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A player has 52 cards and dice. Every time he picks a card (without seeing it) throws the dice. How many cards at least does he have to draw to make sure that discovering them there are at least 4 cards of the same suit and that the same number appears at least 3 times in the sequence of the dice throw?

I'm struggling with this problem. Can anyone help me?

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    $\begingroup$ If the player only had the die, or only the deck of cards, could you solve the problem? $\endgroup$ – Arthur Jan 24 at 9:33
  • $\begingroup$ @Arthur Yes, I suppose $\endgroup$ – Jack Jan 24 at 9:34
  • $\begingroup$ And once you have the solution to both of those, can you see how to use that to solve this combined problem? $\endgroup$ – Arthur Jan 24 at 9:35
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The player wants to be sure to get 3 times the same dice number and 4 times the same suit. Then, we can look for the longest series without 3 times any number nor 4 times the same suit. Next die and next card will add will add a repeated 3rd time number or a repeated 4t time suit.

With a 12 dice series we could get the numbers 1 to 6 tow times each - worst case. We can be sure that 13th die will show one of those 6 numbers that have already appeared.

Similarly, with a 12 cards series we could get at worst three card of each suit. The 13th card will be a repetition of one of those suits.

Therefore, a 13 cards and 13 dice series is enough.

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