Ask The Wizard #333

The answer is e - 1 = approx. 1.7182818..

Typically, we can anticipate that the first light bulb will be switched on within one day.

Subsequent to this, it’s expected to take roughly half a day for the next major occurrence to happen – either a new bulb being lit or the first one burning out. This adds half a day to our initial waiting period. Hence, our total waiting time now stands at 1 + 0.5 = 1.5 days.

There’s a 50% likelihood that the next event involves another bulb being turned on. If that happens, we anticipate a 1/3 day wait before the next notable occurrence, whether that’s one of the first two bulbs burning out or the activation of a new bulb. Therefore, we include the product of 1/2 (the chance of reaching this point) and 1/3, yielding 1/6, which raises the waiting time to 1.5 + 1/6 = 5/3 = approximately 1.66667 days.

The chance of the third significant event being the illumination of a third bulb is (1/2)*(1/3), giving us a 1/6 probability. In this instance, the wait for the next major event is a quarter of a day, accounting for either one of the first three bulbs going out or a new bulb lighting up. Consequently, we add 1/6 (the probability we calculated) multiplied by 1/4, resulting in 1/24, which adjusts our waiting time to 5/3 + 1/24 = 41/24 = about 1.7083 days.

Following this mathematical pattern, the solution can be expressed as the sum of (1/1!) + (1/2!) + (1/3!) + (1/4!) + (1/5!) + ...

It's widely recognized that the mathematical constant e can be represented as (1/0!) + (1/1!) + (1/2!) + (1/3!) + (1/4!) + (1/5!) + ...

The key difference here is that our solution does not include the 1/0! component. Thus, we arrive at the conclusion that the answer is e - 1/0! = e - 1, which is approximately 1.7182818...

The typical number of hands required to see every card in the deck amounts to roughly 16.41217418.

Utilizing a Markov chain in Excel aided me in solving this conundrum. The table below illustrates the probability of revealing all 52 cards within a range of 4 to 100 hands. The leftmost column signifies the number of hands played, the middle column indicates the likelihood of seeing the 52nd card precisely within that count of hands, and the rightmost column reflects the chances of having displayed all 52 cards in that number of hands or fewer. For instance, the probability of requiring exactly 20 hands is 4.64%, while the likelihood of accomplishing this in 20 hands or less is 84.63%.