Question

Probability of getting an ordinary cards

Answer 1

Assuming:

10≠J≠Q≠K≠A10≠J≠Q≠K≠A

There are thirteen different values of card, each with four copies. If we imagine this event *as* it’s unfolding, we can get a better intuition for what’s going on.

The first person draws a card, any card, with probability 100%100%.

The second person draws a card now. There are 51 cards left, 48 of which are different value, so he draws a card different from Person #1 with probability 48/51,48/51.

The third person draws. Now there are 50 cards left, and 44 of them are differently valued then #1 and #2. Probability to get a new card is 44/50,44/50

Finally, the fourth person draws: 40/49,40/49

So, if the odds of being successful are the product of those four values:

1,48/51,44/50,40/49