ASA Connect

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

------------------------------
Mark Martin

Original Message:
Sent: 10-18-2016 09:26
From: Gabriel Farkas
Subject: "The Price Is Right" probability

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

------------------------------
Gabriel Farkas
------------------------------

Mark,

at first, I totally agreed with your math. But then, on second thought, it seems a bit more complicated: you calculated the probability of three contestants getting $1.00 under the implicit assumption that the first contestant always spins the wheel a second time if the first spin results in less than $1.00. But if I remember the rules of the game correctly, the first contestant is free to stop after the first spin.

To get a more realistic model, we need the probabilty that the first contestant spins the wheel a second time if the first spin got less than $1.00. Let's call this probability p. Then we'd get the following:

P( 3 times $1.00)  =  (1/20 + 19/20 * p * 1/20 ) * (1/20 + 19/20 * 1/20 )^2

Since we don't know p, we can't calculate this exactly. Let's assume the first contestant decides to stop whenever she gets 75c or more. Then p would be 5/19, and  P( 3 times $1.00) = 0.00059414 or about 1 in 1,683.  This means: on average we have to watch 1,683 showcase showdowns until we witness a "3 times $1.00"-event. This interpretation would not be possible if we ignore the first contestant's stopping option, imho.

-Hans-

Edit: Sorry, mistake in the last paragraph: p would be 1 - 5/19, and  P( 3 times $1.00) =  about 1 in 1,238

------------------------------
Hans Kiesl
Regensburg University of Applied Sciences
Germany

Original Message:
Sent: 10-19-2016 11:35
From: Mark Martin
Subject: "The Price Is Right" probability

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

------------------------------
Mark Martin

Original Message:
Sent: 10-18-2016 09:26
From: Gabriel Farkas
Subject: "The Price Is Right" probability

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

------------------------------
Gabriel Farkas
------------------------------

Hans,

Good point, and I agree with your assessment.  It truly does depend on how willing the first contestant is to opt for a second spin of the wheel.

Mark,

at first, I totally agreed with your math. But then, on second thought, it seems a bit more complicated: you calculated the probability of three contestants getting $1.00 under the implicit assumption that the first contestant always spins the wheel a second time if the first spin results in less than $1.00. But if I remember the rules of the game correctly, the first contestant is free to stop after the first spin.

To get a more realistic model, we need the probabilty that the first contestant spins the wheel a second time if the first spin got less than $1.00. Let's call this probability p. Then we'd get the following:

P( 3 times $1.00)  =  (1/20 + 19/20 * p * 1/20 ) * (1/20 + 19/20 * 1/20 )^2

Since we don't know p, we can't calculate this exactly. Let's assume the first contestant decides to stop whenever she gets 75c or more. Then p would be 5/19, and  P( 3 times $1.00) = 0.00059414 or about 1 in 1,683.  This means: on average we have to watch 1,683 showcase showdowns until we witness a "3 times $1.00"-event. This interpretation would not be possible if we ignore the first contestant's stopping option, imho.

-Hans-

Edit: Sorry, mistake in the last paragraph: p would be 1 - 5/19, and  P( 3 times $1.00) =  about 1 in 1,238

------------------------------
Hans Kiesl
Regensburg University of Applied Sciences
Germany

Original Message:
Sent: 10-19-2016 11:35
From: Mark Martin
Subject: "The Price Is Right" probability

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

------------------------------
Mark Martin

Original Message:
Sent: 10-18-2016 09:26
From: Gabriel Farkas
Subject: "The Price Is Right" probability

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

------------------------------
Gabriel Farkas
------------------------------

Why would anyone stop at some number less than one dollar if a second spin gives him a 1/20 chance of winning $1,000?

Mark,

at first, I totally agreed with your math. But then, on second thought, it seems a bit more complicated: you calculated the probability of three contestants getting $1.00 under the implicit assumption that the first contestant always spins the wheel a second time if the first spin results in less than $1.00. But if I remember the rules of the game correctly, the first contestant is free to stop after the first spin.

To get a more realistic model, we need the probabilty that the first contestant spins the wheel a second time if the first spin got less than $1.00. Let's call this probability p. Then we'd get the following:

P( 3 times $1.00)  =  (1/20 + 19/20 * p * 1/20 ) * (1/20 + 19/20 * 1/20 )^2

Since we don't know p, we can't calculate this exactly. Let's assume the first contestant decides to stop whenever she gets 75c or more. Then p would be 5/19, and  P( 3 times $1.00) = 0.00059414 or about 1 in 1,683.  This means: on average we have to watch 1,683 showcase showdowns until we witness a "3 times $1.00"-event. This interpretation would not be possible if we ignore the first contestant's stopping option, imho.

-Hans-

Edit: Sorry, mistake in the last paragraph: p would be 1 - 5/19, and  P( 3 times $1.00) =  about 1 in 1,238

------------------------------
Hans Kiesl
Regensburg University of Applied Sciences
Germany

Original Message:
Sent: 10-19-2016 11:35
From: Mark Martin
Subject: "The Price Is Right" probability

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

------------------------------
Mark Martin

Original Message:
Sent: 10-18-2016 09:26
From: Gabriel Farkas
Subject: "The Price Is Right" probability

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

------------------------------
Gabriel Farkas
------------------------------

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

I agree with the result (39/400)^3 for a single game.  However, The Price is Right has been around for a long time.  How many times has the game been played?  The probability of a triple winner never occurring approximately (1-1/1079)^N.  So by the time the game has been played 750 times the odds become better than even that it will happen.

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

Emil,

I looked up the game on Wikipedia.  Since the show's debut in 1972, more than 8000 (!) episodes have aired, roughly 180 episodes annually.  The "Showcase Showdown" began in 1975.  I also learned that the showcase showdown is played twice per episode.  So we could assume about 7500 episodes * 2 showdowns.  Therefore they've aired about 15,000 showcase showdowns.

750 plays of the "Showcase Showdown" would take place in just over 2 years.  So as you've implied, every few years we'd expect all 3 contestants to get an exact $1.00 result, even with Hans' adjustment of the odds to approximately 1 in 1,683 times played.

I agree with the result (39/400)^3 for a single game.  However, The Price is Right has been around for a long time.  How many times has the game been played?  The probability of a triple winner never occurring approximately (1-1/1079)^N.  So by the time the game has been played 750 times the odds become better than even that it will happen.

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

I love this discussion and have been thinking of ways to incorporate all facets into my classroom this year.

I thought of one other follow up question:

If you are the first player to spin (out of three) in this situation, is there an optimal stopping value to maximize your probability of making the showcase showdown?  In other words, what value or above on your first spin should you immediately stop with.  I'm assuming here too that the next two competitors will try to beat your value even if that means they might go over $1.00.

Happy weekend everyone!

Kevin

Emil,

I looked up the game on Wikipedia.  Since the show's debut in 1972, more than 8000 (!) episodes have aired, roughly 180 episodes annually.  The "Showcase Showdown" began in 1975.  I also learned that the showcase showdown is played twice per episode.  So we could assume about 7500 episodes * 2 showdowns.  Therefore they've aired about 15,000 showcase showdowns.

750 plays of the "Showcase Showdown" would take place in just over 2 years.  So as you've implied, every few years we'd expect all 3 contestants to get an exact $1.00 result, even with Hans' adjustment of the odds to approximately 1 in 1,683 times played.

I agree with the result (39/400)^3 for a single game.  However, The Price is Right has been around for a long time.  How many times has the game been played?  The probability of a triple winner never occurring approximately (1-1/1079)^N.  So by the time the game has been played 750 times the odds become better than even that it will happen.

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.

I love this discussion and have been thinking of ways to incorporate all facets into my classroom this year.

I thought of one other follow up question:

If you are the first player to spin (out of three) in this situation, is there an optimal stopping value to maximize your probability of making the showcase showdown?  In other words, what value or above on your first spin should you immediately stop with.  I'm assuming here too that the next two competitors will try to beat your value even if that means they might go over $1.00.

Happy weekend everyone!

Kevin

Emil,

I looked up the game on Wikipedia.  Since the show's debut in 1972, more than 8000 (!) episodes have aired, roughly 180 episodes annually.  The "Showcase Showdown" began in 1975.  I also learned that the showcase showdown is played twice per episode.  So we could assume about 7500 episodes * 2 showdowns.  Therefore they've aired about 15,000 showcase showdowns.

750 plays of the "Showcase Showdown" would take place in just over 2 years.  So as you've implied, every few years we'd expect all 3 contestants to get an exact $1.00 result, even with Hans' adjustment of the odds to approximately 1 in 1,683 times played.

I agree with the result (39/400)^3 for a single game.  However, The Price is Right has been around for a long time.  How many times has the game been played?  The probability of a triple winner never occurring approximately (1-1/1079)^N.  So by the time the game has been played 750 times the odds become better than even that it will happen.

Gabriel, thank you for a fun exercise to do while eating breakfast.  Though I haven't seen "The Price Is Right" in years, I did frequently watch it as a young kid who was interested in how much things cost.

This could be an interesting question to pose in a probability class -- not just to see someone's answer, but to hear their reasoning behind it to assess understanding of probability principles.

I noted that you want the probability of all 3 contestants specifically needing two spins to reach exactly $1.00.  There are 20 possibilities during each spin of the wheel.  Here goes for a single player:

P(1st spin <> $1.00) * P(2nd spin yields $1.00 total)  =  19/20 * 1/20  =  0.0475

For all 3 players:

(19/20 * 1/20) ^ 3  =  0.0475 ^ 3   =   0.0001071718  or about 1 in 9,331.

The logic: there are 19 non-$1.00 possibilities (out of 20) on the first spin.  After that, there is only 1 possibility (out of 20) on the 2nd spin, which would yield a sum of $1.00.

As to the assumptions, yes, I assumed equal likelihood for reaching all 20 spots on the wheel.  The subtle differences due to starting position on the wheel relative to the "target result" is unknowable.  It mostly depends on the strength of the spin that a contestant can give.  The more revolutions given, the closer to reaching our assumption.  For those "weak" contestants who can barely achieve the required minimum of 1 full revolution of the wheel, the odds could be far from the assumption.

By the way, I think many viewers of the show would be more interested in the question: "What's the probability that all 3 contestants get $1.00 for the $1,000 prize?"  (... regardless of whether it took one or two spins for any given contestant.)  Here's how I'd do that:

For a single contestant, P(1 spin) + P(2 spins)  =  1/20 + (19/20 * 1/20)  =  0.05 + 0.0475  =  0.0975

For all 3 contestants, P($1.00)  =  0.0975 ^ 3  =  0.0009268594  or about 1 in 1,079.

I had a fun little question I wanted to toss out to the group, related to something that happened recently on the game show "The Price Is Right". You can get an overview from here: All Three "Price Is Right" Contestants Spun $1 And OMG They Freaked Out.

For those not as familiar with the show, there's a part where contestants spin a large wheel, which is labeled with money amounts from five cents to $1.00 in five-cent increments ($0.05, $0.10, $0.15, etc.). The contestants spin the wheel once and then optionally a second time. Any contestant who spins exactly $1.00, either on the first spin or as a total from two spins, wins an extra $1,000.

What happened is that 3 contestants in a row each spun exactly $1.00 on two spins: $0.35 + $0.65, $0.50 + $0.50, and $0.85 + $0.15.

Naturally, the question that comes to mind is: what is the probability of all three contestants spinning numbers that add up to $1.00, on two spins each?

My first thought was 1 / ( ( 20 ^ 2) ^ 3), or 1 in 64,000,000. But that doesn't take into account relative positioning on the wheel, and assumes that every amount is equally likely to be landed on from every other amount. Thinking about it more, it might just be 1 / (20^3), or 1 in 8,000.

I'd love to read anyone elses' thoughts about how to approach this.