Suppose you have a fair die. (Most of them are these days.)
You roll it once and observe a 3. You roll it again and observe a 1. You roll it yet again and get a 4. In three throws you’ve observed 3 of the possible sides.
What if you keep going?
What are the expected rolls that it takes to see all six sides at least once rounded off to the nearest integer?
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If a coin that flips heads with probability p , q = 1 − p , then the expected trials to get a head will be
E(trials to get head) = p + 2 q p + 3 q 2 p + 4 q 3 p + . . . = p [ ( 1 + q + q 2 + q 3 + . . . ) + ( q + q 2 + q 3 + . . . ) + ( q 2 + q 3 + . . . ) + . . . ] = p [ 1 − q 1 + 1 − q q + 1 − q q 2 + . . . ] = 1 − q p [ 1 + q + q 2 + . . . . ] = 1 − q 1 = p 1
Let X i be the state that total i faces have not been rolled out yet, the probability of getting any i faces 6 i
E ( X 6 + X 5 + X 4 + X 3 + X 2 + X 1 ) = E ( X 6 ) + E ( X 5 ) + E ( X 4 ) + E ( X 3 ) + E ( X 2 ) + E ( X 1 ) = 6 6 + 5 6 + 4 6 + 3 6 + 2 6 + 1 6 = 6 6 0 [ 1 0 + 1 2 + 1 5 + 2 0 + 3 0 + 6 0 ] = 1 0 1 4 7 = 1 4 . 7 ≈ 1 5