Tog

2009-Sep-24, 10:32 AM

Please keep in mind that my two years of algebra 1 were a very long time ago. I want to see if I did somethig right, because in the recent threads on the lottery, I got very different answers to some of the probability questions.

First one:

Let's say that I have a collection of 5 blocks. Each has three white sides and three black sides. They have to go in a certain order though, so Block 1 will always be on top. When stacked, they can appear as any one of 2^5 (32) combinations. Is this correct?

Total options for color raised to the power of the number of items to be colored. Or do I have that backwards?

Next, what if there is one color per side, for a total of 6 colors per block? That should be 6^5 (7776) as I understand it.

Now, let's say that the face of each block has a letter, and that those letters can be any of the 6 colors, on a background of any of the 6 colors.

That makes 6^2 color options for each block, so we have 36^5 (60,466,176) options, right?

Now, we let the letter be any one of the 26 in English. So we have 26 possible letters, and 36 possible color combinations for them. Then, a black "A" on a black background will look the same as a black "O" on a black background, we'd need to subtract 26*6 options from the total, as they would look the same. That would mean that each block has (26^36-(26*6)) options, raised to the power of 5 for the number of blocks. Is this right? I ask because that number is HUGE.

If not, where did I go wrong?

Part two:

In a lottery, there are 50 balls numbered 1 to 50. 6 of these will be chosen to make the winning combination. What are the odds that the winning combination will be 1-2-3-4-5-6? The way I thought is should be done was 50*49*48*47*46*45, which gives me one in 11.44 billion. When i tried to do it that way in the Belgian lottery thread, I got numbers that a lot higher than what everyone else posted.

What did I do wrong there?

First one:

Let's say that I have a collection of 5 blocks. Each has three white sides and three black sides. They have to go in a certain order though, so Block 1 will always be on top. When stacked, they can appear as any one of 2^5 (32) combinations. Is this correct?

Total options for color raised to the power of the number of items to be colored. Or do I have that backwards?

Next, what if there is one color per side, for a total of 6 colors per block? That should be 6^5 (7776) as I understand it.

Now, let's say that the face of each block has a letter, and that those letters can be any of the 6 colors, on a background of any of the 6 colors.

That makes 6^2 color options for each block, so we have 36^5 (60,466,176) options, right?

Now, we let the letter be any one of the 26 in English. So we have 26 possible letters, and 36 possible color combinations for them. Then, a black "A" on a black background will look the same as a black "O" on a black background, we'd need to subtract 26*6 options from the total, as they would look the same. That would mean that each block has (26^36-(26*6)) options, raised to the power of 5 for the number of blocks. Is this right? I ask because that number is HUGE.

If not, where did I go wrong?

Part two:

In a lottery, there are 50 balls numbered 1 to 50. 6 of these will be chosen to make the winning combination. What are the odds that the winning combination will be 1-2-3-4-5-6? The way I thought is should be done was 50*49*48*47*46*45, which gives me one in 11.44 billion. When i tried to do it that way in the Belgian lottery thread, I got numbers that a lot higher than what everyone else posted.

What did I do wrong there?