# Probability: How to play a game when the expected value is always positive

I want to play a coin flipping game. On the first flip, if the coin is heads, I get \$1, if it is tails I lose. On subsequent turns, if the coin is heads I get double the money from last turn added to my current winnings, if it is tails I lose all of my money (so on the$n$th turn I can gain$\$2^{n-1}$ or lose $\$(2^{n-1}-1)$). I can stop any time and keep what I have. When should I stop playing? It seems to me that the expected value is positive before every turn, so I should never stop playing, but this obviously can't be the optimal strategy so I'm confused. • Maybe it isn't the best reference our there, but there was a Numberphile video saying that this strategy si exactly what mathematics does. In the end you will just lose$1. You need to add Economics and maybe opportunity cost to make the decision. – Cehhΐro Oct 18 '16 at 15:28
• In my opinion you loose as much as you win at each turn. At the second turn you have $\$1$. You can get$\$2$ or $\$0$. Thus the expected value is$\$1$ (fair coin). If the criteria is the expected value then it doesn´t matter when you stop. But you can also can consider the variance (uncertainty). In general people do not like uncertainty. On the other hand if you need $\$1000$very urgent, then you should play the game until you have won at least$\$1000$. – callculus Oct 18 '16 at 15:39
• The fact is that anytime you win you should stop, because your next play has 50% chance to loose all. You only need to loose once to loose all. The house only needs to win once to get your first bet and probably the entry fee. The house aims to win your first bet and is willing to risk all on the bet that you will play again after a win. – WindSoul Nov 8 '19 at 20:00

On the first flip, if the coin is heads, I get $1, if it is tails I lose. So 50% you get +1 and 50% you get -1, Net is 0 At the nth round you have 2^n, so with So 50% you get +2^n more and 50% you get -2^n, Net is 0 (In other worlds, at the nth round, you do not win +2^n+1, just +2^n, the rest is your hard won ;) money from previous rounds) So the Expected value is 0. Notice that this is similar to this question, However, in that situation you should start by playing, since you do not have to pay to play, and you may gain something. Furthermore in that question, you are not getting double the rewards in every round, so it always makes sense to stop because at each round you bet more money to get a constant amount. You can introduce something like the value winning money... For example i can do very litle with 1 dollar, so i will pick it if i see it on the street but its value is small... I could go on a trip around the world with 4k, so that would be great. And having 8k would be better, but not doubly so: I would go to more expensive hotels etc. but that would improve my experience only but not double the value. Since at the beginning E=0, you can equally decide to win or not. An interesting question is what mathematics would do if you are were in a situation where p was slightly positive... Say the coin is loaded in your favour and you win 60% of the time... Then the question is can you play the game many times? Because if you can, math would say, as soon as you can, start playing multiple times the same game... For example say you have \$2. Then instead of putting them in one go, start two games and every time you win, stop and start two new games of \$1. Even better if you can bet cents, start 200 games, it becomes very improbable that you ll lose.... If you are not allowed to play as may games as you want, [then you will end up playing until you lose, or until you owe all the money in the world, edit: I am not certain about this part, "the expected value of the lottery only grows logarithmically with the resources of the casino. As a result, the expected value of the lottery, even when played against a casino with the largest resources realistically conceivable, is quite modest.":wikipedia]. If you add value to the mix, then probably you can start removing some part of your winnings at every step, but you should probably keep playing. Now what the game is rigged against you? Mathematics predicts you will lose if you play forever. So the only winning move may be not to play. On the other hand, if you add value to the mix, it might be worth playing but you have to set a goal in advance You should play knowing you are loosing money, but depending on your goal, you might have good chances of achieving it. What is interesting here is that dividing your bets hurts you. If you need to bet \$2, better to bet them at once, not go for two small bets. This strategy is called bold playing though it does not mean you should always bet all your money. For example suppose i have 1000, but for some reason I really wish I could buy this ring that costs \$1001 to offer to my future wife. I can start betting \$1. If I win I stop. If I loose, I will bet \$2 (because i now have 999 and still need to arrive at 1001) and so on. Notice that although its more likely that I will win than not, if i win, i get 1000 times less money than if I loose, so overall the best strategy is to go to the merchant and say: "hey I really need this ring to propose, and I only have 1000\$, but I will give them all of them to you if you only lower your price by \\$1" Chances are he will accept :)