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Written by GrenGrad
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Question
On the Turn
You have an open-ended straight draw, while your opponent has top two
pair. The pot initially contains $100. You have a $2750 stack and your
opponent has $200. You are last to act. Assuming that he will he will
pay you off on the river, what is the largest bet that you COULD call?
The Answer
You have an 18 percent chance of winning on the river.
Pot odds dictate that to call you could only call 18% of the pot.
So a bet of 28 for a pot of 100+28+28=156*.18=28.08 would be the minimum you could bet by pot odds.
We have something different here though. We have implied pot odds.
Implied pot odds are how much we can bet based on our future knowledge.
Usually, they are best calculated when drawing to the nuts, and you
know your opponent will call even if you hit the nuts.
So our current pot(100)+our implied additional pot(200)+the money we are throwing in to call.
(100+200+x)*.1818=x
x=54+.1818x
.8182x=54.54
x=66.66
The most we can call is 66.66
This is rounded to 66 for this answer.
Let's look at why.
Lets say we run the hand 10000 times.
All 10000 times we payed $66 to see the river.
This cost us $660,000.
1818 times we catch the straight and win a profit of 366 dollars. (Our call+the 100 in the pot+ his 200 he had left)
Total gain $665,388
The 8218 times we do not catch the straight we fold.
Total cost $0
$660,000-$665,388= $5388 in profit.
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