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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?

Perikles
2009-Sep-24, 12:23 PM
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?This one is easier to answer. You are going wrong because your probabilities are for drawing the balls in exactly that order.

If the balls may be drawn in any order,

the probability of the first one drawn is 6/50
Next is 5/49
Next is 4/48
.
.
Probability is thus (6*5*4*3*2*1)/(50*49*48*47*46*45)

Notice that the numerator is the number of ways which you can draw all 6 balls.

HenrikOlsen
2009-Sep-24, 05:42 PM
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.

Yes.



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?
Each block only has 6 sides, so each block can only show 1/6 of the total combinations unless you make the blocks as d36's, so it's still 6^5.

If you're allowed to replace blocks and have enough in all combinations, then yes, it's (6*6)^5.


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?
Each block still only has 6 sides:)

And the number is (26*36-26*6+6) for the number of possibly distinguishable faces (if you don't care which way they're turned) 26*36 for the total, minus 26*6 for all with same foreground and background, plus 6 to still have one of each of those as they are distinct from all the others.

But yes, if you make enough blocks to show all the different faces, (26*36-26*6+6)^5 is still rather big (about the number of cents the US debts increased by this year) though not quite so huge as the one you had.
299994111526176 vs. 49568141375015269721181880396619523032064223556864 19283186063131696782488282294831609344330470077634 02606646452892989290651724746590256872093164272502 99687172758162994271869033515775831445409020901810 23226251071267371099573295333307945765663761628832 00000

Tog
2009-Sep-25, 07:32 AM
Ahh. Thanks to you both. I'll look back at my original problem and see how things change. Funny thing is, I think I did it right the first time, then messed up the post above when I tried to summarize how I did it.:)