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You know the drill.

Only one of the doors is a winner; but this time there are four doors.

Behind three of them stand bill collectors.

Behind the other is your high-school sweetheart.

Monty Hall gives you the rules:

  1. You point to a door.
  2. Of the other three doors, at least two have bill collectors. I will open one of them.
  3. You decide whether to stay with your choice or pick another door.
  4. I will then open another bill collector door that is not your current pick.
  5. You then decide whether to stay with your choice or pick the only other unopened door.
What are the best odds for kissing once again your HS sweetie, and how do you get them?

Should you

  • stick / stick,
  • stick / switch,
  • switch / stick, or
  • switch / switch?
    1. You pick door 1.
    2. MH opens door 2: Bill Collector.
    3. You switch your pick to door 3.
    4. MH opens door 1: Bill Collector.
    5. You decide whether to stick with door 3 or switch to door 4.
    1. [*]You pick door 4.

      [*]MH opens door 1: BC.

      [*]You stick with door 4.

      [*]MH opens door 2: BC.

      [*]You decide whether to stick with door 4 or switch to door 3.

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Are you offering to beat me with your whiteboard? I decline. Don't take me wrong, I love violence, but it's not the most effective way to solve probability problems.

Bearing the whiteboard threat in mind, I have to concede my point. After you switch your first choice and AFTER Monty opens second door, the probability of your door to hide the prize becomes 1/3 if Monty opened your originally chosen door second time around and 3/7 if he opened another one.

See my post of another probability problem, which I derived from our exchange.

I suppose there's a lesson in the fact that you only agreed with me after I stopped trying to convince you.

And that lesson is: threats of violence will always solve all problems without any consequences or repercussions.

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I'm so sorry that I came to this thread so late. I sure looks like I missed all the fun. (For what it's worth, I think Chuck had it right on page 1).

Anyway, I have a unique twist to this problem that I'd like some help with...

My old High School Sweetheart IS a bill collector. I need to find only her (and get my ring back). Is a simple 1/4 chance?? The wrong door will ruin everything.

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Yeah, I finally got the message and I am eating crow with a nice dessert of humble pie. So, as a feeble attempt to redeem myself, let me see if I can explain the whole problem in simple terms that I understand.

Let's assume I decide from the beginning that I am going to stick with my original pick all the way (stick/stick). Then I have 1 chance in 4 of picking the right door at first. Probability of win = 1/4.

Now let's assume I change my strategy and go with switch/stick. My first pick has a 1/4 chance of being right and 3/4 chance of being wrong. There are two other doors to choose from when I switch, so each of them has half of the 3/4 chance of being right or 3/8 each. Since I am going to stick with this choice after the other door is opened, my probability of win with this strategy is 3/8.

If my strategy is switch/switch, then after the initial switch I have 3/8 probability of being right. When the next door is opened, the remaining unpicked door has the remaining 5/8 probability of being right so the switch/switch strategy has a 5/8 probability of win.

Finally, if my strategy is stick/switch, I am back to having 1/4 chance of being right after the first door is opened. When the next door is opened, the remaining unpicked door has a 3/4 probability of being right. So stick/switch has a probability of win of 3/4 and is the best strategy.

Also for the n door problem, sticking to the original pick to the last chance to switch and then switching at the end gives the (n-1)/n probability that has been discussed earlier.

I know I am not the first to put in all this and I want to give full credit to all those who put it in earlier. I just wanted to acknowledge my earlier error and summarize the correct results.

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