# Optimal strategy for playing "Deal or No Deal" **URL:** \ **Category:** Cafe Society\ **Created:** [February 28, 2006, 5:50pm UTC](https://boards.straightdope.com/t/optimal-strategy-for-playing-deal-or-no-deal/346481 "2006-02-28T17:50:30Z")\ **Posts on this page:** 5\ **Page:** 4
**Author:** ![StGermain](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/stgermain/32/2868_2.png) [@StGermain](https://boards.straightdope.com/u/StGermain)\ **Post date:** [March 2, 2006, 8:29am UTC](https://boards.straightdope.com/t/optimal-strategy-for-playing-deal-or-no-deal/346481/61 "2006-03-02T08:29:55Z")
You’re all missing the most important strategy - you have to listen to the audience. For surely they _want_ you to win, and they’ve got to be smarter than you, right? So you just look helpless and say “what should I do?” /sarcasm The audience scares me. They sound like I imagine the Romans sounded while watching gladiators fight to the death. StG ---
**Author:** ![Enderw24](https://avatars.discourse-cdn.com/v4/letter/e/ba9def/32.png) [@Enderw24](https://boards.straightdope.com/u/Enderw24)\ **Post date:** [March 3, 2006, 12:30am UTC](https://boards.straightdope.com/t/optimal-strategy-for-playing-deal-or-no-deal/346481/62 "2006-03-03T00:30:19Z")
The math continues to intrigue me. I played another game where I came down to three boxes: $100, 400K, 500K The bank offered me $300,000 which is exactly what my EV should be. Mathematically I could take this offer or not. In the long run, I’d break exactly even regardless of my choice. But when you’re playing the game your brain says “wait a sec…they’re offering me $300,000 when there are two boxes out there greater than $300,000! I should give up this deal.” So what happens if you give up the deal, pick a box, and the banker offers you 50% on your next choice? 1/3 of the time you’d be offered $200,000 1/3 of the time you’d be offered $250,000 1/3 of the time you’d be offered $450,000 For a total average offer of $300,000! So in the long run again, you’ll come up with exactly the same expected value whether you took the first deal or picked a box and went with the second deal. Anyway, my hypothesis is that this game is both always beatable and never beatable. It’s always beatable in that anything you take away from the show is free money. It’s never beatable in that the banker will never offer you true value on your box. But given that, with a little bit of skill, it can always be played optimally. You can always walk out with the maximum amount of money given your completely random guesses if you understand some basic mathematical principles. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [March 4, 2006, 7:27pm UTC](https://boards.straightdope.com/t/optimal-strategy-for-playing-deal-or-no-deal/346481/63 "2006-03-04T19:27:49Z")
Just out of curiousity, how would it effect the odds of the game if instead of offering you a specific amount, the banker offered to trade you the contents of your chosen box for an amount equal to the average of all the unopened boxes left? Using the numbers from the previous post (750,000 - 400,000 - 300,000 - 200,000 - 100,000 - 10,000 - 500 - 10), suppose you took the offer. Amount in your box/Average in other boxes 10/251,500 500/251,430 10,000/250,073 100,000/237,216 200,000/222,930 300,000/208,644 400,000/194,359 750,000/144,359 ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [March 4, 2006, 7:54pm UTC](https://boards.straightdope.com/t/optimal-strategy-for-playing-deal-or-no-deal/346481/64 "2006-03-04T19:54:32Z")
Doesn’t make any difference to the expected value. The average of (the average of all except the first box, the average of all except the second box, …) equals the average of seven sets of all eight boxes, which is the same as the average of just one set of the eight boxes. But since the bank’s offers tend to be below the expected value, it would be a better offer than usual. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [March 4, 2006, 8:32pm UTC](https://boards.straightdope.com/t/optimal-strategy-for-playing-deal-or-no-deal/346481/65 "2006-03-04T20:32:36Z")
> [@](#): > > Doesn’t make any difference to the expected value. The average of (the average of all except the first box, the average of all except the second box, …) equals the average of seven sets of all eight boxes, which is the same as the average of just one set of the eight boxes. It’s not so much a mathematical concept (as you point out the probablities are unchanged) as it is a psychological one. Being offered the average of the unopened boxes means being offered a chance to capitalize on a bad pick. Because the lower the amount in your box, the higher the average amount in the other boxes. Instead of being asked whether you think you picked the million dollar box, you’d be getting asked whether you thought you had picked the one dollar box. [Previous page](https://boards.straightdope.com/t/optimal-strategy-for-playing-deal-or-no-deal/346481.md?page=3)