You can go to a fast food restaurant to buy chicken nuggets in 6-pack, 9-pack or 20-packs. is there such a number N, such that for all numbers bigger than or equal to N, you can buy that number of chicken nuggets?

Solution: nuggest

You can get any number in the right column except 3 by adding 6s and 9s. So cross out the entire right column except 3. You can also add 20 to any crossed-out number and cross that number out. So cross out everything in column two below and including 26. Finally, by the same logic you can add 20 to the crossed-out numbers in column 2 and thereby cross out everything in column one below and including 46.

The largest number that's left over is 43. Incidentally, cross out 20 and 40 and your map is complete.

Admin

28 יוני 2017

42.

You have two identical crystal orbs. you need to figure out how high an orb can fall from a 100 story building before it breaks. you know nothing about the toughness of the orbs: they may be very fragile and break when dropped from the first floor, or they may be so tough that dropping them from the 100th floor doesn't even harm them.

What is the largest number of orb-drops you would ever have to do in order to find the right floor? (i.e. what's the most efficient way you could drop the orbs to find your answer?)

You are allowed to break both orbs, provided that in doing so you uniquely identify the correct floor.

Solution: orbs

14.

Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (ie move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100). Call the first floor at which it breaks n and the previous tested floor n'. Then try the intervening floors (n'+1 .. n'-1) with the other orb.

Worst case is if correct floor is 13,14,26,27, etc which require m drops with the first orb and 14-m drops with the second.

Admin

28 יוני 2017

43.

Three coworkers would like to know their average salary. how can they do it, without disclosing their own salaries?

How about: Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes to A. Now A subtracts her random number (AR), passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.

As has been noted already, there's no way to liar-proof the scheme.

It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries.

Admin

28 יוני 2017

You have $10,000 dollars to place a double-or-nothing bet on the Yankees in the World Series (max 7 games, series is over once a team wins 4 games).

Unfortunately, you can only bet on each individual game, not the series as a whole. how much should you bet on each game, so that, if the yanks win the whole series, you expect to get 20k, and if they lose, you expect 0?

Basically, you know that there may be between 4 and 7 games, and you need to decide on a strategy so that whenever the series is over, your final outcome is the same as an overall double-or-nothing bet on the series.

Solution: world Series

This probably isn't the cleanest solution, but...

a dynamic-programming type solution is:

(1) Create a 5x5 matrix P.

So, P[i,j] holds your pile of money when the yanks have won i games and the mets have won j games.

initialize P[4,j] := 20 for j from 0 to 3 initialize P[i,4] := 0 for i from 0 to 3

fill P in bottom-right to top left by averaging bottom and right adjacent cells:

P[i,j] := (P[i+1,j]+P[i,j+1]) / 2

(2) Make another 5x5 matrix, B, which represets your bet at any-time.

So, B[i,j] represents your bet when the yanks have won i games and the Mets j games.

fill this top-left to bottom right by:

B[i,j] = P[i+1,j] - P[i,j]

(3) Look in matrix B for your bet at any time.

The final matricies are:

Admin

28 יוני 2017

45.

How many trailing zeroes are there in 100! (100 factorial)?

One per factor of 10, and one per factor of 5 (there are more than enough 2's to pair with the 5's), plus one per factor of ten squared (one occurrence) and one per factor of 5 squared (three occurrences).

So if I'm counting correctly, that'd be 10 + 10 + 1 + 3== 24 zeroes.

Assuming the question meant *trailing* zeroes. It'd be much harder to also count the intermingled zero digits in the entire expansion.

Admin

28 יוני 2017

46.

You are an oil mogul considering the purchase of drilling rights to an as yet unexplored tract of land.

The well's expected value to its current owners is uniformly distributed over [$1..$100]. (i.e., a 1% chance it's worth each value b/w $1..$100, inclusive).

Because you have greater economies of scale than the current owners, the well will actually be worth 50% more to you than to them (but they don't know this).

The catch: although you must bid on the well before drilling starts (and hence, before the actual yield of the well is known), the current owner can wait until *after* the well's actual value is ascertained before accepting your bid or not.

What should you bid?

Solution: oil mogul

This problem amounts to properly defining the expected value of the well to you.

The following equation does it:

(1%) * [(1.5 - 1)] +

(1%) * [(3 - 2) + (1.50 - 2)] +

(1%) * [(4.5 - 3) + (3 - 3) + (1.5 - 3)] +

. . .

(1%) * [(150-100) + ... + (3-100) + (1.5-100)]

Each line represents your expected value from a bid of 1$, 2$, ..., 100$, respectively.

eg, consider line 2 above. if you bid $2...

With 98% probability you won't win the contract, so your profit is 0. With 1% probability, you will win something worth (150%*1) = 1.5, for which you paid 2$ With 1% probability, you will something worth worth (150%*2) = 3, for which you paid 2$

So, your goal is to maximize the following function of x, where x is your bid.

f(x) = 1% * Sum_{i = 1 to floor(x)}{ x - 1.5*i }

There's no benefit to non-integer bets, so re-write the maximization function as :

ARGMAX(k) {1% * Sum_{i = 1 to k}{1.5*i - k}}

(=) ARGMAX(k) {Sum_{i=1 to k}{1.5*i - k}} /* 1% isn't a function of k or i, so toss it */

(=) ARGMAX(k) {Sum_{i=1 to k}{1.5*i} - Sum_{i=1 to k}{k}} /* Split the summation */

(=) ARGMAX(k) {(0.75)(K)(K+1) - K^2 }} /* Closed form for summations */

And that function is maximized at k = 1 and k = 2.

When choosing b/w $1 and $2, you should bid $1 because of time-value, reinvestment risk, etc, of the extra dollar.

(ie, if you don't have to spend the extra $$ now, don't)

That's my solution.

Admin

28 יוני 2017

47.

You find yourself in a duel with two other gunmen. You shoot with 33% accuracy, and the other two shoot with 100% and 50% accuracy, respectively. The rules of the duel are one shot per-person per-round. The shooting order is from worst shooter to best shooter, so you go first, the 50% guy goes second, and the 100% guy goes third.

Where or who should you shoot at in round 1?

Solution: duel

You have 3 options to consider:

(1) Shoot at 50% guy. Death comes with: 72% likilhood (2) Shoot at the 100% guy. Death comes with: 63.64% likihood (3) Shoot into the air (purposely miss). Death comew with 58.33% likihood.

So, you shoot into the air and hope for the best.

Calculations:

The p in the probabilities below represents your probability of dying at some point if you MISS either guy in the first round. It'll be filled in later.

(1) if you shoot at the 50% guy and get him, you're guaranteed to get shot the next round.

prob of dying at some point by aiming at the 50% guy:

(=) (33.3%)(100%)+(66.67%)*(p)

(=) (33.3%)+(66.67%)(p)

(2) if you shoot at the 100% guy and get him, you're left with this geometric sum of probability of getting shot by the 50% guy at some point in the future...it converges.

(=) [(50%)] +

[(50%)(66.67%)(50%)] +

[(50%)(66.67%)(50%)(66.67%)(50%)] + ...

(=) 50% * { 1 + (1/3) + (1/3)^2 + ... }

(=) 50% * {3/2)

(=) 75%

so, your prob of getting shot if you shoot at the 100% guy is:

(=) (33.3%)(75%)+(66.67%)(p)

(=) (24.75%) + (66.67%)(p)

still crummy, but it dominates the alternative, for positive p.

(3) What happens if you miss? ie, what's (p)?

if you miss, the second guy has a choice to make...does he shoot at you, or the other guy? his options:

(1) Shoot you: if he gets you, he's guaranteed to die on the next shot.

if he misses, he has q chance of dying (which we'll get later).

(=) (50%)(100%)+(50%)(q)

(=) (50%)+(0.5)(q)

(2) Shoot at the 100% dude:

If he gets the 100% guy, there some infinite sum representing his probability of getting shot by you:

[(33.33%)] +

[(66.67%)(50%)(33.33%)] +

[(66.67%)(50%)(66.67%)(50%)(33.33%)] + ...

(=) 33.33% * [ 1 + (1/3) + (1/3)^2 + ... ]

(=) 33.33% * [(3/2)]

(=) (50.0%)

So, he chances of getting shot if he shoots at the 100% guy are:

(=) (50%)*(50%) + (50%)(q)

(=) (25%)+(.5)q

No matter what q is, he'd prefer shooting the 100% guy to shooting you.

Now, what happens if *he* misses (the 100%) guy? ie, what's q? if he misses, then the 100% guy has to make a decision:

(1) Shoot you:

he's guaranteed to get you, so his chances of dying are just 50%. (game ends after the next round...the 50% gets only one shot)

prob of dying: 50%

(2) Shoot the 50% guy:

by the same logic, he'd have a 33.33% chance of death (getting shot by you).

So, he prefers to shoot the 50% guy.

So, q is 100% (ie, if the 50% guy misses, the 100% guy shoots him immediately). From that, we know the 50% guy's optimal move, if everyone's around on his first shot. He'd prefer shooting at the 100% guy to shooting at you or purposely missing.

Now, we need to get p, which is our probability of dying if we purposely miss either guy (shoot into the air).

We know the 50% guy shoots at the 100% guy if we miss.

(1) With 50% probability, the 50% guy kills the 100% guy, resulting in a shootout b/w us and the 50% guy.

(2) He misses him. Then, we get one shot at the 100% guy (after the 100% guy shoots the 50% guy).

Our chances of death are:

(50%)*(66.67%) = (33.33%)

So, p is (25%)+(33.33%) = 58.3%

So, again, our three choices are

(1) Shoot at 50% guy. Death comes with:

(=) (33.3%)+(66.67%)(p)

(=) 72% likilhood

(2) Shoot at the 100% guy. Death comes with:

(=) (24.75%) + (66.67%)(p)

(=) 63.64% likihood

(3) Shoot into the air (purposely miss).

(=) p

Admin

28 יוני 2017

48.

If a hen and a half lay an egg and a half in a day and a half, how many hens does it take to lay six eggs in six days?

If 1.5 hens lay 1.5 eggs in 1.5 days (or 36 hours) then: 1 hen lays 1 egg in 1,5 days or 4 eggs in six days thus 1.5 hens lay 6 eggs in 6 days

Admin

28 יוני 2017

49.

Arrange the numbers 1 to 8 in the grid below such that adjacent numbers are not in adjacent boxes (horizontally, vertically, or diagonally).

The arrangement above, for example, is wrong because 3 & 4, 4 & 5, 6 & 7, and 7 & 8 are adjacent.

Solution: box o' numbers

The key (as I see it) is putting the 1 & 8 in the centre spots - which is required, because those spots both border all but one of the other spots, and 1 & 8 are the only numbers that are only adjacent to one number.

From there, the 2 & 7 are forced, then you have your choice of the next number, which then forces the rest. Really, though there are only two valid solutions, and they are mirror images of each other.

Admin

28 יוני 2017

50.

"A line of 100 airline passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let's say that the nth passenger in line has a ticket for the seat number n.)

Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random.

What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?"

If any passenger <100 randomly selects seat 1, then each successive passenger can sit in his/her assigned seat including passenger 100.

If any passenger <100 randomly selects seat 100, then passenger 100 is forced to choose the only other remaining seat.

For each passenger (who is forced to choose randomly), the chance of choosing 1 and 100 is equal.
(The EXACT probability depends on the number of other remaining seats, but in each case the chance of choosing 1 and 100 is the same. 1/100 and 1/100 for the 1st passenger, 1/40 and 1/40 for the 60th passenger)
Each choice that doesn't choose seat 1 or 100 simply defers the final decision to a later passenger.

Therefore the chance of passenger 100 getting his own seat is 50%.

41.

You can go to a fast food restaurant to buy chicken nuggets in 6-pack, 9-pack or 20-packs. is there such a number N, such that for all numbers bigger than or equal to N, you can buy that number of chicken nuggets?

Solution: nuggestYou can get any number in the right column except 3 by adding 6s and 9s. So cross out the entire right column except 3. You can also add 20 to any crossed-out number and cross that number out. So cross out everything in column two below and including 26. Finally, by the same logic you can add 20 to the crossed-out numbers in column 2 and thereby cross out everything in column one below and including 46.

The largest number that's left over is 43. Incidentally, cross out 20 and 40 and your map is complete.

42.

You have two identical crystal orbs. you need to figure out how high an orb can fall from a 100 story building before it breaks. you know nothing about the toughness of the orbs: they may be very fragile and break when dropped from the first floor, or they may be so tough that dropping them from the 100th floor doesn't even harm them.

What is the largest number of orb-drops you would ever have to do in order to find the right floor? (i.e. what's the most efficient way you could drop the orbs to find your answer?)

You are allowed to break both orbs, provided that in doing so you uniquely identify the correct floor.

Solution: orbs14.

Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (ie move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100). Call the first floor at which it breaks n and the previous tested floor n'. Then try the intervening floors (n'+1 .. n'-1) with the other orb.

Worst case is if correct floor is 13,14,26,27, etc which require m drops with the first orb and 14-m drops with the second.

43.

Three coworkers would like to know their average salary. how can they do it, without disclosing their own salaries?

How about: Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes to A. Now A subtracts her random number (AR), passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.

As has been noted already, there's no way to liar-proof the scheme.

It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries.

You have $10,000 dollars to place a double-or-nothing bet on the Yankees in the World Series (max 7 games, series is over once a team wins 4 games).

Unfortunately, you can only bet on each individual game, not the series as a whole. how much should you bet on each game, so that, if the yanks win the whole series, you expect to get 20k, and if they lose, you expect 0?

Basically, you know that there may be between 4 and 7 games, and you need to decide on a strategy so that whenever the series is over, your final outcome is the same as an overall double-or-nothing bet on the series.

Solution: world SeriesThis probably isn't the cleanest solution, but...

a dynamic-programming type solution is:

(1) Create a 5x5 matrix P.

So, P[i,j] holds your pile of money when the yanks have won i games and the mets have won j games.

initialize P[4,j] := 20 for j from 0 to 3 initialize P[i,4] := 0 for i from 0 to 3

fill P in bottom-right to top left by averaging bottom and right adjacent cells:

P[i,j] := (P[i+1,j]+P[i,j+1]) / 2

(2) Make another 5x5 matrix, B, which represets your bet at any-time.

So, B[i,j] represents your bet when the yanks have won i games and the Mets j games.

fill this top-left to bottom right by:

B[i,j] = P[i+1,j] - P[i,j]

(3) Look in matrix B for your bet at any time.

The final matricies are:

45.

How many

trailingzeroes are there in 100! (100 factorial)?One per factor of 10, and one per factor of 5 (there are more than enough 2's to pair with the 5's), plus one per factor of ten squared (one occurrence) and one per factor of 5 squared (three occurrences).

So if I'm counting correctly, that'd be 10 + 10 + 1 + 3== 24 zeroes.

Assuming the question meant *trailing* zeroes. It'd be much harder to also count the intermingled zero digits in the entire expansion.

46.

You are an oil mogul considering the purchase of drilling rights to an as yet unexplored tract of land.

The well's expected value to its current owners is uniformly distributed over [$1..$100]. (i.e., a 1% chance it's worth each value b/w $1..$100, inclusive).

Because you have greater economies of scale than the current owners, the well will actually be worth 50% more to you than to them (but they don't know this).

The catch: although you must bid on the well before drilling starts (and hence, before the actual yield of the well is known), the current owner can wait until *after* the well's actual value is ascertained before accepting your bid or not.

What should you bid?

Solution: oil mogulThis problem amounts to properly defining the expected value of the well to you.

The following equation does it:

(1%) * [(1.5 - 1)] +

(1%) * [(3 - 2) + (1.50 - 2)] +

(1%) * [(4.5 - 3) + (3 - 3) + (1.5 - 3)] +

. . .

(1%) * [(150-100) + ... + (3-100) + (1.5-100)]

Each line represents your expected value from a bid of 1$, 2$, ..., 100$, respectively.

eg, consider line 2 above. if you bid $2...

With 98% probability you won't win the contract, so your profit is 0. With 1% probability, you will win something worth (150%*1) = 1.5, for which you paid 2$ With 1% probability, you will something worth worth (150%*2) = 3, for which you paid 2$

So, your goal is to maximize the following function of x, where x is your bid.

f(x) = 1% * Sum_{i = 1 to floor(x)}{ x - 1.5*i }

There's no benefit to non-integer bets, so re-write the maximization function as :

ARGMAX(k) {1% * Sum_{i = 1 to k}{1.5*i - k}}

(=) ARGMAX(k) {Sum_{i=1 to k}{1.5*i - k}} /* 1% isn't a function of k or i, so toss it */

(=) ARGMAX(k) {Sum_{i=1 to k}{1.5*i} - Sum_{i=1 to k}{k}} /* Split the summation */

(=) ARGMAX(k) {(0.75)(K)(K+1) - K^2 }} /* Closed form for summations */

(=) ARGMAX(k) {(0.75)(k)-(0.25)(K^2)}} /* Algebra */

And that function is maximized at k = 1 and k = 2.

When choosing b/w $1 and $2, you should bid $1 because of time-value, reinvestment risk, etc, of the extra dollar.

(ie, if you don't have to spend the extra $$ now, don't)

That's my solution.

47.

You find yourself in a duel with two other gunmen. You shoot with 33% accuracy, and the other two shoot with 100% and 50% accuracy, respectively. The rules of the duel are one shot per-person per-round. The shooting order is from worst shooter to best shooter, so you go first, the 50% guy goes second, and the 100% guy goes third.

Where or who should you shoot at in round 1?

Solution: duelYou have 3 options to consider:

(1) Shoot at 50% guy. Death comes with: 72% likilhood (2) Shoot at the 100% guy. Death comes with: 63.64% likihood (3) Shoot into the air (purposely miss). Death comew with 58.33% likihood.

So, you shoot into the air and hope for the best.

Calculations:

The p in the probabilities below represents your probability of dying at some point if you MISS either guy in the first round. It'll be filled in later.

(1) if you shoot at the 50% guy and get him, you're guaranteed to get shot the next round.

prob of dying at some point by aiming at the 50% guy:

(=) (33.3%)(100%)+(66.67%)*(p)

(=) (33.3%)+(66.67%)(p)

(2) if you shoot at the 100% guy and get him, you're left with this geometric sum of probability of getting shot by the 50% guy at some point in the future...it converges.

(=) [(50%)] +

[(50%)(66.67%)(50%)] +

[(50%)(66.67%)(50%)(66.67%)(50%)] + ...

(=) 50% * { 1 + (1/3) + (1/3)^2 + ... }

(=) 50% * {3/2)

(=) 75%

so, your prob of getting shot if you shoot at the 100% guy is:

(=) (33.3%)(75%)+(66.67%)(p)

(=) (24.75%) + (66.67%)(p)

still crummy, but it dominates the alternative, for positive p.

(3) What happens if you miss? ie, what's (p)?

if you miss, the second guy has a choice to make...does he shoot at you, or the other guy? his options:

(1) Shoot you: if he gets you, he's guaranteed to die on the next shot.

if he misses, he has q chance of dying (which we'll get later).

(=) (50%)(100%)+(50%)(q)

(=) (50%)+(0.5)(q)

(2) Shoot at the 100% dude:

If he gets the 100% guy, there some infinite sum representing his probability of getting shot by you:

[(33.33%)] +

[(66.67%)(50%)(33.33%)] +

[(66.67%)(50%)(66.67%)(50%)(33.33%)] + ...

(=) 33.33% * [ 1 + (1/3) + (1/3)^2 + ... ]

(=) 33.33% * [(3/2)]

(=) (50.0%)

So, he chances of getting shot if he shoots at the 100% guy are:

(=) (50%)*(50%) + (50%)(q)

(=) (25%)+(.5)q

No matter what q is, he'd prefer shooting the 100% guy to shooting you.

Now, what happens if *he* misses (the 100%) guy? ie, what's q? if he misses, then the 100% guy has to make a decision:

(1) Shoot you:

he's guaranteed to get you, so his chances of dying are just 50%. (game ends after the next round...the 50% gets only one shot)

prob of dying: 50%

(2) Shoot the 50% guy:

by the same logic, he'd have a 33.33% chance of death (getting shot by you).

So, he prefers to shoot the 50% guy.

So, q is 100% (ie, if the 50% guy misses, the 100% guy shoots him immediately). From that, we know the 50% guy's optimal move, if everyone's around on his first shot. He'd prefer shooting at the 100% guy to shooting at you or purposely missing.

Now, we need to get p, which is our probability of dying if we purposely miss either guy (shoot into the air).

We know the 50% guy shoots at the 100% guy if we miss.

(1) With 50% probability, the 50% guy kills the 100% guy, resulting in a shootout b/w us and the 50% guy.

(=) (50%) * [ (66.67%)*(50%) + (66.67%)*(50%)*(66.67%)(50%) + ...

(=) (50%) * [ (1/3) + (1/3)^2 + ... ]

(=) (50%) * [0.5]

(=) (25%)

(2) He misses him. Then, we get one shot at the 100% guy (after the 100% guy shoots the 50% guy).

Our chances of death are:

(50%)*(66.67%) = (33.33%)

So, p is (25%)+(33.33%) = 58.3%

So, again, our three choices are

(1) Shoot at 50% guy. Death comes with:

(=) (33.3%)+(66.67%)(p)

(=) 72% likilhood

(2) Shoot at the 100% guy. Death comes with:

(=) (24.75%) + (66.67%)(p)

(=) 63.64% likihood

(3) Shoot into the air (purposely miss).

(=) p

48.

If a hen and a half lay an egg and a half in a day and a half, how many hens does it take to lay six eggs in six days?

If 1.5 hens lay 1.5 eggs in 1.5 days (or 36 hours) then: 1 hen lays 1 egg in 1,5 days or 4 eggs in six days thus 1.5 hens lay 6 eggs in 6 days

49.

Arrange the numbers 1 to 8 in the grid below such that adjacent numbers are not in adjacent boxes (horizontally, vertically, or diagonally).

The arrangement above, for example, is wrong because 3 & 4, 4 & 5, 6 & 7, and 7 & 8 are adjacent.

Solution: box o' numbersThe key (as I see it) is putting the 1 & 8 in the centre spots - which is required, because those spots both border all but one of the other spots, and 1 & 8 are the only numbers that are only adjacent to one number.

From there, the 2 & 7 are forced, then you have your choice of the next number, which then forces the rest. Really, though there are only two valid solutions, and they are mirror images of each other.

50.

"A line of 100 airline passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let's say that the nth passenger in line has a ticket for the seat number n.)

Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random.

What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?"

If any passenger <100 randomly selects seat 1, then each successive passenger can sit in his/her assigned seat including passenger 100.

If any passenger <100 randomly selects seat 100, then passenger 100 is forced to choose the only other remaining seat.

For each passenger (who is forced to choose randomly), the chance of choosing 1 and 100 is equal. (The EXACT probability depends on the number of other remaining seats, but in each case the chance of choosing 1 and 100 is the same. 1/100 and 1/100 for the 1st passenger, 1/40 and 1/40 for the 60th passenger) Each choice that doesn't choose seat 1 or 100 simply defers the final decision to a later passenger.

Therefore the chance of passenger 100 getting his own seat is 50%.