harey Posted April 29, 2022 Report Share Posted April 29, 2022 (edited) To decide who will do the dishes, we put in a jar tickets with prime numbers up to 111. The one who draws the ticket with a larger number wins. I look at my ticket: 3. Maybe you have a low number like 11... So I offer to exchange our tickets before showing them. What is the probability I will do the dishes? Edited April 29, 2022 by harey Quote Link to comment Share on other sites More sharing options...
0 harey Posted May 6, 2022 Author Report Share Posted May 6, 2022 OK, so here the solution. Spoiler If you have the highest number, you will not exchange it. Recursively, we can restart the game without the highest number. You will not agree to exchange as long as you do not have the lowest number. As I have the second lowest number, my only hope is that you have 2. However, I spoiled my last chance because in this case, you trade. So I will 100% do the dishes. Should you have problems with the recursivity: if you have the 2nd highest number, you will not exchange because you will not receive the highest number if you have the 3rd highest number, we have seen that higher numbers will not be exchanged ... Quote Link to comment Share on other sites More sharing options...
0 harey Posted May 2, 2022 Author Report Share Posted May 2, 2022 It seems you need a hint: Spoiler If you have the ticket with the number 109, would you exchange? Quote Link to comment Share on other sites More sharing options...
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harey
To decide who will do the dishes, we put in a jar tickets with prime numbers up to 111. The one who draws the ticket with a larger number wins.
I look at my ticket: 3. Maybe you have a low number like 11... So I offer to exchange our tickets before showing them.
What is the probability I will do the dishes?
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