[jay_shark]

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Ok here is the answer to the first problem . I'll let others think about the second .

1) The sb can either raise an additional 2.5 bb's which means the BB is getting 2:1 on his call . If the sb folds at any point for any specific hand then his EV =0 . I'll show that pushing with any number is positive EV .

If you have card #1 , then it's clearly the lowest number form the deck . However , the bb is not aware of this . He must call you if he believes his hand can beat at least a third of yours . Since he's getting 2:1 on his call , he should call with numbers 34,35,36,...100 . Notice that 34 beats precisely 33 numbers and loses to 66 numbers . So , if the sb pushes with any card , then he actually increases his EV . Since this is the case , there is no bluffing frequency for the sb .

The probability that the BB wins given that he calls you will converge to 2/3 as the numbers approach infinity . In this case , the numbers stop at 100 but it still converges to 2/3 fairly quickly .

Just show that 1/3 + 2/3*1/2 = 2/3 . Simply reason that the BB will beat one third of the hands when he calls and the sb shows 1-33 . However two thirds of the time , he will win half of the hands (2/3*1/2) .

Ev(sb) = 1/3*1.5 + 2/3*(3.5*1/3 - 2.5*2/3)
Ev(sb) = 0.166666666

This shows that raising with any number is better than folding , even if your first card is a 1 .

The second problem is a bit harder and algebra intensive but it is pretty neat . The solution hinges primarily on ideas expressed in the first problem but it's still interesting to work out .

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I gotta disagree with this result.

BB should always fold 1 thru 62, raise 1/8 of the time with 63, and always raise 64 thru 100

By doing this, the SB is indifferent to raising or folding with 1 thru 62, and should raise with 63 thru 100.

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Mykey , you are wrong . The sb should be raising with hands 44 and up . The BB should be calling with hands 63 and up .