Can you apply the logic from the Monty Hall problem to Deal or No Deal? I’m no statistician, but after a conversation this week, I think you can…
For those of you unfamiliar with the Monty Hall problem, imagine a game show where you have three doors. Behind one door lies a shiny new sports car, while the other two doors hide goats (or, if you’re concerned about animal welfare, they can be empty). You’re asked to pick one door. The game show presenter will then open one of the doors that does not hide the car and then ask you if you’d like to change you choice.
Many people think that this is a 50/50 choice. In fact, if you change your choice, the odds are 2:1 in your favour of winning the car. Yes, it’s counterintuitive, but if you take the time to play out all the combinations you’ll see that this is correct.
Some people assume that you cannot apply the same principle of swapping your selection to Deal or No Deal. Firstly, the Monty Hall problem relies on a single item of value and multiple worthless items, and secondly, in the Monty Hall problem, the presenter always knows where the prize is located.
Although the Monty Hall problem features only one item of value and multiple low-value items, the game can be extrapolated. Provided there are always more low-value than high-value items to choose from, and the low-value items are always removed, you are more likely to win by swapping at the end of the game. Therefore, the Monty Hall problem can be used to consider games featuring multiple high-value items.
How can this be applied to Deal or No Deal? I suspect that most contestants on the game show don’t come on just hoping to win the top prize (£250,000 in the UK). Most people will be happy to win, or at least want to win, a range of values, from a minimum amount all the way up to the £250,000.
So the game will start and prize money will be lost as different boxes are selected by the contestant. Now, in the Monty Hall problem, the presenter knows the location of the high value prize(s) so these aren’t removed. This isn’t the case on Deal or No Deal – some of the bigger prizes will be removed by chance. However, some prizes in this range will still remain.
I wrote a short computer program to measure the likelihood of winning your minimum target or greater if you chose to swap when there was only one prize left in this range. As you can see, the greater your minimum amount, the more likely you are to win that value or greater if you swap; this mirrors what we see in the Monty Hall problem.
|Minimum desired winnings||Successes if swap||Successes if stay|
Obviously, you don’t have this option in the real game but you could apply its logic to your decision making when offered a deal by the banker: if there are fewer prizes in your target range than outside it, and the banker offers you a deal greater than your minimum amount, you should take it, as this is tantamount to swapping. From a purely rational point of view, you should take it anyway (and now you have a statistical reason why) although humans clearly aren’t rational and contestants clearly enter the show either with grandly over ambitious goals, or get overwhelmed with excitement (or possibly greed?) as the game progresses.
I haven’t quite worked out what to do if you get given an offer below you target prize, though. Statistically, the winning box(es) are less likely to be picked so by playing a little longer the deal could shift in your favour. This could be especially beneficial if there are some very high value boxes left, and anecdotal evidence suggest that the banker’s offers get more generous as the game goes on.
And, for the sake of completeness, if you came to the final two boxes, with one containing a penny and the other £250,000, you should swap (but don’t re-swap). However, I’d say you’d be very foolish not to take the banker’s offer at that point!