Satellite bubble decisions

When a satellite awards identical seats, a larger finishing stack does not buy a larger seat. That changes how a chip gamble can affect prize equity near the qualification bubble. A transfer can increase expected chips while reducing the modeled chance of receiving a ticket. The prize structure and possible resulting stacks determine the comparison.

This lesson uses four fictional players and three identical seats. A separate cash ladder shows what changes when first, second and third receive different prizes. The branch desk uses a stated static ICM profile and a win probability you supply. It does not read cards, estimate that probability or turn a stack snapshot into a real fold-or-call verdict.

Ace: “More chips must mean a bigger prize, right?” Victoria: “Look at the prizes first. Three identical seats give us a different question from three different cash awards.”

Confirm the qualification and prize rules

An ordinary equal-seat satellite can end when enough players have been eliminated to award the remaining seats. Check the event’s actual structure before using that description. Leftover prize money may create an additional cash award, and some events have different ticket values or other qualification procedures. A flat prize vector must come from the event terms, not from the word “satellite” alone.

A target-stack format asks a different qualification question from an event played down to a fixed number of survivors. This desk has no target-stack trigger and no rule for moving a player out of play after reaching a target. Its three-seat example is specifically the ordinary surviving-player model described here. Read any event-specific qualification and hand-for-hand rules separately.

The satellite prize vector is [1, 1, 1, 0]. One means one identical seat; zero means no prize. The comparison cash vector is [50, 30, 20, 0] in hypothetical units. It totals 100, while the seat vector totals three. Those scales are intentionally different. Compare changes within each profile rather than subtracting a cash-unit result from a seat probability.

A ticket’s practical value also depends on its terms. Usability, scheduling, transfer rules and any alternative award are outside this calculation. No ticket is treated as automatically exchangeable for cash. The exercise studies prize chances under stated rules and does not recommend entering a particular event or spending money to qualify.

Define a transfer without hiding a pot

Players are A, B, C and D. Hero is always B, and the opponent in the transfer is A. The starting stacks are the unchanged reference state. In the win branch, the transfer amount moves from A to Hero. In the loss branch, it moves from Hero to A. C and D keep their starting chips in both outcomes.

The transfer cannot exceed either participant’s starting stack. Its win and loss amounts are equal, and every branch must conserve the total chips. There is no current pot, dead money, posted blind, ante, outstanding call or side pot inside the reference. Those omissions make the arithmetic clear and keep it distinct from reconstructing a real hand.

In a live decision, folding may already leave you without a posted blind or earlier bet. Calling may contest money contributed by several players. Those resulting stacks can differ from the symmetric transfer here. A call amount pasted into this form does not create the actual fold and call branches. Build the real outcomes from the hand history before applying a prize model.

The only possible transfer outcomes here are win and loss. Ties, split pots, multiple opponents, partial refunds and simultaneous eliminations are excluded. A single supplied win probability cannot describe extra branches that have been left out. Zero transfer is permitted as a control: it leaves both branches equal to the starting reference.

State the ICM ranking assumption

The desk’s stack-proportional ICM profile assigns first-place probability according to each player’s share of all chips. It then removes that possible first-place player and assigns the next place in proportion to the remaining stacks. Repeating that step produces finish-order probabilities. Multiply them by the stated finishing prizes and add the contributions for each player.

This is a model of a static stack distribution. It does not include differences in skill, position, upcoming blinds or future betting decisions. It also does not inspect hole cards or infer an opponent’s range. The model can price the supplied states consistently while still being an incomplete description of a real satellite. Keep that distinction with every result.

The same ranking calculation is used for the seat and cash profiles, so their difference comes from the prizes. In the satellite profile, the first three finishing positions each earn one unit. A player’s expected seat units therefore equal its modeled probability of receiving a seat. Across the four players, those probabilities sum to three, because exactly three seats are awarded.

Start with four equal stacks

The default stacks are 25 / 25 / 25 / 25, for a total of 100. Symmetry gives every player the same chance to occupy each finishing position. Each has a 25% chance of missing the three seats and a 75% model seat probability. Four players each at 75% account for three expected seats in total; the probabilities are not supposed to sum to 100%.

Hero’s cash-ladder equity in that unchanged state is 25 units. One quarter of 50, plus one quarter of 30, plus one quarter of 20, plus one quarter of zero equals 25. Every other player has the same value, and the four values sum to the entire 100-unit cash prize pool.

These are weighted model values. An actual outcome awards a particular prize; it does not divide a ticket into fractions or pay the model average after every hand.

Build the win and loss states first

Let A and Hero B contest a 25-chip transfer. If Hero wins, the stacks become 0 / 50 / 25 / 25. If Hero loses, they become 50 / 0 / 25 / 25. Each state still totals 100. Hero’s own stack doubles or falls to zero, while the other participant supplies or receives the same 25 chips.

In the win state, three players remain with chips and the model awards each one seat. Hero’s seat probability is one. In the loss state, Hero has zero chips and receives no seat; the other three each receive one. These are terminal results under this stated ordinary three-seat structure, not a promise that any positive stack is safe while four players remain.

The unchanged seat probability is 75%, the win value is 100%, and the loss value is zero. Winning gains 25 percentage points relative to unchanged. Losing costs 75 percentage points. The chip transfer is symmetric, but the prize-equity change is not. Extra chips cannot give Hero more than one of these identical seats.

Weight the outcomes with a supplied probability

Use a fictional 70% win probability. Hero’s expected chips are 0.70 × 50 + 0.30 × 0 = 35, above the unchanged 25 by ten chips. Its weighted seat probability is 0.70 × 1 + 0.30 × 0 = 0.70, below the unchanged 0.75 by five percentage points. Both calculations are correct under the inputs.

Positive chip expectation does not cancel negative seat-equity change. Nor does lower seat expectation predict a lost hand: Hero can win this particular transfer despite the lower weighted comparison.

The 70% input comes from the exercise. Changing it explores a dependency; it does not establish a real matchup probability or show that the reference matches folding.

Ace: “The transfer adds expected chips, so why did the seat value fall?” Victoria: “The model gain is only a quarter of a seat chance. The loss gives up the whole three-quarter chance already in the reference.”

Apply the cash ladder to the same states

Keep the chips and 70% probability unchanged, but price finishing positions at 50 / 30 / 20 / 0 units. In Hero’s win state, the remaining positive stacks are 50 / 25 / 25. Hero has a one-half model chance to finish first. It finishes second with probability one third and third with probability one sixth.

Hero’s win-branch cash equity is therefore 0.50 × 50 + (1/3) × 30 + (1/6) × 20 = 115/3 units, approximately 38.3333. Its loss-branch cash equity is zero. Weighting at 70% gives 161/6 units, approximately 26.8333, above the unchanged 25 by about 1.8333 units.

The cash ladder leaves a benefit to finishing first rather than third after surviving. Three identical seats do not. This is why the same chip branches and supplied probability can move cash-ladder and seat equity in different directions. No real cash event or satellite is being forecast, and the hypothetical cash units are not an estimate of a ticket’s monetary value.

A 70% seat chance is not 70 cash units. Keep each profile’s headings and units attached when comparing or copying its results.

Derive each break-even reference

Let U be Hero’s model equity in the unchanged state, W in the win state, and L in the loss state. With supplied win probability q, the weighted value is qW + (1 – q)L. Setting it equal to U gives q = (U – L) / (W – L), provided W and L differ. Use one prize profile and one unit consistently in that equation.

For the default seat profile, the threshold is 0.75 / 1 = 75%. At exactly that supplied probability, the weighted seat value matches unchanged. For the cash ladder, the threshold is 25 / (115/3) = 15/23, approximately 65.2174%. The default 70% lies between them, which explains the opposite directions in the two profiles.

These are thresholds against the unchanged reference, not pot-odds thresholds for a live call. The form has no pot or call price, and its reference has no posted cost. Displayed decimals are rounded. Use the underlying fraction when exact equality matters; entering a nearby rounded percentage can leave a small difference.

Use the branch desk within its frame

Enter four positive whole-chip starting stacks, a transfer no larger than A or B, and a win probability from zero through 100%. Hero remains B, so changing the stack named A does not silently move the Hero label. C and D are spectators to the transfer but remain part of the prize calculation.

First reproduce the equal-stack example. Read the unchanged, win and loss stack rows before comparing prize values. Then check the weighted seat result of 70%, cash result of approximately 26.8333 units, and expected Hero stack of 35. The total chips must remain 100 in all three states.

Try a smaller transfer while keeping the starting stacks fixed. Both participants may survive both outcomes, so the branch seat probabilities need not be zero and one. Next change the supplied probability to zero and then 100%. The weighted value must equal the loss branch and win branch respectively, under both prize profiles.

Input changes clear the old result. Reset restores the defaults and focuses A’s field. The examples remain readable without JavaScript. No account, identity or hand-history input is requested.

Check the boundaries rather than forcing a threshold

Set the transfer to zero. All three states match, so there is no unique break-even probability and dividing by W – L would divide by zero. The desk also treats branch differences below its numerical tolerance as too small to report a stable unique threshold. It labels that qualification explicitly.

A zero-stack player created by this transfer receives the fourth-place zero award. Only A or Hero can be eliminated, and a positive transfer cannot eliminate both. The other players start positive. This avoids inventing a tie-break procedure for simultaneous bust-outs or assigning a paid finishing position to an unresolved eliminated player.

Starting stacks are limited to one million each and win probabilities to two decimal places. Scaling every stack and the transfer by the same factor preserves the model probabilities. Scaling all prizes scales their equity values.

Separate a model endpoint from a safe stack

While four positive stacks compete for three seats, a large stack still has a nonzero modeled chance of missing. The desk does not mark it “locked” because it looks comfortable. After a specified branch leaves exactly three survivors, each receives one seat under this exercise’s terminal rule. Those statements concern different states.

The model cannot establish how long someone can wait, whether another stack busts first, or which hand can be folded without consequence. Actual event rules and reconstructed branches still matter.

Ace: “Can I write that the model says to fold?” Victoria: “Write the stacks, prizes, branches and supplied probability. We compared those states. We did not reconstruct a live fold or establish your hand’s chance to win.”

A study note should keep the four stacks, prizes, transfer and supplied probability beside the result. Here seat equity is 70% against 75% unchanged, while the stated cash ladder moves in the other direction. Another reader can check those inputs.

For a real review, keep the actual prize schedule and qualification conditions beside the hand history. Account for chips already committed, all relevant outcomes and any remaining action. If the supplied win probability or a branch is unresolved, leave it unresolved. A precise model answer to a fictional question does not fill in missing evidence about the next hand.

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