You and Me are playing bets. I take $10 from you and that begins our game. I shuffle the deck of cards (52 cards normal deck) and take out four random cards out of it. Now I place the cards in front of you with face down. Then I turn over the first card. You can either keep it or you can ask me to open the second card. When I turn the second card, you have the same options again till I open the third card. If you deny to keep it, you must take the fourth card no matter what the card is.
Now if the card you choose has n value, I will give you $n and we will keep on playing like this till the end of the day. You may feel that this is completely over the chance. But there is a nice strategy which will allow you to win.
Can you find out the strategy so that by the end of the day you have won a decent amount?
Solution:
It is not as tough as it seems. The strategy is quite simple. You just have to pick up a card out of first three whenever you find a 9 or higher number.
Let us offer you a strong valid reason for the strategy. The probability that the four card kept below are below the number nine is 32*31*30*29 / 52*51*50*49 = 0.133
Thus in such case, you will lose $2 to $9 with equal probability = 0.133/8 = 0.0166
Let us now calculate the probability that one of the four cards is nine or even higher.
1 – 0.133 = 0.867
Now you will surely stop at the first sight of 9 or any value above 9 as per the strategy. Thus you have a chance of winning -1, 0, 1, 2, 3 with an equal probability = 0.867/5 = 0.173
In such case you stand a chance of winning overall amount of 0.14 every game you play.
Let us consider the other strategies. If you want to stop at the sight of 10 or above 10, the expected amount you will win per game will be 0.09 which is way below the present strategy. If you want to stop at above ten only, you will have an expected winning in negative.
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