?

Furthermore, we're comparing a % fact against unknown %, therefore cannot be answered.

Just because the choice is only between red ball v black ball doesn't mean this 50% applies to the both boxes. What is more important is from which box it comes from.....

Furthermore, we're comparing a % fact against unknown %, therefore cannot be answered.

Just because the choice is only between red ball v black ball doesn't mean this 50% applies to the both boxes. What is more important is from which box it comes from.....

If box B is filled with only red and black balls on a completely random selection then in the long term box B will be a 50/50 shot so you have no edge in picking one box over the other

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Its a study to show how people often prefer to gamble with a known chance of winning as opposed to not knowing what the chance of winning is ..

So whoever votes for Box A is more risk averse preferring the safer option even though the Box B has the same chance

If box B is filled with only red and black balls on a completely random selection then in the long term box B will be a 50/50 shot so you have no edge in picking one box over the other

I guess you only have one chance to choose.....right? so "long term" doesn't apply because this will be an average over time for box B.

Also you cannot define a concrete % for box B, therefore you cannot compare between box A and B together.....because they are different.

To say no edge isn't true....to say there is an edge isn't true either as box B is not known.

I see ISIL's UK frontman Abu ibn Mr. Charts has been in here with his Shar'ia obsession with removing testicles and painting em. I dunno about the rest of yous, but if the allies started doing videos of us removing captured ISIL militia's balls, I think we get a much quicker resolution to the problem.

I guess you only have one chance to choose.....right? so "long term" doesn't apply because this will be an average over time for box B.

Also you cannot define a concrete % for box B, therefore you cannot compare between box A and B together.....because they are different.

To say no edge isn't true....to say there is an edge isn't true either as box B is not known.

Yes but you're looking at it after the event, for example saying if I'd gone with box B I would have had 70% chance of winning so box B was the best choice.

If you were ladbrokes, what odds would you give someone else on pulling a red ball from box B (providing you didn't have inside info)?

Its a study to show how people often prefer to gamble with a known chance of winning as opposed to not knowing what the chance of winning is ..

So whoever votes for Box A is more risk averse preferring the safer option even though the Box B has the same chance

Best answer on here so far imo.

...........If you were ladbrokes, what odds would you give someone else on pulling a red ball from box B ................

If I were Ladbrokes I'd offer 4:5 on a red from box A and a mouthwatering 20:1 on a red from box B in the knowledge that all the balls in it were black

Yes but you're looking at it after the event, for example saying if I'd gone with box B I would have had 70% chance of winning so box B was the best choice.

If you were ladbrokes, what odds would you give someone else on pulling a red ball from box B (providing you didn't have inside info)?

No! not after the event but actually before the event.

Box B 70% is a made up number!....an assumption which isn't true before the event.

ladbrokes.....sorry, am not a gambler nor do I play dice!

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No! not after the event but actually before the event.

Box B 70% is a made up number!....an assumption which isn't true before the event.

ladbrokes.....sorry, am not a gambler nor do I play dice!

The first point is that if one is thinking in terms of the probability of a favourable outcome, then there is no reason to prefer a ball to be drawn from one pot rather than the other in either of the two variants of Pat`s game. In the case of Pot A, we know that the probability of drawing a red ball (or a black ball) is 1 in 2 (50%). This is also the case for Pot B, even though the precise number of red balls (and black balls) it contains is unknown.

To understand why it is necessary to consider what follows from the fact that all combinations of red and black balls are equally likely. This means that the probability that Pot B will contain (50 - n) red balls is identical to the probability that it will contain (50 + n) red balls (where n is any number between 0 and 50). The n’s here cancel each other out, leaving an overall probability of 50 out of 100 (or 1 in 2).

If that’s not clear, think in terms of particular numbers. For example, the probability that the pot will contain 49 red balls (and 51 black balls) is identical to the probability that it will contain 51 red balls (and 49 black balls); the probability of 48 red balls (and 52 black balls) is the same as it is for 52 red balls (and 48 black balls) and and so on.

This means that the average probability across all possible combinations of red and black balls that a red ball (or black ball) will be drawn (given that the pot contains 100 balls in total, and all combinations of red and black balls are equally likely) is 50 out of 100 - that is, 1 in 2 (50%).

This means that whichever pot was selected made no difference to the chance of a red ball being drawn,

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