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My friends and I play a game with sixteen individual numbered balls. Only 13 people are allowed to play at any given time and 13 people have played all 3 games. It takes a 16 to win.

The same person has drawn the 16 3 times in a row from varies spots in the circle. First time drawing 3rd. Second drawing 8th. Third drawing 10th. Out of a possible 13 draws each time.

If I understand the basics there is a 15 to 1 chance of drawing the 16, but the odds would change by the position of the draw with fewer numbers to choose from if my understanding is correct.

It's too much of a brain tickle for my little mind to figure out. Once it went beyond 16x16x16 I knew I'd need help.

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  • $\begingroup$ You haven't quite made the rules of your game clear. Suppose all thirteen players pick once, without anyone guessing the location of the 16 ball. Do they continue to choose balls in the same order until someone gets it? $\endgroup$ – hardmath Nov 4 '15 at 13:08
  • $\begingroup$ If no one draws the 16 then the game starts over. Order is different every time since it's a once a day thing. $\endgroup$ – Blackthirteen Nov 4 '15 at 15:24
  • $\begingroup$ Sorry to belabor the point, but do you mean that if no one wins in the first start, the game starts over again that same day? Or does no one win that day? $\endgroup$ – hardmath Nov 5 '15 at 2:29
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Someone was likely to draw $16$ in the first round. The chance they drew it again is one in sixteen; the chance they drew it all three times is one in $256$. This would happen once a year if you played it every night.
But there are other coincidences that might have come up instead, (for example the first person to draw in each round gets the 16), so the chance of a coincidence is much better than one in $256$.
If you already had your eye on this player, the chance of them drawing three sixteens is one in $4096$

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