Ask The Wizard #333

On average, it will take one day for the first light bulb to be turned on.

From there, it will take half a day, on average, until the next significant event, either a new bulb being turned on or the first bulb burning out. We add 1/2 a day to the waiting time until that event. So, we're now at 1 + (1/2) = 1.5 days.

There is a 1/2 chance the second event was a second bulb being turned on. In that case, there is a 1/3 day waiting time until the next significant event (either one of the first two bulbs burning out or a new bulb being turned on). So, add the product of 1/2 (the probability of getting this far) and 1/3, which equals 1/6, to the waiting time. So, we're not at 1.5 + 1/6 = 5/3 = 1.66667 days.

There is a (1/2)*(1/3) = 1/6 chance the third signficant event was a third bulb being turned on. In that case, there is a 1/4 day waiting time until the next significant event (either one of the first three bulbs burning out or a new bulb being turned on). So, add the product of 1/6 (the probability of getting this far) and 1/4, which equals 1/24, to the waiting time. So, we're not at 5/3 + 1/24 = 41/24 = 1.7083 days.

Following this pattern, the answer is (1/1!) + (1/2!) + (1/3!) + (1/4!) + (1/5!) + ...

It should be common knowledge that e = (1/0!) + (1/1!) + (1/2!) + (1/3!) + (1/4!) + (1/5!) + ...

The only difference is our answer lacks the 1/0! factor. Thus, the answer is e - 1/0! = e - 1 = apx. 1.7182818...

I used a Markov chain in Excel to solve this problem. The following table shows the probability of seeing all 52 cards in 4 to 100 hands. The left column shows the number of hands. The middle column shows the probability the player sees the 52nd card in this many hands exactly. The right column shows the probability the player sees all 52 cards in this many hands of less. For example, the probability it takes exactly 20 hands is 4.64% and the probability it takes 20 hands or less is 84.63%.