Maths question with a festive twist

[cst]

Evening all,

Four sales reps visit you and each bring you a box of chocolates. Assuming there are four options to choose from (Quality Street, Roses, Miniature Heroes and Celebrations), what are the odds on all four bringing the same tin of chocolates.

Thanks in advance

[john2443]

cst said:

Evening all,

Four sales reps visit you and each bring you a box of chocolates. Assuming there are four options to choose from (Quality Street, Roses, Miniature Heroes and Celebrations), what are the odds on all four bringing the same tin of chocolates.

Thanks in advance

4x4x4x4= 1 in 256

[CoolHands]

Well, that was fun.

[GT119]

1in 64 as the first rep can bring any box so it's only the next three that need to follow suit

[GT119]

Ah, but is a tin the same as a box?

[cst]

That's what I thought, thanks for clearing that up. Others were suggesting other options and making me question if I was missing something, maybe the sugar so I just thought I'd check with pistonheads as they know everything. On a side note they were all celebrations and now we have one tub left and it has just Topics in it.

[AdamIndy]

Send them my way, I like topics!

In our office all we seem to have is quality street, the only ones left are those long toffees which I am slowly knawing my way through!

[john2443]

GT119 said:

1in 64 as the first rep can bring any box so it's only the next three that need to follow suit

Oh yeh!

[Skyrat]

john2443 said:

GT119 said:

1in 64 as the first rep can bring any box so it's only the next three that need to follow suit

Oh yeh!

Nope. It's an independent process, isn't it? John was correct.

[AmitG]

Skyrat said:

john2443 said:

GT119 said:

1in 64 as the first rep can bring any box so it's only the next three that need to follow suit

Oh yeh!

Nope. It's an independent process, isn't it? John was correct.

Surely it is 1 in 64.

There are 4x4x4x4 = 256 possible combinations.

Of these, there are 4 possible combinations where all the boxes are the same (because there are 4 types of box).

Therefore, 4 out of 256 combinations meet the criteria.

Therefore, odds are 4/256 = 1/64.

[CoolHands]

AmitG said:

Surely it is 1 in 64.

There are 4x4x4x4 = 256 possible combinations.

Of these, there are 4 possible combinations where all the boxes are the same (because there are 4 types of box).

Therefore, 4 out of 256 combinations meet the criteria.

Therefore, odds are 4/256 = 1/64.

To give a simpler example. The odds of throwing a (1) on a fair dice is 1/6. The odds of throwing the same number again is 1/6 * 1/6 = 1/36. it is not 2/36 which is what you are saying in your method above, (which would cancel to 1/18).

If we think of the boxes as 4-sided dice then it would be 1/4*1/4*1/4*1/4

So, in summary, you're wrong

[Sway]

CoolHands said:

AmitG said:

Surely it is 1 in 64.

There are 4x4x4x4 = 256 possible combinations.

Of these, there are 4 possible combinations where all the boxes are the same (because there are 4 types of box).

Therefore, 4 out of 256 combinations meet the criteria.

Therefore, odds are 4/256 = 1/64.

To give a simpler example. The odds of throwing a (1) on a fair dice is 1/6. The odds of throwing the same number again is 1/6 * 1/6 = 1/36. it is not 2/36 which is what you are saying in your method above, (which would cancel to 1/18).

If we think of the boxes as 4-sided dice then it would be 1/4*1/4*1/4*1/4

So, in summary, you're wrong

In your example, you're requiring the first dice throw to be a specific number. In the OP's example you don't care what box is brought, it only matters that subsequent boxes are the same as the first. Which means the calc is 4/4*1/4*1/4*1/4.

If you wanted the odds that all four were Quality Street, then the calc would be as per your example.

[AmitG]

I'm not good at maths so if I'm beyond help then tell me and I'll drop it

In your example (with dice), if the question was "what the odds of throwing two 3's" then the answer is indeed 1/36.

But if the question was "what are the odds of throwing the same number, regardless of what it is" the answer is surely 1/6.

Would you agree that in the two examples above, the probabilities must be different, because the second example there are more matching combinations than the first?

In the OP's example, if the question was "what are the odds that all 4 sales reps bring Quality Street" the answer would be 1/256. But any sweet is allowed.

[MethylatedSpirit]

CoolHands said:

AmitG said:

Surely it is 1 in 64.

There are 4x4x4x4 = 256 possible combinations.

Of these, there are 4 possible combinations where all the boxes are the same (because there are 4 types of box).

Therefore, 4 out of 256 combinations meet the criteria.

Therefore, odds are 4/256 = 1/64.

To give a simpler example. The odds of throwing a (1) on a fair dice is 1/6. The odds of throwing the same number again is 1/6 * 1/6 = 1/36. it is not 2/36 which is what you are saying in your method above, (which would cancel to 1/18).

If we think of the boxes as 4-sided dice then it would be 1/4*1/4*1/4*1/4

So, in summary, you're wrong

Depends on when you start:

If you have already rolled a (1) The odds of rolling the number again are 1/6

If you haven't rolled any dice, The odds of rolling (1) twice is 1/6 * 1/6

[CoolHands]

No, I would not agree (Amit).

Let's make it even simpler with two coins.

If you want to get two heads the odds (whether you wanted it or not) of getting a head on the first occasion is 1/2. The odds on the second occasion of getting a head, is again 1/2. The odds of theowing one head, then another head, are 1/2*1/2 =1/4.

If, on the other hand you said to me what are the odds of getting two coins the same, it would still be

1/2*1/2=1/4. (Because it could be a head followed by a head, which is 1/2*1/2, or a tail followed by a tail which is also 1/2*1/2).

Whether you specify it must be two heads, or just two the same, makes no difference.

Edited by CoolHands on Thursday 22 December 20:37

[Fastpedeller]

Ah, but the odds are skewed due to the local supermarket offering one of the 4 at a special price!

[Vizsla]

Sway said:

CoolHands said:

AmitG said:

Surely it is 1 in 64.

There are 4x4x4x4 = 256 possible combinations.

Of these, there are 4 possible combinations where all the boxes are the same (because there are 4 types of box).

Therefore, 4 out of 256 combinations meet the criteria.

Therefore, odds are 4/256 = 1/64.

To give a simpler example. The odds of throwing a (1) on a fair dice is 1/6. The odds of throwing the same number again is 1/6 * 1/6 = 1/36. it is not 2/36 which is what you are saying in your method above, (which would cancel to 1/18).

If we think of the boxes as 4-sided dice then it would be 1/4*1/4*1/4*1/4

So, in summary, you're wrong

In your example, you're requiring the first dice throw to be a specific number. In the OP's example you don't care what box is brought, it only matters that subsequent boxes are the same as the first. Which means the calc is 4/4*1/4*1/4*1/4.

If you wanted the odds that all four were Quality Street, then the calc would be as per your example.

Correct.

[a7x88]

CoolHands said:

No, I would not agree (Amit).

Let's make it even simpler with two coins.

If you want to get two heads the odds (whether you wanted it or not) of getting a head on the first occasion is 1/2. The odds on the second occasion of getting a head, is again 1/2. The odds of theowing one head, then another head, are 1/2*1/2 =1/4.

If, on the other hand you said to me what are the odds of getting two coins the same, it would still be

1/2*1/2=1/4. (Because it could be a head followed by a head, which is 1/2*1/2, or a tail followed by a tail which is also 1/2*1/2).

Whether you specify it must be two heads, or just two the same, makes no difference.

Edited by CoolHands on Thursday 22 December 20:37

Wrong

The odds of getting two coins the same is 1/2. Play out each scenario;

1 head and 1 tail
1 head and 1 head
1 tail and 1 head
1 tail and 1 tail

Half of those scenarios are the same coin. Not one quarter.

If you asked for 2 head then yes, it would indeed be 1/4.

Ergo - it does matter wether you specify two heads or two the same

[stefd]

Probability of all 4 being Quality Street = 1/4*1/4*1/4*1/4 = 1/256
Probability of all 4 being Roses = 1/256
Probability of all 4 being Miniature Heroes = 1/256
Probability of all 4 being Celebrations - 1/256

Any of the 4 outcomes above would satisfy the questions "what are the odds on all four bringing the same tin of chocolates?" so the odds are therefore the sum of the above:

Probability - 1/256+1/256+1/256+1/256 = 4/256 = 1/64

[Sway]

CoolHands said:

No, I would not agree (Amit).

Let's make it even simpler with two coins.

If you want to get two heads the odds (whether you wanted it or not) of getting a head on the first occasion is 1/2. The odds on the second occasion of getting a head, is again 1/2. The odds of theowing one head, then another head, are 1/2*1/2 =1/4.

If, on the other hand you said to me what are the odds of getting two coins the same, it would still be

1/2*1/2=1/4. (Because it could be a head followed by a head, which is 1/2*1/2, or a tail followed by a tail which is also 1/2*1/2).

Whether you specify it must be two heads, or just two the same, makes no difference.

Edited by CoolHands on Thursday 22 December 20:37

The first throw, all you're looking for is for the coin to land with a side up. That's guaranteed. So the odds are 2/2, or 1.

The permutations for two coins being tossed are:

1) h 2) h
1) h 2) t
1) t 2) h
1) t 2) t

So there are two desirable outcomes, out of four. So 2/4 or 1/2 is the correct odds of throwing two coins with the same result.