Probabilities in the form of a logarithmic scale? Confused!

[Simonsky]

Here's the question:

A spinner is constructed so that the numbers 1 to 10 are possible and the probability of seeing a number k+1 is twice as likely as seeing the number k (for k=1,2,....9).

Find the probability of getting A) 1 B) 10 (answers correct to 3 decimal places).

I had to read this a few time before I got some inkling of it but I'm not sure I'm correct. I thought if the k+1th term is twice as likely as the kth term then we are dealing with a geometric series which increases by a ration of 2:1 so that would mean that the probability of 10 would be 2^9 x the first probability. All the probabilities must add up to 1. So I then thought I could divide 1 by 2^9 to get the first value which came out as 0.001953...which I left as 0.001 (should I have rounded it up to 0.002?) . I then multiplied this number by 2^5 to get 0.512. The answers in the book were 0.001 and 0,500 which looks they've rounded to significant figures rather than three decimal places but I might be wrong.

Not all all sure I'm correct here, so help appreciated!

 

[Dr.Peterson]

Here's the question:

A spinner is constructed so that the numbers 1 to 10 are possible and the probability of seeing a number k+1 is twice as likely as seeing the number k (for k=1,2,....9).

Find the probability of getting A) 1 B) 10 (answers correct to 3 decimal places).

I had to read this a few time before I got some inkling of it but I'm not sure I'm correct. I thought if the k+1th term is twice as likely as the kth term then we are dealing with a geometric series which increases by a ration of 2:1 so that would mean that the probability of 10 would be 2^9 x the first probability. All the probabilities must add up to 1. So I then thought I could divide 1 by 2^9 to get the first value which came out as 0.001953...which I left as 0.001 (should I have rounded it up to 0.002?) . I then multiplied this number by 2^5 to get 0.512. The answers in the book were 0.001 and 0,500 which looks they've rounded to significant figures rather than three decimal places but I might be wrong.

Not all all sure I'm correct here, so help appreciated!

My first thought is that this will turn out to be very unrealistic. But let's see.

Rather than start with probabilities, I'd work with something simpler. Suppose that the probability of 1 is x. Then the probability of 2 is 2x, and so on. The the sum of all 10 of these has to be 1. What is x? What do the terms you got add up to?

This amounts to what you said you would do, but I don't see what you actually did -- why divide by 2^9, and multiply by 2^5?

Their answers are rounded to the nearest thousandth.

And I was right that you couldn't really make this spinner! In fact, you couldn't possibly tell if you get a 1 or a 2.

 

[Simonsky]

My first thought is that this will turn out to be very unrealistic. But let's see.

Rather than start with probabilities, I'd work with something simpler. Suppose that the probability of 1 is x. Then the probability of 2 is 2x, and so on. The the sum of all 10 of these has to be 1. What is x? What do the terms you got add up to?

This amounts to what you said you would do, but I don't see what you actually did -- why divide by 2^9, and multiply by 2^5?

Their answers are rounded to the nearest thousandth.

And I was right that you couldn't really make this spinner! In fact, you couldn't possibly tell if you get a 1 or a 2.

Sorry.. I made a stupid typo, I multiplied the first term I got by 2^9 not 2^5. But was my method correct? Am I right in saying the probabilities form a logarithmic scale? using the method I used you get 0.001,0.002,0.004,0.008,0.016,0.032,0.064, 0.128, 0.256, 0.512 I which adds to 1.023 so slightly out. I was thinking there could be a method using logs but I can't remember it or I might be talking nonsense.

Using you method x+2x+3x....10x =55x so x= 1/55 =0.018 then the values would be: 0.018,0.036,0.072,0.144,0.288,0.576 then it gets to big???

 

[Dr.Peterson]

Sorry.. I made a stupid typo, I multiplied the first term I got by 2^9 not 2^5. But was my method correct? Am I right in saying the probabilities form a logarithmic scale? using the method I used you get 0.001,0.002,0.004,0.008,0.016,0.032,0.064, 0.128, 0.256, 0.512 I which adds to 1.023 so slightly out. I was thinking there could be a method using logs but I can't remember it or I might be talking nonsense.

Using you method x+2x+3x....10x =55x so x= 1/55 =0.018 then the values would be: 0.018,0.036,0.072,0.144,0.288,0.576 then it gets to big???

I wouldn't call it logarithmic; exponential, maybe. I don't think the name matters.

I asked why you divided 1 by 10^9; can you explain? If what you did had been right, I wouldn't have rounded at all; if I rounded, I would round to 0.002, not 0.001, and preferably to 0.00195. But you have the wrong divisor.

Note that when you multiplied by 10^9, you should have gotten 1 back, but your bad rounding gave you something close to 1/2 because 0.001 is close to half of 0.00195!

Following my method, it isn't x+2x+3x....10x =55x, is it? Didn't you mean x+2x+4x+...+512x? Add those up. (The result will look almost familiar.)

 

[Simonsky]

I wouldn't call it logarithmic; exponential, maybe. I don't think the name matters.

I asked why you divided 1 by 10^9; can you explain? If what you did had been right, I wouldn't have rounded at all; if I rounded, I would round to 0.002, not 0.001, and preferably to 0.00195. But you have the wrong divisor.

Note that when you multiplied by 10^9, you should have gotten 1 back, but your bad rounding gave you something close to 1/2 because 0.001 is close to half of 0.00195!

Following my method, it isn't x+2x+3x....10x =55x, is it? Didn't you mean x+2x+4x+...+512x? Add those up. (The result will look almost familiar.)

Ah, I see I stupidly added the x's without the doubling each time. What I'm still not clear about is why my dividing 1 by 2^9 is wrong, it seems to produce something very similar. Your x's now add to 1023x so x =0.0009775...which is close to 0.001. 2^9 = 512 so I've divided by a number that is half of yours!

Hmmm....I think my brain needs a rest-bedtime over here ! I'll ponder it tomorrow-thanks for help.

 

[Dr.Peterson]

Ah, I see I stupidly added the x's without the doubling each time. What I'm still not clear about is why my dividing 1 by 2^9 is wrong, it seems to produce something very similar. Your x's now add to 1023x so x =0.0009775...which is close to 0.001. 2^9 = 512 so I've divided by a number that is half of yours!

Hmmm....I think my brain needs a rest-bedtime over here ! I'll ponder it tomorrow-thanks for help.

One thing to ponder: that sum, 1023, is 2^10 - 1. If you've learned about geometric series (or other things like binary numbers) you might see why that is. So if you had used 2^10, you would have been very close.