[Closed] UTG shoving range

Here is a link to the Nash calculator for that situation, which gives you a pushing range of 14.9%, 22+ A9s+ ATo+ K9s+ KQo Q9s+ J9s+ T9s.

The pairs in the calling range shouldn't be just down to 88. That range punishes 77 too much. UTG+1 might not call with 88, while the big blind may call with weaker pairs.

If you hold 77, the chance that a particular opponent has one of the 78 combinations of 88+ AJs+ AQo+ is 78/1225 or about 6.37% because of the card removal effects. 77 does not block any of the calling hands.

The chance to be called is closer to 1-(1-6.37%)^8 ~ 41% rather than 8x6.37% ~ 51%. This is because sometimes more than one opponent will have a calling hand. 8x6.37% is the expected number of players with calling hands, but the probability of getting called is lower.

There appear to be big problems with your EV equation. You do not lose 22,500 chips when you get called and lose. You lose 10,000 chips compared with the start of the hand, or 9,900 chips compared with posting the ante and then folding. To be consistent, you might want everything to be compared with losing the ante. For simplicity you could assume that the caller is outside the blinds, but note that sometimes the caller will be in the blinds which decreases the amount you can win.

EV push ~ 59% x 2,300 blinds and antes + 41% (35% x 12,200 - 65% x 9,900) = +469 chips.

Accounting for the blinds better (but not overcalls) reduces that by about 41% x 35% x 1500/8 = 27 chips, for a gain of 442 chips by pushing.

thank you for the great reply , i felt that my math had some problems , funny but nash had the same pushing range as mine but only from feeling and not math

One question : why do you use this formula 1-(1-6.37%)^8 ~ 41% instead of 8*6.37? is it because each fold reduces the probabilty of getting called? so like if utg+1 folds then the chances of getting called are reduced and the 8*6.37 is not accunting for that? .

The ev calculations were also wrong , but that part i understood where my mistake was. And thanks again

P(A or B) = P(A) + P(B) - P(A and B).

If A and B are close to independent and have small probabilities, then P(A and B) will be much smaller, and you can approximate P(A or B) by P(A) + P(B) (or 2 x P(A) if P(A) = P(B)), but this is just an approximation. It assumes A and B are disjoint, that they can't happen at the same time. If P(A) + P(B) is large (particularly when it is greater than 1) or when the events are correlated, you might not be able to ignore the P(A and B) term.

Suppose you toss two fair coins. The probability that at least one is a head is 1/2 + 1/2 - 1/4 = 3/4. Another way to get this is to say that to avoid getting at least one head, the first coin must be a tail (probability 1/2) and the second coin must be a tail (probability 1/2). These are both tails with probability 1/2 x 1/2 = 1/4. The complement is 1- 1/4 = 3/4.

Similarly, a reasonably good approximation to the probability that at least one player has a calling hand is 1 - P(player doesn't have a calling hand)^(# players). That's off by a little because whether players have calling hands are not independent. If the first two players have AA, the next player can't have AA or AK since there are no more aces in the deck. This is not a large effect for ranges like this, so the independent approximation is quite good.

Similarly, suppose you deal out two cards, and ask for the probability that at least one card was a spade. The chance that the first card is a spade is 1/4. The chance that the second card is a spade is 1/4. The chance that both are spades is 1/4 x 12/51 = 1/17. So, the probability that at least one was a spade was 1/4 + 1/4 - 1/17 = 44.12%. The disjoint approximation was 2 x 1/4 = 50%. The independent approximation was 1-(3/4)^2 = 1/4 + 1/4 - 1/16 = 43.75%.

How about three cards? The probability at least one is a spade is 1-(39 choose 3)/52 choose 3) = 997/1700 = 58.65%. The disjoint approximation is 3*1/4 = 75%. The independent approximation is 1-(3/4)^3 = 57.81%.