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This question already has an answer here:

A variation of the Monty Hall, where the values inside the choices are changed

You were just out gambling and lost 5000 dollars! Of course being the gambler that you are, you pay 2500 to play this game known as the "three boxes" You decide to play this because you're chances of getting the money back is 2/3! One box contains 7500 dollars, and another contains 20000! Of course the other contains a machine that you must insert 3000 dollars.

This could be the best risk of your life, or the worst. You get to pick a box, the host will then show you what is in one of the other boxes, he knows what's in there, and won't show you the 20000, once he does this you're allowed to keep, or change your answer, (even to the one he opens), explorer the 2 possible cases where he shows you the -3000 and where he reveals the 7500.

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marked as duplicate by xnor, McMagister, mdc32, A E, Tryth Jan 1 '15 at 23:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ whats the question? $\endgroup$ – Elgert Oct 21 '14 at 13:30
  • $\begingroup$ explorer the 2 possible cases where he shows you the -3000 and where he reveals the 7500 $\endgroup$ – warspyking Oct 21 '14 at 13:33
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    $\begingroup$ I like this variation because the answer depends on the utility of the money: If you need to win your money back, the strategy is different from when you can play this game 100 times for maximum winnings. $\endgroup$ – oerkelens Oct 21 '14 at 14:01
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    $\begingroup$ @warspyking Just a personal opinion: If you are going to do a minor variation, it might be better to use a more general one. This problem is simple as you always win and should always switch. The host can try to minimize you wins but not by much and you strategy is still fine. If you have A>B>C instead of specifically 20000>7500>-3000, you would see if there are other regions with potentially different behaviors. $\endgroup$ – kaine Oct 21 '14 at 15:02
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    $\begingroup$ @warspyking because they're minor variations of the original. Stack Exchange sites aren't forums or message boards; it isn't here to be "active", but to have high quality questions and answers. This might be an entertaining question, but it doesn't add much when there are very similar questions elsewhere. $\endgroup$ – TheRubberDuck Oct 21 '14 at 18:50
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If I want, I can always win my money back!

Not switching is academic: I have a 1/3 chance of picking the 20k, 1/3 of picking the -3k and 1/3 of picking the +7.5k. My expected win is 6.6k-1k+2.5k = 8.1k.

Blindly switching to the closed door raises my chances. The original Month Hall problem teaches me that I have 2/3 chance of picking the 20k by switching to the closed door.

If the opened door shows +7.5k, my expected win is 13.3k-(1/3*-3k)=12.3k
If the opened door shows -3k, mys expected win is 13.3k+(1/3*7.5k)=15.8k.

If I choose the opened door, obviously I'm dumb to take 100% chance for -3k.
If the opened door shows +7.5k, I am sure to win 7.5 k.

So, if I am going for the maximum win, I always switch to the closed door, which gives me the highest expected win.

However, here utility comes into play again!

If my goal is to get my money back, seeing as I put in 7.5k (I lost 5k, and paid 2.5k to play), I should choose the 7.5k if it shows behind the opened door. Otherwise, I switch to the closed door (100% chance of winning my money back (the -3k is behind the opened door then!) and 2/3 chance for 20k).

So, in case winning back my money is important (because the guy that lent me the 7.5 is behind me with a gun...):

1) If the open door shows -3k, I pick the closed door. Minimal win (1/3) = 7.5k, maximum win (2/3) = 20k. Expected = 15.8k.

2) If the opened door shows +7.5k, I pick that. Minimum, maximum, expected win is always 7.5k (sticking with my choice gives me an expected win of 12.3k, but I could loose 3k, and the guy behind me will shoot me!)

In case I want to maximize my expected winnings, I follow a different strategy:

1) If the open door shows -3k, I pick the closed door. Minimal win (1/3) = 7.5k, maximum win (2/3) = 20k. Expected = 15.8k.

2) If the opened door shows +7.5k, I still pick the closed door. Minimum win (1/3) = -3k, maximum win (2/3) = 20k, expected win is 12.3k.


As @kaine points out correctly, if we assume that the host has his own strategy, trying to minimize our win, the situation changes slightly.
If we play to recover our loss, nothing changes (we always win a minimum of $7500).

If we play to maximize our winnings, the host knows we will always switch to the closed door (unless he reveals the 20k, which therefore he will not do).

So in 2/3 of the cases (when we do not pick 20k originally), switching still gives us the 20k.

In the remaining 1/3 of the cases, the host will show us the 7.5k, knowing we will choose the -3k.

Our expected winnings are then easily calculated as 2/3 * 20k + 1/3 * -3k = $12333.

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I always start with box 1. I'm assuming he picks randomly if neither box 2 or 3 contains the 20,000.

If he shows the -3000 box, you're better off switching.

Case 1: Probability 33% - Box 1 contains 20k, remaining box contains 7500. I win 7500 by switching.

Case 2: Probability 66% - Box 1 contains 7500, remaining box contains 20k. I win 20,000 by switching.

Expected outcome from switching = $7500\times 0.33+20000\times 0.66 = 15833.\overline{3}$

Expected outcome from not switching = $20000\times 0.33+7500\times 0.66 = 11666.\overline{6}$

If he shows the 7500 box, you're better off switching.

Case 1: Probability 33% - Box 1 contains 20k, remaining box contains -3000. I lose 3000 by switching.

Case 2: Probability 66% - Box 1 contains -3000, remaining box contains 20k. I win 20,000 by switching.

Expected outcome from switching = $-3000\times 0.33+20000\times 0.66 = 12333.\overline{3}$

Expected outcome from not switching = $20000\times 0.33 -3000\times 0.66 = 4666.\overline{6}$

This is exactly like the Monty Hall Problem. Always switch.

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  • $\begingroup$ You can also take the one he opened, would that change your answer? $\endgroup$ – warspyking Oct 21 '14 at 13:54
  • $\begingroup$ What if I need to win my money back, regardless of any chances of winning more money? What if losing 3k means I get killed by the guy that lent be 7.5k to play and who doesn't want to give me another 3k to pay my new debt? $\endgroup$ – oerkelens Oct 21 '14 at 13:59
  • $\begingroup$ @oerkelens Thanks for the correction. :) The problem doesn't specify any of that (or really ask any question at all...) If the ultimate goal is to recover your money, then you'd pick the 7.5k if it were visible and switch otherwise, a 100% chance of recovering your money, I believe. $\endgroup$ – Jason Patterson Oct 21 '14 at 14:04
  • $\begingroup$ "I'm assuming he picks randomly if neither box 2 or 3 contains the 20,000." That is the crux. Should you trust that he is not playing you when chosing whether to should you -3000 or 7500? What (assuming you play perfectly) would he do? $\endgroup$ – kaine Oct 21 '14 at 14:05
  • $\begingroup$ @warspyking You tell me... Is the expected outcome of switching ever lower than the best possible outcome of taking an open door? $\endgroup$ – Jason Patterson Oct 21 '14 at 14:06
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If you pick \$-3K first, he will show you \$7.5K. You winnings are $\$23000x_{7.5}-\$3000$ where x in this equation is probability you will switch after seeing $7.5K.

If you pick \$7.5K first, he will show you \$-3K. You winnings are $\$12500x_{-3}+\$7500$ where x in this equation is probability you will switch after seeing \$-3K.

Obviously if those were the only two cases, both x's should be 1.

Lets assume that in the case wher you are currently sitting on 20K, he shows you 7.5K with a probability of $p$.

He intellegently chooses $p$ based on the assumption that you are a perfect player. Your winnings will be $$(-\$23000x_{7.5}+\$20000)p+(-\$12500x_{-3}+\$20000)(1-p)$$ $$=(\$12500x_{-3}-\$23000x_{7.5})p-\$12500x_{-3}+\$20000$$.

If we assume you should always switch, that quickly simplifies to $(-\$10500)p-\$7500$. So $p$ should equal $1$. You will always win \$20K if you see \$-3K (and picked $7.5K to start).

If you see \$7.5K, therefore, the full calculation is: $$.5*(\$23000x_{7.5}-\$3000)+.5*(/$20000-\$23000x_{7.5}) = /$8500)$$

This means, of course, that his best strategy is to assume you will always switch and $2/3$ of the time you will win on average \$8500 and $1/3$ of the time you will always win \$20000. On average you will win \$12333. Obviously you still win money but less than stated by the other answers.

Please note that you should still never pick the one he opens. \$20000 is large enough that $7500 is never appealing.

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  • $\begingroup$ Indeed, you are right that if the host tries to minimize your win, your expected win is slightly lower. In short, original Monty Hall proves you win 20k 2/3 of the time, and the remaining 1/3 the host will always show the 7.5k, meaning I win -3k, leasing to an overall expected win of (2/3*20)+(1/3*3)=12.3k in case of a maximized winnings-strategy. $\endgroup$ – oerkelens Oct 22 '14 at 6:34

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