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Six tosses are needed on average to get a 6.

Each of the first five tosses gives (1+2+3+4+5)/5 = 3

spots on average, plust 6 spots on last toss,

so total 5*3 + 6 = 21 spots on averate are expected.

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Nice Witzar. I had another approach which was to consider any expected dot rather than only 6...

Since the expected number is also added, the expected number of dots would be same for any number from 1 to 6.

Probability of each is 1/6

That means in 6 tosses all the numbers appear once (expected)... therefore, total dots = 1 + 2 + 3 + 4 + 5 + 6 = 21

This means that not counting the last dot, the expected numbers of dots for 1 to 6 are 20,19,18,17,16,15 respectively!

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Six tosses are needed on average to get a 6.

Each of the first five tosses gives (1+2+3+4+5)/5 = 3

spots on average, plust 6 spots on last toss,

so total 5*3 + 6 = 21 spots on averate are expected.

Nice Witzar. I had another approach which was to consider any expected dot rather than only 6...

Since the expected number is also added, the expected number of dots would be same for any number from 1 to 6.

Probability of each is 1/6

That means in 6 tosses all the numbers appear once (expected)... therefore, total dots = 1 + 2 + 3 + 4 + 5 + 6 = 21

This means that not counting the last dot, the expected numbers of dots for 1 to 6 are 20,19,18,17,16,15 respectively!

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