Poker. 26 red, 26 black. take one every time, you can choose to guess whether it’s red. you have only one chance. if you are right, you get 1 dollar. what’s the strategy? and what’s the expected earn?

339/8788= 0.039

Step-by-step explanation:

Let black card be represented with B and the red card be represented with r. Therefore, P(B,r) is the expected value.

Hence, we say, (b,r) is not equal to (0,0) for the first state, we then, have a probability of B/r+B(the probability of drawing out a black card and loosing a point.

Also, (B-1,r) state with the probability of r/r+B(which is the probability of drawing out a red card and gaining a point).

The expected value is therefore;

B/r+B(-1+(B-1,r) + r/r+B(1+P(B,r-1))

Therefore, if we have negative, then;

P(B,r)=0,B/r+B(-1,r)) + r/r+B(1+P(B,r-1))...

P(0,0)=0; P(26,26) = 339/8788

=.039

Note: we say that the expectation at the start if we draw red,gaining +1 and if we draw black and then draw two reds you end +1. That is; 1/26 × 2/26 × 1/26 = 1+ 338/8788

= 339/8788.

step-by-step explanation:

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step-by-step explanation: